iCON Education - Modern Physics

Q51 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Q52 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Q53 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Q54 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Correct Answer: D

Explanation:

Radius $R \propto n^2$
Speed $v \propto \frac{1}{n}$
Total energy $E \propto -\frac{1}{n^2}$
Therefore $Rv \propto n^2 \times \frac{1}{n}$
$Rv \propto n$

Q55 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

$E_n/2$

Q56 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

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

Correct Answer: B

Explanation:

Wavelength of radiation in hydrogen atom is given by
$\frac{1}{\lambda} = R \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right]$
$\frac{1}{\lambda_0} = R \left[\frac{1}{2^2} - \frac{1}{3^2}\right] = R \left[\frac{1}{4} - \frac{1}{9}\right] = \frac{5R}{36}$
and $\frac{1}{\lambda'} = R \left[\frac{1}{2^2} - \frac{1}{4^2}\right] = R \left[\frac{1}{4} - \frac{1}{16}\right] = \frac{3R}{16}$
From equation (i) and (ii)
$\frac{\lambda'}{\lambda_0} = \frac{5/36}{3/16} = \frac{5}{36} \times \frac{16}{3} = \frac{20}{27}$
$\lambda' = \frac{20}{27} \lambda_0$

Q57 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Q58 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

$3.2E$

Q59 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Q60 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Correct Answer: D

Explanation:

Energy of an electron in nth orbit of a hydrogen-like atom is
$E_n = \frac{-13.6Z^2}{n^2}$ eV
For $Li^{++}$, $Z = 3$
Required energy for transition from first to third orbit
$\Delta E = 13.6Z^2\left(\frac{1}{1^2} - \frac{1}{3^2}\right)$
$\Delta E = 13.6 \times 9\left(1 - \frac{1}{9}\right)$
$\Delta E = 108.8$ eV
$\Delta E = 108.8 \times 1.6 \times 10^{-19}$ J
Using $\Delta E = \frac{hc}{\lambda}$
$\lambda = \frac{hc}{\Delta E}$
$\lambda = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{108.8 \times 1.6 \times 10^{-19}}$
$\lambda = 1.1374 \times 10^{-8}$ m
$\lambda = 113.74$ Å

Q61 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Q62 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

$3\lambda$

Correct Answer: D

Explanation:

For transition $3 \to 1$:
$\Delta E = 2E - E = E$
$E = \frac{hc}{\lambda}$
$hc = E\lambda$
For transition $2 \to 1$:
$\Delta E' = \frac{4E}{3} - E = \frac{E}{3}$
$\frac{E}{3} = \frac{hc}{\lambda'}$
$\lambda' = \frac{3hc}{E}$
Using $hc = E\lambda$, $\lambda' = 3\lambda$

Q63 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Q64 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Q65 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Correct Answer: C

Explanation:

First member of Paschen series means
$\lambda_{max}=\frac{144}{7R}$
Short wavelength limit of Paschen series means
$\lambda_{min}=\frac{9}{R}$
Therefore
$\frac{\lambda_{max}}{\lambda_{min}}=\frac{144}{7R}\times\frac{R}{9}$
$\frac{\lambda_{max}}{\lambda_{min}}=\frac{16}{7}$
$\lambda_{min}=\frac{7}{16}\lambda_{max}$
$\lambda_{min}=\frac{7}{16}\times18800$ Å
$\lambda_{min}=8225$ Å

Q66 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Correct Answer: C

Explanation:

For Lyman series first line:
$\frac{1}{\lambda_L}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$
$\frac{1}{\lambda_L}=R\left(1-\frac{1}{4}\right)=\frac{3R}{4}$
$\lambda_L=\frac{4}{3R}$
For Balmer series first line:
$\frac{1}{\lambda_B}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$
$\frac{1}{\lambda_B}=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5R}{36}$
$\lambda_B=\frac{36}{5R}$
$\frac{\lambda_L}{\lambda_B}=\frac{4}{3R}\times\frac{5R}{36}$
$\frac{\lambda_L}{\lambda_B}=\frac{5}{27}$

Q67 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Correct Answer: C

Explanation:

Using $\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
For the third line of Balmer series, $n_1 = 2$ and $n_2 = 5$
$\frac{1}{108.5 \times 10^{-9}} = 1.1 \times 10^7 Z^2\left(\frac{1}{2^2} - \frac{1}{5^2}\right)$
$\frac{1}{108.5 \times 10^{-9}} = 1.1 \times 10^7 Z^2\left(\frac{1}{4} - \frac{1}{25}\right)$
$\frac{1}{108.5 \times 10^{-9}} = 1.1 \times 10^7 Z^2\left(\frac{21}{100}\right)$
$Z^2 \approx 4$
$Z = 2$
Ground state energy magnitude $= 13.6Z^2 = 13.6 \times 4 = 54.4$ eV
The actual ground state energy is $-54.4$ eV, while the option gives its magnitude.

Q68 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

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

Correct Answer: A

Explanation:

Using $\Delta E \propto Z^2$ since $n_1$ and $n_2$ are the same
$\frac{hc}{\lambda} \propto Z^2$
$\lambda Z^2 = \text{constant}$
$\lambda_1Z_1^2 = \lambda_2Z_2^2 = \lambda_3Z_3^2 = \lambda_4Z_4^2$
For hydrogen, deuterium, $He^+$ and $Li^{++}$, the values of $Z$ are 1, 1, 2 and 3 respectively.
$\lambda_1 \times 1^2 = \lambda_2 \times 1^2 = \lambda_3 \times 2^2 = \lambda_4 \times 3^2$
$\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$

Q69 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Q70 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

$n = \infty$

Q71 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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$^{-1}$

Correct Answer: D

Explanation:

Using $\bar{\nu} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
For the same transition, $\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ is constant.
Therefore $\bar{\nu} \propto Z^2$
$\frac{\bar{\nu}_{He^+}}{\bar{\nu}_H} = \left(\frac{Z_{He^+}}{Z_H}\right)^2$
$\frac{\bar{\nu}_{He^+}}{\bar{\nu}_H} = \left(\frac{2}{1}\right)^2 = 4$
$\bar{\nu}_{He^+} = 4 \times 20397$ cm$^{-1}$
$\bar{\nu}_{He^+} = 81588$ cm$^{-1}$

Q72 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

$1/8$

Correct Answer: D

Explanation:

Time period $T \propto n^3$
Radius $R \propto n^2$
Therefore $T \propto R^{3/2}$
$\frac{T_1}{T_2} = \left(\frac{R_1}{R_2}\right)^{3/2}$
$\frac{T_1}{T_2} = \left(\frac{R}{4R}\right)^{3/2}$
$\frac{T_1}{T_2} = \left(\frac{1}{4}\right)^{3/2}$
$\frac{T_1}{T_2} = \frac{1}{8}$

Q73 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

$9E$

Q74 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Q75 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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

Q76 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

$38.2$

Correct Answer: A

Explanation:

After the removal of first electron remaining atom will be hydrogen like atom.
So energy required to remove second electron from the atom $E = 13.6 \times \frac{2^2}{1} = 54.4\text{ eV}$
Total energy required $= 24.6 + 54.4 = 79\text{ eV}$

Q77 DPT Bohr's Model MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

$182.2$

Correct Answer: B

Explanation:

Energy required to remove first electron $= 14.6\text{ eV}$
For removing second electron from $\text{Li}^{+}$ ($Z = 3, n = 2$): $E_2 = 13.6 \times \frac{3^2}{2^2} = 13.6 \times \frac{9}{4} = 30.6\text{ eV}$
For removing third electron from $\text{Li}^{2+}$ ($Z = 3, n = 1$): $E_3 = 13.6 \times \frac{3^2}{1^2} = 13.6 \times 9 = 122.4\text{ eV}$
Total energy required $= 14.6 + 30.6 + 122.4 = 203.0\text{ eV}$

Q78 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

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.

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

Q79 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

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.

$2^{1/3}N$

Q80 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

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.

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

Q81 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

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.

$_2\text{He}^4$

Correct Answer: A

Explanation:

Binding energy per nucleon for $_1\text{H}^2 = 2.22 / 2 = 1.11\text{ MeV/nucleon}$.
Binding energy per nucleon for $_2\text{He}^4 = 28.3 / 4 = 7.075\text{ MeV/nucleon}$.
Binding energy per nucleon for $_{25}\text{Fe}^{56}$ = $492 / 56 = 8.786\text{ MeV/nucleon}$.
Binding energy per nucleon for $_{92}\text{U}^{235} = 1786 / 235 = 7.6\text{ MeV/nucleon}$.
Since $_{25}\text{Fe}^{56}$ has the highest binding energy per nucleon, it is the most stable nucleus.

Q82 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

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

Correct Answer: A

Explanation:

The nuclear density $\rho$ is given by the ratio of the mass of the nucleus to its volume:
$\rho = \frac{m}{\frac{4}{3}\pi R^3}$
The radius of the nucleus is $R = R_0 A^{1/3}$, which gives $R^3 = R_0^3 A$.
Substituting this into the density formula:
$\rho = \frac{3m}{4\pi R_0^3 A}$
Rearranging for the mass number $A$:
$A = \frac{3m}{4\pi R_0^3 \rho}$
Substituting the given values:
$A = \frac{3 \times 19.926 \times 10^{-27}}{12.56 \times (1.2 \times 10^{-15})^3 \times 2.29 \times 10^{17}}$
$A = \frac{59.778 \times 10^{-27}}{12.56 \times 1.728 \times 10^{-45} \times 2.29 \times 10^{17}}$
$A = \frac{59.778 \times 10^{-27}}{49.7 \times 10^{-28}}$
$A = \frac{597.78}{49.7} \approx 12$

Q83 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

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 $13:1$

Q84 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

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.

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

Q85 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

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

Q86 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

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

Q87 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

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

Q88 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

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

Q89 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

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

Q90 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

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

Q91 DPT Nuclear Physics MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

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

Q92 DPT Radioactivity MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Oct 2026

Concept

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

Q93 DPT Radioactivity MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Oct 2026

Concept

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

Q94 DPT Radioactivity MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Oct 2026

Concept

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.

$1/(6\lambda)$

Q95 DPT Radioactivity MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Oct 2026

Concept

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.

$2/3$

Q96 DPT Radioactivity MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Oct 2026

Concept

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%

Q97 DPT Radioactivity MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Oct 2026

Concept

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.

$T_B / T_A$

Correct Answer: A

Explanation:

The rate of change of the number of nuclei of B is $dN_B/dt = \lambda_A N_A - \lambda_B N_B$.
After a long time, secular equilibrium is reached, meaning the number of nuclei of B becomes constant, so $dN_B/dt = 0$.
Therefore, $\lambda_A N_A - \lambda_B N_B = 0$, which gives $\lambda_A N_A = \lambda_B N_B$.
Since activity $A = \lambda N$, we get $A_A = A_B$.
Thus, the ratio $A_A / A_B = 1$.

Q98 DPT Radioactivity MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Oct 2026

Concept

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

Q99 DPT Radioactivity MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Oct 2026

Concept

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

Q100 DPT Radioactivity MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Oct 2026

Concept

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

Physics-II Chapters

Modern Physics

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