Question 1: Palette 1/104

DPT MCQ

The rate of photon emission per second is given by $n = (5 \times 10^{24} \text{ J}^{-1}\text{m}^{-1}) P \times \lambda$. Total number of photons emitted in time $t$ is $N = n \times t$.

Power of a radiator is 100 watt and its $\lambda$ is 400 nm. Calculate No. of photon emitted in 10 hrs.

A

$7.2 \times 10^{24}$

B

$2.0 \times 10^{20}$

C

$3.6 \times 10^{22}$

D

$5.0 \times 10^{24}$

Question 2: Palette 2/104

DPT MCQ

The rate of emission of photons $n$ is given by $n = \frac{P\lambda}{hc}$, which gives power as $P = \frac{nhc}{\lambda}$. Thus, the ratio of power of two sources is $\frac{P_{2}}{P_{1}} = \frac{n_{2}}{\lambda_{2}} \times \frac{\lambda_{1}}{n_{1}}$.

A source $S_{1}$ is producing $10^{15}$ photon/s of wavelength 5000 Å. Another source $S_{2}$ is producing $1.02 \times 10^{15}$ photon/s of wavelength 5100 Å. Calculate ratio of power of $S_{2}$ to power of $S_{1}$.

A

$1 : 2$

B

$1 : 1$

C

$2 : 1$

D

$1.02 : 1$

Question 3: Palette 3/104

DPT MCQ

Intensity $I$ is related to frequency $\nu$ and photon rate $n$ by $I = nh\nu$. The photoelectric current $i$ is directly proportional to $n$.

In a photo-electric experiment frequency of incident light is doubled and its intensity is tripled. How much time the photon current is obtained?

A

$3/2$ times

B

$2/3$ times

C

$3$ times

D

$6$ times

Question 4: Palette 4/104

DPT MCQ

The number of photons incident per unit area per second $n$ is given by $n = \frac{IA}{E} = \frac{IA\lambda}{hc}$.

The energy flux of sunlight reaching the Earth's surface is $1.388 \times 10^{3} \text{ W m}^{-2}$. How many photons (nearly) per square meter are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of $1550 \text{ nm}$.

A

$1.08 \times 10^{22}$

B

$3.83 \times 10^{21}$

C

$2.50 \times 10^{20}$

D

$1.38 \times 10^{23}$

Question 5: Palette 5/104

DPT MCQ

Energy released per second equals the power of the bulb ($P$). Energy of a single photon is $E = \frac{hc}{\lambda}$. The number of photons emitted per second is given by $n = \frac{P\lambda}{hc}$.

The power of a bulb is 60 milliwatt and the wavelength of light is 6000 Å. Calculate the number of photons/second emitted by the bulb?

A

$1.8 \times 10^{17} \text{ photon/sec}$

B

$3.2 \times 10^{18} \text{ photon/sec}$

C

$5.4 \times 10^{16} \text{ photon/sec}$

D

$2.5 \times 10^{17} \text{ photon/sec}$

Question 6: Palette 6/104

DPT MCQ

Intensity ($I$) is defined as power emitted per unit area ($I = \frac{P}{A}$). The number of photons per square meter per second ($n$) is calculated by dividing total intensity by energy of a single photon ($n = \frac{I}{E}$, where $E = \frac{hc}{\lambda}$).

The energy flux of sunlight reaching on the Earth's surface is $1.388 \times 10^{3} \text{ W/m}^{2}$. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average length of 550 nm.

A

$1.50 \times 10^{20} \text{ photon/m}^{2}\text{s}$

B

$3.83 \times 10^{21} \text{ photon/m}^{2}\text{s}$

C

$7.20 \times 10^{22} \text{ photon/m}^{2}\text{s}$

D

$4.12 \times 10^{19} \text{ photon/m}^{2}\text{s}$

Question 7: Palette 7/104

DPT MCQ

When radiation falls normally on a surface, the absorbed power $P_{a}$ exerts a force $F_{a} = \frac{P_{a}}{c}$ and the reflected power $P_{r}$ exerts a force $F_{r} = \frac{2P_{r}}{c}$. The total force exerted on the surface is $F = \frac{P_{i} + P_{r}}{c}$, where $P_{i}$ is the total incident power.

A beam of electromagnetic wave of power 10 watt is incident normally on a surface, which absorb 40% power and remaining power is reflected. Calculate force exerted on a surface.

A

$5.33 \times 10^{-8} \text{ N}$

B

$3.33 \times 10^{-8} \text{ N}$

C

$1.60 \times 10^{-7} \text{ N}$

D

$2.67 \times 10^{-8} \text{ N}$

Question 8: Palette 8/104

DPT MCQ

Energy per photon is $E = \frac{hc}{\lambda}$. The rate at which photons are delivered is $n = \frac{P}{\lambda hc} = \frac{P}{E}$, where $P$ is the power of the lamp.

A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light, which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?

A

$E = 3.38 \times 10^{-19} \text{ J}, n = 2.96 \times 10^{20} \text{ photons/s}$

B

$E = 2.11 \times 10^{-19} \text{ J}, n = 5.00 \times 10^{20} \text{ photons/s}$

C

$E = 4.12 \times 10^{-19} \text{ J}, n = 1.50 \times 10^{20} \text{ photons/s}$

D

$E = 1.80 \times 10^{-19} \text{ J}, n = 3.20 \times 10^{20} \text{ photons/s}$

Question 9: Palette 9/104

DPT MCQ

Each photon reflected normally transfers a momentum change of $\Delta p = \frac{2h}{\lambda}$. The net force exerted by $n$ photons per second is $F = \frac{2nh}{\lambda}$, which gives $n = \frac{F\lambda}{2h}$.

How many photons of wavelength $\lambda = 6600 \text{ nm}$ must strike a totally reflecting screen per second at normal incidence so as to exert a force of 1 N?

A

$5 \times 10^{27} \text{ photons/s}$

B

$2.5 \times 10^{27} \text{ photons/s}$

C

$1 \times 10^{28} \text{ photons/s}$

D

$3.3 \times 10^{26} \text{ photons/s}$

Question 10: Palette 10/104

DPT MCQ

According to Einstein's photoelectric equation, $K_{\text{max}} = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}}$, where $\lambda$ is the wavelength of incident photon and $\lambda_{0}$ is the threshold wavelength. Numerically, $K_{\text{max}} = \frac{12400}{\lambda(\text{Å})} - \frac{12400}{\lambda_{0}(\text{Å})} \text{ eV}$.

The threshold wavelength of a metal is 400 nm. Photo electrons have kinetic energy maximum 1.5 eV. Find the wavelength of incident photon.

A

2696 Å

B

4000 Å

C

3100 Å

D

1850 Å

Question 11: Palette 11/104

DPT MCQ

Einstein's photoelectric equation gives maximum kinetic energy as $K_{\text{max}} = \frac{hc}{\lambda} - \phi = \frac{12400}{\lambda(\text{Å})} - \phi(\text{eV})$.

The work function of a metal is 2.3 eV and the wavelength of incident photon is $4.8 \times 10^{-7}\text{ m}$. Find maximum kinetic energy of photo electrons.

A

0.28 eV

B

1.50 eV

C

2.30 eV

D

0.58 eV

Question 12: Palette 12/104

DPT MCQ

By conservation of momentum, total impulse transmitted to the surface is the sum of photon momentum and maximum photoelectron momentum: $\text{Impulse} = p_{\text{electron}} + p_{\text{photon}} = \sqrt{2m K_{\text{max}}} + \frac{E}{c}$.

Light quanta with an energy 4.9 eV eject photoelectrons from metal with work function 4.5 eV. Find the maximum impulse transmitted to the surface of the metal when each electron flies out.

A

$3.43 \times 10^{-25} \text{ kg m/s}$

B

$1.60 \times 10^{-25} \text{ kg m/s}$

C

$5.20 \times 10^{-25} \text{ kg m/s}$

D

$2.80 \times 10^{-24} \text{ kg m/s}$

Question 13: Palette 13/104

DPT MCQ

For photoelectric effect to occur, energy of incident photon $E = \frac{hc}{\lambda} = \frac{12400}{\lambda(\text{Å})} \text{ eV}$ must be greater than or equal to the work function $\phi$ of the metal ($E \geq \phi$).

The work function of a metal is 4 eV if 5000 Å wavelength of light is incident on the metal. Is there any photo electric effect?

A

Yes, with maximum kinetic energy 1.52 eV

B

No, because $E < \phi$

C

Yes, with maximum kinetic energy 2.48 eV

D

Data insufficient

Question 14: Palette 14/104

DPT MCQ

Intensity of light from a point source varies inversely with the square of distance ($I \propto \frac{1}{d^{2}}$). Photoelectric current $i$ is directly proportional to intensity ($i \propto I \propto \frac{1}{d^{2}}$).

In a photo cell 4 unit photo electric current is flowing, the distance between source and cathode is 4 unit. Now distance between source and cathode becomes 1 unit. What will be photo electric current now?

A

16 units

B

32 units

C

64 units

D

4 units

Question 15: Palette 15/104

DPT MCQ

Stopping potential $V_{0}$ is related to photon wavelength by $e V_{0} = \frac{hc}{\lambda} - \phi = \frac{12400}{\lambda(\text{Å})} - \phi$. The difference in stopping potential is $\Delta V_{0} = \frac{12400}{e} \left(\frac{1}{\lambda_{2}} - \frac{1}{\lambda_{1}}\right)$.

The wavelength of photons in two cases are 4000 Å and 3600 Å respectively what is difference in stopping potential for these two?

A

0.34 V

B

0.68 V

C

1.25 V

D

0.15 V

Question 16: Palette 16/104

DPT MCQ

According to Einstein's photoelectric equation, the maximum kinetic energy is related to incident frequency by $\frac{1}{2}m V_{\text{max}}^{2} = h(\nu - \nu_{0})$.

Threshold frequency of a surface is $\nu_{0}$. It is illuminated by $3\nu_{0}$ frequency, then maximum speed of photo electrons is $V \text{ m/sec}$. What will be maximum speed if incident frequency is $9\nu_{0}$?

A

$2V$

B

$V/2$

C

$4V$

D

$3V$

Question 17: Palette 17/104

DPT MCQ

The relation between stopping potential $V_{s}$ and incident wavelength $\lambda$ is given by $e V_{s} = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}}$, where $\lambda_{0}$ is the threshold wavelength.

When incident wavelength is $\lambda$, stopping potential is $3V_{0}$. If incident wavelength is $2\lambda$ then stopping potential is $V_{0}$. Find out threshold wavelength in terms of $\lambda$.

A

$2\lambda$

B

$4\lambda$

C

$3\lambda$

D

$6\lambda$

Question 18: Palette 18/104

DPT MCQ

The rate of incident photons is $n_{\text{ph}} = \frac{P\lambda}{hc}$. The rate of emitted photoelectrons is $n_{e} = \eta \cdot n_{\text{ph}}$, where $\eta$ is quantum efficiency. The photoelectric current is $i = n_{e} e$.

A light beam of power 1.5 mW and 400 nm wavelength incident on a cathode. If quantum efficiency is 0.1% then, find out obtained photo current and number of photoelectron per second.

A

$i = 0.24 \ \mu\text{A}, n_{e} = 1.5 \times 10^{12} \text{ s}^{-1}$

B

$i = 0.96 \ \mu\text{A}, n_{e} = 6.0 \times 10^{12} \text{ s}^{-1}$

C

$i = 0.48 \ \mu\text{A}, n_{e} = 3.0 \times 10^{12} \text{ s}^{-1}$

D

$i = 1.20 \ \mu\text{A}, n_{e} = 4.5 \times 10^{12} \text{ s}^{-1}$

Question 19: Palette 19/104

DPT MCQ

Einstein's photoelectric equation relates stopping potential $V_{0}$ to incident frequency $\nu'$ and work function $\phi$: $e V_{0} = h\nu' - \phi$.

A metallic surface of work function $h\nu$ is illuminated by a radiation beam of frequency $5\nu$. Stopping potential observed is $X$. What will be stopping potential if the surface is illuminated by radiations of $7\nu$ frequency?

A

$1.5 X$

B

$2.5 X$

C

$3.0 X$

D

$1.2 X$

Question 20: Palette 20/104

DPT MCQ

According to Einstein's photoelectric equation, total incident photon energy is $E = K_{\text{max}} + \phi$. Doubling the frequency doubles the incident photon energy ($E' = 2E$), and the new maximum kinetic energy is $K_{\text{max}}' = E' - \phi$.

The kinetic energy of the fastest moving photo electron from a metal of work function 2.8 eV is 2 eV. If the frequency of light is doubled, then find the maximum kinetic energy of photo electron.

A

4.0 eV

B

6.8 eV

C

5.6 eV

D

9.6 eV

Question 21: Palette 21/104

DPT MCQ

Einstein's photoelectric equation is $K = h\nu - h\nu_{0}$. For a new metal with threshold frequency $\nu_{0}'$ and incident frequency $\nu'$, the maximum kinetic energy is $K' = h\nu' - h\nu_{0}'$.

A monochromatic light incident on metal 'A' having threshold frequency $\nu_{0}$. It emits photo electrons of maximum kinetic energy $K$. Now incident frequency is made three times and fall on a metal 'B' having threshold frequency $2\nu_{0}$. What will maximum kinetic energy of photo electrons emitted by metal 'B'?

A

$3K + h\nu_{0}$

B

$2K + h\nu_{0}$

C

$3K - h\nu_{0}$

D

$K + 2h\nu_{0}$

Question 22: Palette 22/104

DPT MCQ

Einstein's photoelectric equation in terms of stopping potential $V_{0}$ and wavelength $\lambda$ is $e V_{0} = \frac{hc}{\lambda} - \phi \implies \phi = \frac{hc}{\lambda} - e V_{0}$.

If light of wavelength 4000 Å falls on a metal which has a stopping potential 1.4 volt against photoelectric emission then what is the work function of the metal. [Take $h = 6.6 \times 10^{-34} \text{ Js}$ and $c = 3 \times 10^{8} \text{ ms}^{-1}$]

A

1.7 eV

B

3.1 eV

C

1.4 eV

D

2.5 eV

Question 23: Palette 23/104

DPT MCQ

For a charged particle accelerated through potential $V$, $\lambda = \frac{h}{\sqrt{2mqV}}$.
For an electron, the practical formula is $\lambda = \sqrt{\frac{150}{V}} \text{ Å}$.

With what potential an electron should be accelerated so that its de-Broglie wavelength becomes $10 \text{ Å}$?

A

$1.0 \text{ volt}$

B

$2.0 \text{ volt}$

C

$1.5 \text{ volt}$

D

$15 \text{ volt}$

Question 24: Palette 24/104

DPT MCQ

De-Broglie wavelength is $\lambda = \frac{h}{mv}$.
If speed is same, $\lambda \propto \frac{1}{m}$.
For alpha particle, $m_\alpha \approx 4m_p$.

Find the ratio of de-Broglie wavelengths for a proton and an alpha particle, if both have same speed.

A

$4:1$

B

$1:4$

C

$2:1$

D

$1:2$

Question 25: Palette 25/104

DPT MCQ

De-Broglie wavelength in terms of kinetic energy is $\lambda = \frac{h}{\sqrt{2mK}}$.
If kinetic energy is same, $\lambda \propto \frac{1}{\sqrt{m}}$.
For deutron, $m_d \approx 2m_p$.

Find the ratio of de-Broglie wavelength for a proton and a deutron, if both have same kinetic energy.

A

$1:\sqrt{2}$

B

$2:1$

C

$1:2$

D

$\sqrt{2}:1$

Question 26: Palette 26/104

DPT MCQ

For particles in thermal equilibrium, the average translational kinetic energy is $\frac{3}{2}kT$. The de-Broglie wavelength associated with this translational motion is given by the formula $\lambda = \frac{h}{\sqrt{3mkT}}$, where $h$ is Planck's constant, $m$ is the mass of the particle, $k$ is the Boltzmann constant, and $T$ is the absolute temperature.

Compute the typical de-Broglie wavelength of an electron in a metal at 27°C and compare it with the mean separation between two electrons in a metal, which is given to be about $2 \times 10^{-10} \text{ m}$.

A

$6.2 \times 10^{-10} \text{ m}$

B

$2.0 \times 10^{-10} \text{ m}$

C

$6.2 \times 10^{-9} \text{ m}$

D

$3.1 \times 10^{-9} \text{ m}$

Question 27: Palette 27/104

DPT MCQ

For a revolving electron in a Bohr orbit, the stationary wave condition is $2\pi r = n\lambda$.
The radius of the nth Bohr orbit is $r = 0.53\frac{n^2}{Z} \text{ Å}$.
For the first excited state, $n = 2$, and for hydrogen, $Z = 1$.

Calculate de-Broglie wavelength of an electron revolving in first excited state in hydrogen atom.

A

$2.116 \text{ Å}$

B

$1.058 \text{ Å}$

C

$3.325 \text{ Å}$

D

$6.65 \text{ Å}$

Question 28: Palette 28/104

DPT MCQ

According to the Bohr-de Broglie condition, the circumference of the nth orbit is an integral multiple of the de-Broglie wavelength: $2\pi r_n = n\lambda$.
For the third orbit, $n = 3$, so the circumference is $3\lambda$.

In third orbit of hydrogen atom wavelength of revolving electron is $\lambda$. Calculate the circumference of third Bohr orbit.

A

$\lambda$

B

$3\lambda$

C

$2\lambda$

D

$4\lambda$

Question 29: Palette 29/104

DPT MCQ

For a charged particle accelerated through potential $V$, de-Broglie wavelength is $\lambda = \frac{h}{\sqrt{2mqV}}$.
For an electron, $\lambda = \frac{h}{\sqrt{2meV}}$.
Therefore, $\lambda \propto \frac{1}{\sqrt{V}}$.

De-Broglie wavelength of an electron, accelerated by potential $V$ is $\lambda$. What will be de-Broglie wavelength of the electron which is accelerated by $4V$ potential?

A

$2\lambda$

B

$4\lambda$

C

$\frac{\lambda}{2}$

D

$\frac{\lambda}{4}$

Question 30: Palette 30/104

DPT MCQ

For a gas molecule, de-Broglie wavelength is $\lambda = \frac{h}{\sqrt{3mkT}}$.
Therefore, $\lambda \propto \frac{1}{\sqrt{mT}}$.
For $H_2$, molecular mass is $2$ and temperature is $27 + 273 = 300 K$.
For $He$, atomic mass is $4$ and temperature is $127 + 273 = 400 K$.

Find the ratio of de Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^\circ C$ and $127^\circ C$ respectively.

A

$\sqrt{8}:\sqrt{3}$

B

$8:3$

C

$\sqrt{3}:\sqrt{8}$

D

$3:8$

Question 31: Palette 31/104

DPT MCQ

Initial momentum is zero because the parent particle is at rest.
By conservation of momentum, the two product particles have equal and opposite momenta.
Since de-Broglie wavelength is $\lambda = \frac{h}{p}$, equal momentum magnitudes give equal wavelengths.

A particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$ having non zero velocities. Find out the ratio of the de-Broglie wavelengths of the two particles.

A

$m_1:m_2$

B

$1:1$

C

$m_2:m_1$

D

$\sqrt{m_1}:\sqrt{m_2}$

Question 32: Palette 32/104

DPT MCQ

Momentum of a photon is $p = \frac{h}{\lambda}$.
Momentum of an electron is $p = m_e v$.
For equal momentum, $m_e v = \frac{h}{\lambda}$.
Therefore, $v = \frac{h}{m_e \lambda}$.

Find out velocity of an electron so that its momentum is equal to that of photon with a wavelength of $\lambda = 5200 \text{ Å}$.

A

$700 \text{ m/s}$

B

$1400 \text{ m/s}$

C

$2800 \text{ m/s}$

D

$1.4 \times 10^6 \text{ m/s}$

Question 33: Palette 33/104

DPT MCQ

For an electron accelerated through potential $V$, de-Broglie wavelength is
$\lambda = \sqrt{\frac{150}{V}} \text{ Å}$.
Squaring both sides gives $V = \frac{150}{\lambda^2}$, where $\lambda$ is in Å.

Find out voltage applied to an electron microscope to produce electron of wavelength $0.6 \text{ Å}$.

A

$150 \text{ volt}$

B

$250 \text{ volt}$

C

$300 \text{ volt}$

D

$416.6 \text{ volt}$

Question 34: Palette 34/104

DPT MCQ

Energy of a photon is $E = \frac{hc}{\lambda}$.
Using Einstein mass-energy relation, $E = mc^2$.
Therefore, effective mass of photon is $m = \frac{E}{c^2} = \frac{h}{c\lambda}$.
The rest mass of a photon is zero, but its effective mass is not zero.

What is the effective mass of a photon having wavelength $\lambda$?

A

$\frac{h}{c\lambda}$

B

$\frac{hc}{\lambda}$

C

$\frac{h\lambda}{c}$

D

$\frac{c\lambda}{h}$

Question 35: Palette 35/104

DPT MCQ

The de-Broglie wavelength of a particle is $\lambda = \frac{h}{p}$.
For an electron accelerated through potential $V$, $\lambda = \sqrt{\frac{150}{V}}$ Å.

An electron is accelerated by a potential difference of $50$ V. Find the de-Broglie wavelength associated with it.

A

$0.86$ Å

B

$2.43$ Å

C

$1.73$ Å

D

$3.46$ Å

Question 36: Palette 36/104

DPT MCQ

For a gas molecule at temperature $T$, $\lambda = \frac{h}{\sqrt{3mkT}}$.
For two gases, $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2 T_2}{m_1 T_1}}$.

Find the ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^\circ C$ and $127^\circ C$ respectively.

A

$\sqrt{\frac{3}{8}}$

B

$\sqrt{\frac{8}{3}}$

C

$\frac{8}{3}$

D

$\frac{3}{8}$

Question 37: Palette 37/104

DPT MCQ

The de Broglie wavelength of a particle is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$, where $h$ is Planck's constant, $p$ is momentum, $m$ is the mass of the particle, and $E$ is its kinetic energy. For an electron accelerated through a potential difference $V$ in volts, this simplifies to $\lambda \approx \frac{12.27}{\sqrt{V}} \text{ Å} = \frac{1}{\sqrt{V}} \text{ nm}$.

The de Broglie wavelength $\lambda$ for an electron of energy $150 \text{ eV}$ is

A

$10^{-8} \text{ m}$

B

$10^{-10} \text{ m}$

C

$10^{-12} \text{ m}$

D

$10^{-14} \text{ m}$

Question 38: Palette 38/104

DPT MCQ

In three dimensions, a particle in thermal equilibrium has $3$ degrees of freedom. The average kinetic energy is $KE = \frac{3}{2} k_B T$. The de Broglie wavelength is related to momentum by $\lambda = \frac{h}{p}$, where $p = \sqrt{2 m KE}$.

The de Broglie wavelength of particles of mass $m$ with average momentum $p$ at a temperature $T$ in three dimensions is given by

A

$\lambda = \frac{h}{\sqrt{2 m k_B T}}$

B

$\lambda = \frac{h}{\sqrt{3 m k_B T}}$

C

$\lambda = \frac{h}{\sqrt{2 k_B T}}$

D

$\lambda = \frac{h}{\sqrt{3 m}}$

Question 39: Palette 39/104

DPT MCQ

According to Bohr's quantization rule and de Broglie's hypothesis, an electron can revolve only in those orbits where the circumference of the orbit is an integral multiple of its de Broglie wavelength.
Formula: $2 \pi r = n \lambda$, where $n$ is the principal quantum number (orbit number), $r$ is the orbit radius, and $\lambda$ is the de Broglie wavelength.

An electron is moving in Bohr's fourth orbit. Its de Broglie wavelength is $\lambda$. What is the circumference of the 4th orbit?

A

$2 / \lambda$

B

$2 \lambda$

C

$4 \lambda$

D

$-4 / \lambda$

Question 40: Palette 40/104

DPT MCQ

The de Broglie wavelength of a particle is inversely proportional to its linear momentum.
Formula: $\lambda = \frac{h}{p} = \frac{h}{m v}$, where $h$ is Planck's constant, $m$ is the mass of the particle, and $v$ is its velocity.

The velocities of two particles A & B are $0.05 \text{ m/s}$ & $0.02 \text{ m/s}$ respectively. The mass of B is five times of mass of A. The ratio of their de-Broglie wavelength is

A

$2 : 1$

B

$1 : 4$

C

$1 : 1$

D

$4 : 1$

Question 41: Palette 41/104

DPT MCQ

The de-Broglie wavelength of a particle is given by $\lambda = \frac{h}{p} = \frac{h}{m v}$. For gas molecules such as $O_2$, the root mean square velocity is given by kinetic theory of gases as $v_{rms} = \sqrt{\frac{3 k_B T}{m}}$, where $k_B$ is the Boltzmann constant, $T$ is temperature, and $m$ is the mass of a single molecule.

de-Broglie wavelength of oxygen molecules corresponding to rms velocity

A

$\frac{h}{\sqrt{m k_B T}}$

B

$\frac{h}{\sqrt{2 m k_B T}}$

C

$\frac{h}{\sqrt{3 m k_B T}}$

D

$\frac{h}{\sqrt{k_B T}}$

Question 42: Palette 42/104

DPT MCQ

The de Broglie wavelength of a particle is given by $\lambda = \frac{h}{\sqrt{2mE}}$, where $h$ is Planck's constant, $m$ is the mass, and $E$ is the kinetic energy of the particle. For two particles, the ratio of their wavelengths depends on their masses and energies according to $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2 E_2}{m_1 E_1}}$.

The de Broglie wavelength of a proton of energy $E_p$ is twice the de Broglie wavelength of an alpha particle of energy $E_\alpha$. The ratio of $\frac{E_p}{E_\alpha}$ is

A

$16$

B

$4$

C

$1$

D

$1/4$

Question 43: Palette 43/104

DPT MCQ

For a massive particle in thermal equilibrium at temperature $T$, moving with root-mean-square (rms) thermal speed, the de Broglie wavelength is given by the formula $\lambda = \frac{h}{\sqrt{3 m k_B T}}$, where $h$ is Planck's constant, $m$ is the mass, $k_B$ is the Boltzmann constant, and $T$ is the absolute temperature.

A dust particle of mass $10^{-8} \text{ kg}$ is at room temperature ($300 \text{ K}$). Its de Broglie wavelength is nearest to

A

$10^{-19} \text{ m}$

B

$10^{-15} \text{ m}$

C

$10^{-6} \text{ m}$

D

$10^{-9} \text{ m}$

Question 44: Palette 44/104

DPT MCQ

The de Broglie wavelength of a non-relativistic particle is inversely proportional to its momentum, which in turn depends on the square root of its kinetic energy $E$.
Formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \implies \lambda \propto \frac{1}{\sqrt{E}}$

The wavelength of a particle of a particle with energy E is proportional to

A

$1/E$

B

$1/E^{1/2}$

C

$E^{1/2}$

D

$E^2$

Question 45: Palette 45/104

DPT MCQ

The de Broglie wavelength of an electron accelerated through a potential difference $V$ in volts is given by $\lambda = \sqrt{\frac{150}{V}} \text{ Å} = \frac{12.27}{\sqrt{V}} \text{ Å}$, where kinetic energy $E = qV = 120 \text{ eV}$ implies $V = 120 \text{ V}$.

The de-Broglie wavelength of an electron with a kinetic energy of $120 \text{ eV}$ is nearly

A

$1000 \text{ pm}$

B

$100 \text{ pm}$

C

$10 \text{ pm}$

D

$1 \text{ pm}$

Question 46: Palette 46/104

DPT MCQ

According to the law of conservation of linear momentum, the total momentum before decay equals the total momentum after decay. For a particle at rest decaying into two fragments, the two fragments move with equal and opposite momenta.
Formula: $p_1 = p_2$, and the de Broglie wavelength is given by $\lambda = \frac{h}{p}$.

A particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$ having nonzero velocities. The ratio of de-Broglie wavelengths ($\lambda_1 / \lambda_2$) of particles is

A

$m_1 / m_2$

B

$\sqrt{m_2 / m_1}$

C

$1$

D

$\sqrt{m_1 / m_2}$

Question 47: Palette 47/104

DPT MCQ

The de Broglie wavelength of an electron accelerated through a potential difference $V$ (or having kinetic energy $E = eV$) is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m eV}}$. Using standard physical constants, the wavelength in Angstroms is given by the formula $\lambda = \sqrt{\frac{150}{V}}\ \text{\AA}$.

The de Broglie wavelength $\lambda$ for an electron of energy 150 eV is

A

$10^{-8}\text{ m}$

B

$10^{-10}\text{ m}$

C

$10^{-12}\text{ m}$

D

$10^{-14}\text{ m}$

Question 48: Palette 48/104

DPT MCQ

According to Bohr's model, the radius of the nth orbit is proportional to the square of the principal quantum number n. The area of the circular orbit depends on the square of the radius.
Formula involved: $r \propto n^2$ and $A = \pi r^2$

The ratio of areas within the electron orbits for the first excited state to the ground state for hydrogen atom is

A

16: 1

B

18: 1

C

4: 1

D

2: 1

Question 49: Palette 49/104

DPT MCQ

In Bohr model, the radius of the nth orbit of hydrogen is proportional to the square of the principal quantum number. Since diameter is twice the radius, the diameter also varies as the square of n.
Formula involved: $r_n \propto n^2$ and $d_n \propto n^2$

The innermost orbit of the hydrogen atom has a diameter 1.06 Å. The diameter of tenth orbit is

A

5.3 Å

B

10.6 Å

C

53 Å

D

106 Å

Question 50: Palette 50/104

DPT MCQ

The energy of an electron in the nth orbit of hydrogen is negative and its magnitude decreases as n increases. Energy required to transfer an electron from one orbit to another is the difference in the energies of the two orbits.
Formula involved: $E_n = \frac{-13.6}{n^2}$ eV and $\Delta E = E_f - E_i$

Energy of the electron in nth orbit of hydrogen atom is given by $E_n = \frac{-13.6}{n^2}$ eV. The amount of energy needed to transfer electron from first orbit to third orbit is

A

13.6 eV

B

3.4 eV

C

12.09 eV

D

1.51 eV

Question 51: Palette 51/104

DPT MCQ

For hydrogen-like atoms, the energy of the nth orbit depends on the square of the atomic number Z and inversely on the square of the principal quantum number n. The energy required to remove an electron from a given orbit is the positive magnitude of the energy of that orbit.
Formula involved: $E_n = \frac{-13.6 Z^2}{n^2}$ eV and ionization energy $= \frac{13.6 Z^2}{n^2}$ eV

If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of $Li^{++}$ is

A

122.4 eV

B

30.6 eV

C

13.6 eV

D

3.4 eV

Question 52: Palette 52/104

DPT MCQ

For hydrogen-like atoms, the wavelength of emitted radiation for a transition depends on the atomic number. For the same transition, wavelength is inversely proportional to the square of atomic number.
Formula involved: $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ and $\lambda \propto \frac{1}{Z^2}$

The ratio of the wavelengths for $2 \to 1$ transition in $Li^{++}$, $He^+$ and $H$ is

A

1: 2: 3

B

1: 4: 9

C

4: 9: 36

D

3: 2: 1

Question 53: Palette 53/104

DPT MCQ

When an electron jumps from a higher energy orbit to a lower energy orbit, the energy of the emitted photon is equal to the difference in the energies of the two orbits.
Formula involved: $\Delta E = E_i - E_f$ and $E_n = \frac{-13.6}{n^2}$ eV

Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = \frac{-13.6}{n^2}$ eV. The energy of a photon ejected when the electron jumps from $n=3$ state to $n=2$ state of hydrogen is approximately

A

1.9 eV

B

1.5 eV

C

0.85 eV

D

3.4 eV

Question 54: Palette 54/104

DPT MCQ

In the Bohr model for hydrogen, the radius of the orbit varies as $n^2$, the speed of the electron varies as $\frac{1}{n}$, and the total energy varies as $-\frac{1}{n^2}$. By combining these dependencies, we can identify which expression is proportional to $n$.
Formula involved: $R \propto n^2$, $v \propto \frac{1}{n}$ and $E \propto -\frac{1}{n^2}$

In the Bohr model of the hydrogen atom, let $R$, $v$ and $E$ represent the radius of the orbit, the speed of electron and the total energy of the electron respectively. Which of the following quantity is proportional to the quantum number $n$

A

R/E

B

E/v

C

RE

D

vR

Question 55: Palette 55/104

DPT MCQ

The energy of an electron in the nth orbit of a hydrogen-like atom is proportional to the square of the atomic number Z and inversely proportional to the square of the principal quantum number n.
Formula involved: $E_n = -13.6 \frac{Z^2}{n^2}$ eV and $E \propto Z^2$

The energy of hydrogen atom in nth orbit is $E_n$, then the energy in nth orbit of singly ionised helium atom will be

A

$4E_n$

B

$E_n/4$

C

$2E_n$

D

$E_n/2$

Question 56: Palette 56/104

DPT MCQ

The wavelength of radiation emitted during an electron transition in a hydrogen atom is given by the Rydberg formula. It depends on the initial and final principal quantum numbers.
Formula involved: $\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

The wavelength of radiation emitted is $\lambda_0$ when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be

A

$\frac{16}{25} \lambda_0$

B

$\frac{20}{27} \lambda_0$

C

$\frac{27}{20} \lambda_0$

D

$\frac{25}{16} \lambda_0$

Question 57: Palette 57/104

DPT MCQ

According to Bohr’s quantization condition, the angular momentum of an electron in the nth orbit is an integral multiple of $\frac{h}{2\pi}$. Therefore, the change in angular momentum during a transition is the difference between the angular momenta of the two orbits.
Formula involved: $L_n = \frac{nh}{2\pi}$ and $\Delta L = \frac{(n_2 - n_1)h}{2\pi}$

When an electron in hydrogen atom is excited from its 4th to 5th stationary orbit, the change in angular momentum of electron is. Planck’s constant is $h = 6.6 \times 10^{-34}$ J-s

A

$4.16 \times 10^{-34}$ J-s

B

$3.32 \times 10^{-34}$ J-s

C

$1.05 \times 10^{-34}$ J-s

D

$2.08 \times 10^{-34}$ J-s

Question 58: Palette 58/104

DPT MCQ

The energy difference between two Bohr orbits is proportional to the difference of the reciprocal squares of their principal quantum numbers. The ionization energy of hydrogen corresponds to the energy required to take the electron from $n = 1$ to $n = \infty$.
Formula involved: $\Delta E \propto \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

In hydrogen atom, if the difference in the energy of the electron in $n = 2$ and $n = 3$ orbits is $E$, the ionization energy of hydrogen atom is

A

$13.2E$

B

$7.2E$

C

$5.6E$

D

$3.2E$

Question 59: Palette 59/104

DPT MCQ

In the Bohr model, the radius of an orbit is proportional to $n^2$ and the speed of the electron is proportional to $\frac{1}{n}$. Since the time period is $T = \frac{2\pi r}{v}$, the time period is proportional to $n^3$ for a hydrogen atom.
Formula involved: $r \propto n^2$, $v \propto \frac{1}{n}$ and $T \propto n^3$

In Bohr model of hydrogen atom, the ratio of periods of revolution of an electron in $n = 2$ and $n = 1$ orbits is

A

2: 1

B

4: 1

C

8: 1

D

16: 1

Question 60: Palette 60/104

DPT MCQ

For hydrogen-like atoms, the energy of the nth orbit is proportional to $\frac{Z^2}{n^2}$. The energy required for a transition between two orbits is obtained from the difference in their energies, and the corresponding wavelength is calculated using the photon energy relation.
Formula involved: $E_n = \frac{-13.6Z^2}{n^2}$ eV, $\Delta E = 13.6Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ eV and $\lambda = \frac{hc}{\Delta E}$

A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of required radiation for emitting electron from first to third Bohr orbit in $Li^{++}$ will be. Ionisation energy of hydrogen atom is 13.6 eV

A

182.51 Å

B

177.17 Å

C

142.25 Å

D

113.74 Å

Question 61: Palette 61/104

DPT MCQ

For hydrogen atom, if absorption lines are observed from the ground state to higher energy levels, the number of absorption lines is $n-1$, where $n$ is the highest excited state reached. The number of possible emission lines from the nth state to lower states is $\frac{n(n-1)}{2}$.
Formula involved: $N_{absorption} = n-1$ and $N_{emission} = \frac{n(n-1)}{2}$

The absorption transitions between two energy states of hydrogen atom are 3. The emission transitions between these states will be

A

3

B

4

C

5

D

6

Question 62: Palette 62/104

DPT MCQ

The energy of an emitted photon is equal to the difference between the two energy levels involved in the transition. Since photon energy is inversely proportional to wavelength, a smaller energy difference corresponds to a larger wavelength.
Formula involved: $\Delta E = \frac{hc}{\lambda}$

The energy levels of a certain atom for 1st, 2nd and 3rd levels are $E$, $\frac{4E}{3}$ and $2E$ respectively. A photon of wavelength $\lambda$ is emitted for a transition $3 \to 1$. What will be the wavelength of emissions for transition $2 \to 1$

A

$\lambda/3$

B

$4\lambda/3$

C

$3\lambda/4$

D

$3\lambda$

Question 63: Palette 63/104

DPT MCQ

For transitions ending at the same lower energy level, a transition from a higher initial level releases more energy. Since photon energy is inversely proportional to wavelength, higher energy corresponds to shorter wavelength. In the visible region, violet light has higher energy than blue light.
Formula involved: $E = \frac{hc}{\lambda}$

Hydrogen atom emits blue light when it changes from $n = 4$ energy level to $n = 2$ level. Which colour of light would the atom emit when it changes from $n = 5$ level to $n = 2$ level

A

Red

B

Yellow

C

Green

D

Violet

Question 64: Palette 64/104

DPT MCQ

For hydrogen-like atoms, the energy required for an electron transition between two Bohr orbits is proportional to $Z^2$ and depends on the difference of reciprocal squares of orbit numbers.
Formula involved: $\Delta E = 13.6 Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ eV

A single electron orbits a stationary nucleus of charge $+Ze$, where $Z$ is a constant. It requires 47.2 eV to excite the electron from second Bohr orbit to third Bohr orbit. Find the value of $Z$.

A

2

B

5

C

3

D

4

Question 65: Palette 65/104

DPT MCQ

In a spectral series, the first member corresponds to transition from $n_2=n_1+1$ to $n_1$, giving maximum wavelength. The short wavelength limit corresponds to $n_2=\infty$. For Paschen series, $n_1=3$.
Formula involved: $\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, $\lambda_{max}=\frac{144}{7R}$ and $\lambda_{min}=\frac{9}{R}$

The first member of the Paschen series in hydrogen spectrum is of wavelength 18800 Å. The short wavelength limit of Paschen series is

A

1215 Å

B

6560 Å

C

8225 Å

D

12850 Å

Question 66: Palette 66/104

DPT MCQ

The first line of Lyman series corresponds to transition $2 \to 1$. The first line of Balmer series corresponds to transition $3 \to 2$. Wavelengths are obtained from Rydberg formula.
Formula involved: $\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is

A

1: 3

B

27: 5

C

5: 27

D

4: 9

Question 67: Palette 67/104

DPT MCQ

For hydrogen-like ions, the wavelength of a spectral line depends on the atomic number $Z$ through the Rydberg formula. The third line of the Balmer series corresponds to the transition from $n=5$ to $n=2$. The ground state energy of a hydrogen-like ion is proportional to $Z^2$.
Formula involved: $\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ and $E_1 = -13.6Z^2$ eV

The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength 108.5 nm. The ground state energy of an electron of this ion will be

A

3.4 eV

B

13.6 eV

C

54.4 eV

D

122.4 eV

Question 68: Palette 68/104

DPT MCQ

For hydrogen-like atoms, the energy difference for the same transition is proportional to $Z^2$. Since photon energy is $E = \frac{hc}{\lambda}$, the wavelength is inversely proportional to $Z^2$.
Formula involved: $\Delta E \propto Z^2$, $\Delta E = \frac{hc}{\lambda}$ and $\lambda Z^2 = \text{constant}$

Hydrogen, deuterium, singly ionized helium and doubly ionized lithium all have one electron around the nucleus. Consider the transition $n = 2$ to $n = 1$. The wavelengths of emitted radiations are $\lambda_1$, $\lambda_2$, $\lambda_3$ and $\lambda_4$ respectively. Then approximately

A

$\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$

B

$\lambda_1 = \lambda_2 = 2\lambda_3 = 4\lambda_4$

C

$\lambda_1 = 2\lambda_2 = 3\lambda_3 = 4\lambda_4$

D

$\lambda_1 = 3\lambda_2 = 2\lambda_3 = \lambda_4$

Question 69: Palette 69/104

DPT MCQ

A hydrogen atom in the ground state absorbs a photon and is excited to a higher orbit. If the electron reaches the nth orbit, the total number of possible emission lines while returning to lower orbits is $\frac{n(n-1)}{2}$.
Formula involved: $\frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{n^2}\right)$ and $N = \frac{n(n-1)}{2}$

Hydrogen atom in its ground state is excited by radiation of wavelength 975 Å. How many lines will be there in the emission spectrum

A

2

B

4

C

6

D

8

Question 70: Palette 70/104

DPT MCQ

When a hydrogen atom in the ground state absorbs a photon, the photon energy is equal to the difference between the final excited state energy and the initial ground state energy. The energy of the nth orbit of hydrogen is $E_n = \frac{-13.6}{n^2}$ eV.
Formula involved: $E_{photon} = E_n - E_1$ and $E_n = \frac{-13.6}{n^2}$ eV

A photon of energy 12.4 eV is completely absorbed by a hydrogen atom initially in the ground state so that it is excited. The quantum number of the excited state is

A

$n = 1$

B

$n = 3$

C

$n = 4$

D

$n = \infty$

Question 71: Palette 71/104

DPT MCQ

For hydrogen-like atoms, the wave number of emitted radiation is proportional to $Z^2$ for the same transition. Since $He^+$ has $Z = 2$, its wave number becomes four times that of hydrogen.
Formula involved: $\bar{\nu} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ and $\bar{\nu} \propto Z^2$

The wave number of the energy emitted when an electron comes from fourth orbit to second orbit in hydrogen is $20397$ cm$^{-1}$. The wave number of the energy for the same transition in $He^+$ is

A

$5099$ cm$^{-1}$

B

$20497$ cm$^{-1}$

C

$40994$ cm$^{-1}$

D

$81588$ cm$^{-1}$

Question 72: Palette 72/104

DPT MCQ

In the Bohr model, for a given atom, the time period of revolution is proportional to $n^3$, while the radius is proportional to $n^2$. Therefore, the time period is proportional to $R^{3/2}$.
Formula involved: $T \propto n^3$, $R \propto n^2$ and $T \propto R^{3/2}$

In an atom, two electrons move round the nucleus in circular orbits of radii $R$ and $4R$. The ratio of the time taken by them to complete one revolution is

A

$1/4$

B

$4/1$

C

$8/1$

D

$1/8$

Question 73: Palette 73/104

DPT MCQ

For hydrogen-like atoms, the ionisation energy from the nth state is proportional to $Z^2$ and inversely proportional to $n^2$. If the ionisation energy of hydrogen in the ground state is $E$, then for an atom with atomic number $Z$ in the nth state it is $E \frac{Z^2}{n^2}$.
Formula involved: $E_{ion} = E \frac{Z^2}{n^2}$

Ionisation energy for hydrogen atom in the ground state is $E$. What is the ionisation energy of $Li^{++}$ atom in the 2nd excited state

A

$E$

B

$3E$

C

$6E$

D

$9E$

Question 74: Palette 74/104

DPT MCQ

Before emission, the hydrogen atom is at rest. By conservation of momentum, the recoil momentum of the atom is equal in magnitude to the momentum of the emitted photon. Photon momentum is energy divided by the speed of light.
Formula involved: $p = \frac{E}{c}$ and $\Delta E = 13.6\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ eV

An electron jumps from $n = 4$ to $n = 1$ state in a hydrogen atom. The recoil momentum of the hydrogen atom in eV/c is

A

12.75 eV/c

B

6.75 eV/c

C

14.45 eV/c

D

0.85 eV/c

Question 75: Palette 75/104

DPT MCQ

The maximum number of electrons that can be accommodated in the nth shell is $2n^2$. If nature allows shells only up to $n = 4$, the maximum number of electrons, and hence possible elements in this simplified model, is the sum of capacities of the first four shells.
Formula involved: $N_{max} = 2n^2$

If elements with principal quantum number $n > 4$ were not allowed in nature, the number of possible elements would be

A

60

B

32

C

4

D

64

Question 76: Palette 76/104

DPT MCQ

To remove the first electron from a neutral helium atom, an ionization energy of $24.6\text{ eV}$ is supplied. The remaining system becomes a single-electron helium ion ($\text{He}^+$), which is a hydrogen-like atom. The energy required to remove the second electron from a hydrogen-like ion with atomic number $Z$ from its ground state ($n=1$) is given by the formula:
$E = 13.6 \times \frac{Z^2}{n^2}\text{ eV}$
The total energy required to remove both electrons is the sum of the energy required to remove the first electron and the energy required to remove the second electron:
$E_{\text{total}} = E_{\text{first}} + E_{\text{second}}$

An energy of $24.6\text{ eV}$ is required to remove one of the electrons from a neutral helium atom. The energy (in $\text{eV}$) required to remove both the electrons from a neutral helium atom is

A

$79.0$

B

$51.8$

C

$49.2$

D

$38.2$

Question 77: Palette 77/104

DPT MCQ

To remove the first electron from a neutral lithium atom, an ionization energy of $14.6\text{ eV}$ is supplied. The remaining system becomes a hydrogen-like lithium ion ($\text{Li}^{+}$ and then $\text{Li}^{2+}$).
The energy required to remove an electron from a single-electron hydrogen-like ion with atomic number $Z$ from its energy level $n$ is given by the formula:
$E = 13.6 \times \frac{Z^2}{n^2}\text{ eV}$
For a lithium atom ($Z = 3$), the energy needed to remove the remaining two electrons in their respective ground states ($n = 2$ for the second electron, $n = 1$ for the last electron) are calculated using this formula.
The total energy required to remove all three electrons is:
$E_{\text{total}} = E_{\text{first}} + E_{\text{second}} + E_{\text{third}}$

An energy of $14.6\text{ eV}$ is required to remove one of the electrons from a neutral lithium atom. The energy (in $\text{eV}$) required to remove all three electrons from a neutral lithium atom is

A

$157.6$

B

$203.0$

C

$122.4$

D

$182.2$

Question 78: Palette 78/104

DPT MCQ

Concept

+

The nuclear radius $R$ depends on the mass number (nucleon number) $A$ according to the relation $R = R_0 A^{1/3}$, where $R_0$ is a constant. For two nuclei with mass numbers $A_1$ and $A_2$, the ratio of their radii is given by $R_2 / R_1 = (A_2 / A_1)^{1/3}$.

The radius of the nucleus with nucleon number 2 is $1.5 \times 10^{-15}\text{ m}$, then the radius of nucleus with nucleon number 54 will be:

A

$3 \times 10^{-15}\text{ m}$

B

$4.5 \times 10^{-15}\text{ m}$

C

$6 \times 10^{-15}\text{ m}$

D

$9.5 \times 10^{-15}\text{ m}$

Question 79: Palette 79/104

DPT MCQ

Concept

+

The radius $R$ of a nucleus is related to its nucleon number $A$ by the relation $R = R_0 A^{1/3}$, where $R_0$ is a constant. Rearranging this gives $A = (R / R_0)^3$, which means the number of nucleons is directly proportional to the cube of the nuclear radius ($A \propto R^3$).

If there are $N$ nucleons in a nucleus of radius $R$, then the number of nucleons in a nucleus of radius $2R$ will be:

A

$N$

B

$2N$

C

$8N$

D

$2^{1/3}N$

Question 80: Palette 80/104

DPT MCQ

Concept

+

The radius of an atomic nucleus $R$ is given by the relation $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \times 10^{-15}\text{ m}$ (or $1.2\text{ fm}$) and $A$ is the mass number (total number of nucleons).

The radius of $_{29}\text{Cu}^{64}$ nucleus will be:

A

$1.2 \times 10^{-15}\text{ m}$

B

$-2.4 \times 10^{-15}\text{ m}$

C

$3.6 \times 10^{-15}\text{ m}$

D

$4.8 \times 10^{-15}\text{ m}$

Question 81: Palette 81/104

DPT MCQ

Concept

+

The stability of a nucleus is determined by its binding energy per nucleon ($\text{BE}/A$), not total binding energy. The nucleus with the highest binding energy per nucleon is the most stable.

Binding energies of nuclei $_1\text{H}^2$, $_2\text{He}^4$, $_{25}\text{Fe}^{56}$ and $_{92}\text{U}^{235}$ are $2.22\text{ MeV}$, $28.3\text{ MeV}$, $492\text{ MeV}$ and $1786\text{ MeV}$ respectively. Most stable nucleus is:

A

$_{26}\text{Fe}^{56}$

B

$_1\text{H}^2$

C

$_{92}\text{U}^{235}$

D

$_2\text{He}^4$

Question 82: Palette 82/104

DPT MCQ

Concept

+

Nuclear Density $\rho = \frac{m}{V} = \frac{m}{\frac{4}{3}\pi R^3}$ where $R = R_0 A^{1/3}$.
This leads to $\rho = \frac{3m}{4\pi R_0^3 A}$ and rearranging for mass number gives $A = \frac{3m}{4\pi R_0^3 \rho}$.

An unknown nucleus has a nuclear density of $2.29 \times 10^{17}\text{ kg/m}^3$ and mass of $19.926 \times 10^{-27}\text{ kg}$. Its mass number is approximately: (take $R_0 = 1.2 \times 10^{-15}\text{ m}$, $4\pi = 12.56$)

A

12

B

19

C

20

D

16

Question 83: Palette 83/104

DPT MCQ

Concept

+

Nuclear radius is given by $R = R_0 A^{1/3}$, where $A$ is the mass number.
Nuclear volume is given by $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A \implies V \propto A$.
Nuclear density is $\rho = \frac{m}{V} = \frac{m_u A}{\frac{4}{3}\pi R_0^3 A} = \frac{3 m_u}{4\pi R_0^3}$, which is independent of mass number $A$ ($\rho \propto A^0$).

The ratio of nuclear densities and nuclear volumes of $_{26}^{56}\text{Fe}$ and $_{2}^{4}\text{He}$ are, respectively: (NEET 2024)

A

$13:1$ and $14:1$

B

$14:1$ and $1:1$

C

$1:1$ and $14:1$

D

$1:1$ and $13:1$

Question 84: Palette 84/104

DPT MCQ

Concept

+

Nuclear radius is proportional to the cube root of its mass number $A$:
$R = R_0 A^{1/3}$ where $A$ is mass number

If radius of the $_{13}^{27}\text{Al}$ nucleus is taken to be $R_{\text{Al}}$, then the radius of $_{53}^{125}\text{Te}$ nucleus is nearly

A

$\frac{3}{5} R_{\text{Al}}$

B

$(\frac{13}{53})^{1/3} R_{\text{Al}}$

C

$(\frac{53}{13})^{1/3} R_{\text{Al}}$

D

$\frac{5}{3} R_{\text{Al}}$

Question 85: Palette 85/104

DPT MCQ

Concept

+

Stability of a nucleus is decided by its binding energy per nucleon, not by its total binding energy.
Formula: B.E per nucleon = $\frac{B.E}{A}$, where A is the mass number (number of nucleons).
The nucleus with higher $\frac{B.E}{A}$ is more stable.

Which is more stable, $^{24}_{12}X$ with binding energy 24 MeV or $^{16}_{8}Y$ with binding energy 18 MeV?

A

$^{24}_{12}X$ is more stable

B

$^{16}_{8}Y$ is more stable

C

Both are equally stable

D

Stability cannot be compared

Question 86: Palette 86/104

DPT MCQ

Concept

+

Q value of a nuclear reaction A + B → C + D is Q = $(m_R - m_P)c^2 = (m_A + m_B - m_C - m_D)c^2$ = total energy released in the nuclear reaction.
When masses are taken in amu: Q = $(M_R - M_P) \times 931.5$ MeV.
If Q > 0 the reaction is exothermic; if Q < 0 the reaction is endothermic.

In the nuclear reaction $^{2}_{1}H + ^{14}_{7}N \rightarrow ^{12}_{6}C + ^{4}_{2}He$, the masses (in amu) are 2.014, 14.00, 12.00 and 4.00 respectively. What is the Q value of the reaction and its nature?

A

13.041 MeV, endothermic

B

13.041 MeV, exothermic

C

0.014 MeV, exothermic

D

1.3041 MeV, endothermic

Question 87: Palette 87/104

DPT MCQ

Concept

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Total binding energy of a nucleus = (binding energy per nucleon) $\times$ (mass number A).
Gain in binding energy = $(BE)_f - (BE)_i$, where $(BE)_f$ is total binding energy of fragments and $(BE)_i$ is total binding energy of parent nucleus.
Formula: Gain = $[2 \times (A_f \times (BE/A)_f)] - [A_i \times (BE/A)_i]$

A nucleus with mass number 242 and binding energy per nucleon as 7.6 MeV breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as 8.1 MeV, the total gain in binding energy is ___ MeV. [08 April 2023 - Shift 1]

A

121

B

242

C

60.5

D

1839.2

Question 88: Palette 88/104

DPT MCQ

Concept

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In alpha decay the parent nucleus emits a $^{4}_{2}He$ nucleus, and the energy released is Q = $(m_{parent} - m_{daughter} - m_{He})c^2$.
With masses in atomic mass units: Q = $\Delta m \times 931.5$ MeV, where $\Delta m = m_{parent} - m_{daughter} - m_{He}$.

A common example of alpha decay is $^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He + Q$. Given: $^{238}_{92}U = 238.05060$ u, $^{234}_{90}Th = 234.04360$ u, $^{4}_{2}He = 4.00260$ u and $1 u = 931.5 \frac{MeV}{c^2}$. The energy released (Q) during the alpha decay of $^{238}_{92}U$ is ___ MeV. [12 April 2023 - Shift 1]

A

41.0 MeV

B

0.41 MeV

C

40.99 MeV

D

4.1 MeV

Question 89: Palette 89/104

DPT MCQ

Concept

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Energy released in fission = total binding energy of products minus total binding energy of parent nucleus.
Formula: $\Delta E = [n \times A_f \times (BE/A)_f] - [A_i \times (BE/A)_i]$, where n = number of fragments (here n = 2).
Since fragments have higher binding energy per nucleon, the difference appears as released energy.

In a nuclear fission process, a high mass nuclide (A ≈ 236) with binding energy 7.6 MeV/Nucleon dissociated into middle mass nuclides (A_B = 118) having binding energy of 8.6 MeV/Nucleon. The energy released in the process would be ___ MeV. [27 Jan. 2024 - Shift 1]

A

118

B

236

C

472

D

1793.6

Question 90: Palette 90/104

DPT MCQ

Concept

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Neutron separation energy: the energy required to remove a neutron equals the mass defect of the reaction (parent nucleus → residual nucleus + neutron) converted to energy.
Formula: $\Delta E = (M_{parent} - M_{residual} - m_n) \times 931.5$ MeV (magnitude gives required energy), using $1 u = 931.5 \frac{MeV}{c^2}$.

The atomic mass of $_{6}C^{12}$ is 12.000000 u and that of $_{6}C^{13}$ is 13.003354 u. The required energy to remove a neutron from $_{6}C^{13}$, if mass of neutron is 1.008665 u, will be: [27 Jan. 2024 - Shift 2]

A

49.5 MeV

B

0.495 MeV

C

4.95 MeV

D

495 MeV

Question 91: Palette 91/104

DPT MCQ

Concept

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Energy released in fusion = total binding energy of product nucleus minus total binding energy of reactant nuclei.
Formula: $\Delta E = [(BE/A)_Y \times A_Y] - [(BE/A)_X \times A_X \times n]$, where n = number of reactant nuclei.
Total binding energy = (binding energy per nucleon) $\times$ (mass number).

Two lighter nuclei combine to form a comparative heavier nucleus by the relation given below: $^{2}_{1}X + ^{2}_{1}X = ^{4}_{2}Y$. The binding energies per nucleon of $^{2}_{1}X$ and $^{4}_{2}Y$ are 1.1 MeV and 7.6 MeV respectively. The energy released in this process is ___ MeV. (JEE Main-2022)

A

26

B

13

C

30.4

D

4.4

Question 92: Palette 92/104

DPT MCQ

Concept

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The activity of a radioactive substance decreases exponentially with time, given by $A = A_0 (1/2)^{t/T_{1/2}}$.

The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to 1/16 th of its initial value? (NEET 2023)

A

40 minutes

B

60 minutes

C

80 minutes

D

20 minutes

Question 93: Palette 93/104

DPT MCQ

Concept

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According to the law of radioactive decay, rate of decay is directly proportional to the number of undecayed nuclei, $-dN/dt = \lambda N$. Half-life is the time for half the sample to decay, $T_{1/2} = \ln(2)/\lambda$. Mean life is the average lifetime of all nuclei, $\tau = 1/\lambda$.

Given below are two statements
Statement I : The law of radioactive decay states that the number of nuclei undergoing the decay per unit time is inversely proportional to the total number of nuclei in the sample.
Statement II : The half of a radionuclide is the sum of the life time of all nuclei, divided by the initial concentration of the nuclei at time t = 0.
In the light of the above statements, choose the most appropriate answer from the options given below. (NEET 2022)

A

Statement I is incorrect but statement II is correct

B

Both statement I and statement II are correct

C

Both statement I and statement II are incorrect

D

Statement I is correct but statement II is incorrect

Question 94: Palette 94/104

DPT MCQ

Concept

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The ratio of remaining atoms for two substances with different decay constants is given by $N_1/N_2 = e^{-(\lambda_1 - \lambda_2)t}$.

At any instant, two elements X1 and X2 have same number of radioactive atoms. If the decay constant of X1 and X2 are $10\lambda$ and $\lambda$ respectively, then the time when the ratio of their atoms becomes $1/e$ respectively will be : (NEET Re 2022)

A

$1/(5\lambda)$

B

$1/(11\lambda)$

C

$1/(9\lambda)$

D

$1/(6\lambda)$

Question 95: Palette 95/104

DPT MCQ

Concept

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Fraction remaining is $(1/2)^{t/T_{1/2}}$. For non-integer half-lives, evaluate the fractional power.

The half life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be : (NEET 2021)

A

$1/\sqrt{2}$

B

$1/2$

C

$1/(2\sqrt{2})$

D

$2/3$

Question 96: Palette 96/104

DPT MCQ

Concept

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The fraction decayed in a short interval $\Delta t$ is $1 - e^{-\lambda \Delta t} \approx \lambda \Delta t$.

If half life of an element is 69.3h, then how much of its percent will decay in 10th to 11th h. Initial activity = 50 $\mu$Ci. (AIIMS)

A

1%

B

2%

C

3%

D

4%

Question 97: Palette 97/104

DPT MCQ

Concept

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In a successive radioactive decay chain where the parent has a much longer half-life than the daughter ($T_A \gg T_B$), secular equilibrium is established after a sufficiently long time. The rate of production of B equals its rate of decay.

A radioactive element A decays into a radioactive element B. The half-life of A is very large compared to the half-life of B ($T_A \gg T_B$). After a long time, the ratio of their activities $A_A / A_B$ will be: (JEE Main / NEET standard)

A

1

B

0

C

$T_A / T_B$

D

$T_B / T_A$

Question 98: Palette 98/104

DPT MCQ

Concept

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Conservation of mass number and atomic number in radioactive decay series.

A radioactive nucleus $_{90}^{232}Th$ decays into a stable nucleus $_{82}^{208}Pb$ by emitting $\alpha$ and $\beta$ particles. The total number of $\alpha$ and $\beta$ particles emitted are respectively: (NEET PYQ)

A

6, 4

B

4, 6

C

8, 6

D

6, 8

Question 99: Palette 99/104

DPT MCQ

Concept

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The law of radioactive decay and half-life relationship is given by the formula $N = N_0 (1 / 2)^{t/T}$, where $N$ is the remaining amount of the substance, $N_0$ is the initial amount, $t$ is the total elapsed time, and $T$ is the half-life period.

How long will it take for a radioactive sample to decrease to 10%, if its half life is 22 years?

A

55 years

B

65 years

C

73 years

D

82 years

Question 100: Palette 100/104

DPT MCQ

Concept

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The relationship between the decay constant $\lambda$ and the half-life $T$ is given by the formula $\lambda = \frac{0.693}{T}$, where $T$ is converted into seconds for per-second units.

What is the decay constant of a radioactive substance whose half life is $5$ hours?

A

$2.15 \times 10^{-5}$ per sec

B

$3.85 \times 10^{-5}$ per sec

C

$4.62 \times 10^{-5}$ per sec

D

$5.10 \times 10^{-5}$ per sec

Question 101: Palette 101/104

DPT MCQ

Concept

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The fraction of remaining active substance after time $t$ is given by the law of radioactive decay formula $N / N_0 = (1 / 2)^{t/T}$, where $N$ is the remaining amount, $N_0$ is the initial amount, $t$ is the total elapsed time, and $T$ is the half-life period.

The amount of active substance reduces to $1 / 64$ of its initial value in $15$ hours. What is the half life?

A

$1.5$ hours

B

$2.0$ hours

C

$2.5$ hours

D

$3.0$ hours

Question 102: Palette 102/104

DPT MCQ

Concept

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Each $\alpha$ particle emitted decreases the mass number by $4$ and atomic number by $2$, while each $\beta^-$ particle emitted leaves the mass number unchanged and increases the atomic number by $1$. The number of $\alpha$ particles is found from the change in mass number $\Delta A / 4$, and the number of $\beta$ particles is found by balancing the atomic number $Z$.

How many $\alpha$ and $\beta$ particles are emitted when $_{90}\text{Th}^{232}$ converts to $_{82}\text{Pb}^{208}$?

A

$5 \alpha$ and $3 \beta^-$

B

$6 \alpha$ and $4 \beta^-$

C

$4 \alpha$ and $6 \beta^-$

D

$7 \alpha$ and $2 \beta^-$

Question 103: Palette 103/104

DPT MCQ

Concept

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An $\alpha$ decay decreases the mass number by $4$ and the charge number (atomic number) by $2$. A $\beta^-$ decay leaves the mass number unchanged and increases the charge number by $1$. A $\gamma$ decay leaves both the mass number and the charge number completely unchanged.

A radioactive nucleus decays as $X \xrightarrow{\alpha} X_1 \xrightarrow{\beta^-} X_2 \xrightarrow{\alpha} X_3 \xrightarrow{\gamma} X_4$. If mass number and charge number of $X$ are $180$ and $72$ then find these values of $X_4$.

A

Mass number $170$, Charge number $68$

B

Mass number $172$, Charge number $69$

C

Mass number $174$, Charge number $70$

D

Mass number $176$, Charge number $71$

Question 104: Palette 104/104

DPT MCQ

Concept

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When a radioactive nucleus decays simultaneously through multiple independent channels (such as $\alpha$ and $\beta$), the effective mean life $\tau$ is given by $\frac{1}{\tau} = \frac{1}{\tau_\alpha} + \frac{1}{\tau_\beta}$, and the half-life $T$ is related to mean life by $T = 0.693\tau$.

The mean life of a radioactive material for $\alpha$ and $\beta$ decay are $1620$ years and $520$ years. What is the half life of sample.

A

$215$ years

B

$273$ years

C

$320$ years

D

$394$ years