Atoms and Nuclei

In classical mechanics, the total energy (E) of a particle moving under an inverse square law force (like the electrostatic force between an electron and a nucleus) determines the nature of its orbit:

1. If E < 0 (Negative Total Energy): The particle is in a bound state. Its orbit is closed and elliptical (with a circular orbit being a special case of an ellipse). The particle is trapped by the central force.

2. If E = 0 (Zero Total Energy): The particle is at the boundary between bound and unbound states. Its orbit is open and parabolic. The particle just barely escapes the influence of the central force (it reaches infinity with zero kinetic energy).

3. If E > 0 (Positive Total Energy): The particle is in an unbound state. Its orbit is open and hyperbolic. The particle approaches the central force, is deflected, and then moves away to infinity, never to return. It has enough kinetic energy to overcome the attractive potential and escape.

Given that the total energy of the electron is positive (E > 0), the electron is in an unbound state. This means it will not be trapped in a closed orbit around the nucleus. Instead, it will follow an open trajectory (a hyperbola).

Let's evaluate the options:

• A. electron will revolve in a circular orbit. Circular orbits are closed orbits and correspond to E < 0. Incorrect.

• B. electron will revolve in an elliptical orbit. Elliptical orbits are closed orbits and correspond to E < 0. Incorrect.

• C. electron will not follow a closed orbit. This matches our understanding for E > 0, where the orbit is open (hyperbolic). Correct.

• D. electron will fall into the nucleus. Falling into the nucleus implies a highly negative energy or zero angular momentum, leading to a collision, or continuous energy loss through radiation. A positive energy means the electron has more than enough energy to escape, not fall in. Incorrect.

The final answer is $\boxed{C}$

Percentage of atoms left

$ \begin{aligned} & =100-87.5 \\ & =12.5 \% \\ \text { Using } \frac{N}{N_0} & =\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}} \\ \frac{125}{100} & =\left(\frac{1}{2}\right)^{\frac{6}{T_{1 / 2}}} \\ \left(\frac{1}{2}\right)^3 & =\left(\frac{1}{2}\right)^{\frac{6}{T_{1 / 2}}} \Rightarrow 3=\frac{6}{T_{1 / 2}} \\ \Rightarrow \quad T_{1 / 2} & =2 \text { day } \end{aligned} $

Now, in 8 days, 4 half-life get completed

⇒ Fraction of atoms left

$ =\frac{100}{2^4 \times 100}=\frac{6.25}{100} $

⇒ Fraction of atoms decay $=\frac{100-6.25}{100}$

$ =\frac{93.75}{100}=\frac{15}{16} $

The mass number of a nucleus is shown by $A$.

The radius of a nucleus depends on the mass number as $R = R_0 A^{1/3}$, where $R_0$ is a constant.

To compare the radii of two nuclei, use the formula:

$\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1 / 3}$

Here, $\frac{A_1}{A_2} = \frac{27}{125}$.

So,

$\frac{R_1}{R_2} = \left(\frac{27}{125}\right)^{1/3} = \frac{3}{5}$

because $27^{1/3} = 3$ and $125^{1/3} = 5$.

The surface area of a sphere is $S = 4\pi R^2$.

The ratio of their surface areas is:

$\frac{S_1}{S_2} = \frac{4\pi R_1^2}{4\pi R_2^2} = \left(\frac{R_1}{R_2}\right)^2 = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$

So, the ratio of their surface areas is $S_1 : S_2 = 9 : 25$.

The range of weak nuclear force is the order of $10^{-16} \mathrm{~m}$. This force is responsible for radioactive decay and acte over very short distances within the nucleus.

Step 1: Find the Total Energy (TE) from the Given Potential Energy (PE)

The potential energy of the electron is given as -6.8 eV.

We use the relation: $ \mathrm{PE} = 2(\mathrm{TE}) $

This means: $ \mathrm{TE} = \frac{\mathrm{PE}}{2} = \frac{-6.8}{2} = -3.4\,\mathrm{eV} $

Step 2: Find the Value of n (Principal Quantum Number)

For a hydrogen atom, the total energy formula is: $ \mathrm{TE} = \frac{-13.6}{n^2} $

Set this equal to the TE we found: $ -3.4 = \frac{-13.6}{n^2} $

Solve for $ n^2 $: $ n^2 = 4 $

So, $ n = 2 $

Step 3: The Electron is in the Second Orbit

Since $ n = 2 $, the electron is in the second Bohr orbit.

Step 4: Use Bohr's Model and de-Broglie Hypothesis

According to Bohr and de-Broglie:

$ 2\pi r_n = n\lambda $

This gives: $ \lambda = \frac{2\pi r_n}{n} \tag{i} $

Step 5: Use the Formula for the Radius of Orbit

Radius of the nth orbit: $ r_n = n^2 r_0 $

For $ n = 2 $ (second orbit): $ r_2 = 2^2 r_0 = 4r_0 $

Step 6: Calculate the de-Broglie Wavelength

Plug $ r_2 $ into the earlier formula:

$ \lambda = \frac{2\pi r_2}{2} = \frac{2\pi}{2} \times 4r_0 = 4\pi r_0 $

Total time, $t=2$ days

$ \begin{aligned} & =2 \times 24 \text { hours } \\ & =48 \text { hours } \end{aligned} $

Number of intervals of 16 hours $=\frac{48}{16}=3$

Decays per interval = 10\%

Remaining fraction $=0.9$

Remaining fraction after 3 intervals

$ \begin{aligned} f^n & =(0.9)^3 \\ & =0.729=72.9 \% \end{aligned} $

Let $X$ and $Y$ be the mass number and atomic number of nucleus $P$.

$ \therefore{ }_Y^X P \longrightarrow Q+\alpha+2 \beta^{-} $

∴ Mass number of $Q=X-4$

Atomic number of $Q=Y-2+2=Y$

Thus, we see that nuclei $P$ and $Q$ have same atomic number but different mass number. Thus, $P$ and $Q$ are isotopes.

Microscopic domain of physics is the phenomenon that deals with the constitution and structure of matter at the minute scales of atoms and nuclei.

$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For the second line of Balmer series,

$ \begin{aligned} n_1 & =2 \text { and } n_2=4 \\ \therefore \frac{1}{\lambda_B} & =R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\ & =R\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3 R}{16} \end{aligned} $

The safe level of radiation is 32 times less than what the material gives off now.

Each time a half-life (2.5 hours) passes, the radiation becomes half of what it was before.

We can write this as: $\left(\frac{1}{2}\right)^n = \frac{1}{32}$ where $ n $ is the number of half-lives needed until the radiation is safe.

We know $\frac{1}{32} = \left(\frac{1}{2}\right)^5$. So, $ n = 5 $.

This means it takes 5 half-lives to reach a safe level.

Calculating total time:

$ \frac{t}{T_{1 / 2}} = 5 \Rightarrow t = 5 \times T_{1/2} $

Each half-life is 2.5 hours, so total time is:

$ t = 5 \times 2.5 = 12.5 \text{ hours} $

$ \begin{aligned} &\\ &\text { } \begin{aligned} E & =P \times t \\ = & 64 \times 10^3 \times(60 \times 60) \\ = & 64 \times 10^3 \times 3.6 \times 10^3 \\ = & 2.3 \times 10^8 \mathrm{~J} \end{aligned}\\ &\begin{aligned} & \therefore \text { Energy released per fission }=\frac{E}{N} \\ & \qquad=\frac{2.304 \times 10^8}{7.2 \times 10^{18}}=0.32 \times 10^{-10} \mathrm{~J} \end{aligned} \end{aligned} $

Bose-Einstein statics describes the behaviour of a system of identical, indistinguishable particles with integer spin, known as bosons.

The kinetic energy (KE) of an electron in a hydrogen atom depends on the energy level ($n$) it is in.

Here is the formula for kinetic energy at level $n$:

$ \mathrm{KE}_n = \frac{13.6}{n^2} \ \mathrm{eV} $

Step 1: Identify the energy levels

The "third excited state" means $n = 4$, and the "fourth excited state" means $n = 5$.

Step 2: Calculate the kinetic energy for $n = 4$

$ \mathrm{KE}_3 = \frac{13.6}{4^2} = \frac{13.6}{16} \ \mathrm{eV} $

Step 3: Calculate the kinetic energy for $n = 5$

$ \mathrm{KE}_4 = \frac{13.6}{5^2} = \frac{13.6}{25} \ \mathrm{eV} $

Step 4: Find the ratio of kinetic energies

$ \frac{\mathrm{KE}_3}{\mathrm{KE}_4} = \frac{\frac{13.6}{16}}{\frac{13.6}{25}} = \frac{25}{16} $

So, the ratio of kinetic energies is $25:16$.

In $\beta^{-}$-decay, a neutron transforms into a proton, an electron (which is the $\beta^{-}$particle) and an antinutrino $(\overline{\mathrm{v}})$

Step 1: Formula for Remaining Nuclei

We use the formula $ N = N_0 \left(\frac{1}{2}\right)^n $, where:

• $N$ = number of nuclei left
• $N_0$ = starting number of nuclei
• $n$ = number of half-lives passed

Step 2: Setting Up the Ratio

The question asks for the ratio of nuclei left in $A$ and $B$, so: $ \frac{N_A}{N_B} = \frac{\left( \frac{1}{2} \right)^{n_A}}{\left( \frac{1}{2} \right)^{n_B}} $

Step 3: Simplifying the Ratio

This simplifies to: $ \frac{N_A}{N_B} = \left( \frac{1}{2} \right)^{n_A - n_B} $

Step 4: Finding Number of Half-lives for Each Substance

The number of half-lives $ n $ is found by dividing total time by half-life:

• For $A$: $ n_A = \frac{9}{1.5} = 6 $
• For $B$: $ n_B = \frac{9}{4.5} = 2 $

Step 5: Plugging In the Values

$ \frac{N_A}{N_B} = \left(\frac{1}{2}\right)^{6 - 2} = \left(\frac{1}{2}\right)^4 $

Step 6: Final Answer

$ \left(\frac{1}{2}\right)^4 = \frac{1}{16} $ So, the ratio of the remaining nuclei in $A$ and $B$ after $9$ days is $\frac{1}{16}$.

The frequency for hydrogen spectral lines is found using:

$ f = RC\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $

Lyman Series (moving to $n_1 = 1$):

First line: Electron goes from $n_2 = 2$ to $n_1 = 1$

$ f_1^{(L)} = RC\left(1 - \frac{1}{4}\right) = RC \frac{3}{4} $

Second line: Electron goes from $n_2 = 3$ to $n_1 = 1$

$ f_2^{(L)} = RC\left(1 - \frac{1}{9}\right) = RC \frac{8}{9} $

Difference between the two Lyman frequencies:

$ \begin{aligned} f &= f_2^{(L)} - f_1^{(L)} \\ &= RC\left(\frac{8}{9} - \frac{3}{4}\right) \\ &= RC\left(\frac{32 - 27}{36}\right) = RC \frac{5}{36} \end{aligned} $

Balmer Series (moving to $n_1 = 2$):

First line: Electron goes from $n_2 = 3$ to $n_1 = 2$

$ f_1^{(B)} = RC\left(\frac{1}{4} - \frac{1}{9}\right) = RC \frac{5}{36} $

Second line: Electron goes from $n_2 = 4$ to $n_1 = 2$

$ f_2^{(B)} = RC\left(\frac{1}{4} - \frac{1}{16}\right) = RC \frac{3}{16} $

Difference between the two Balmer frequencies:

$ \begin{aligned} f' &= f_2^{(B)} - f_1^{(B)} \\ &= RC\left(\frac{3}{16} - \frac{5}{36}\right) \\ &= RC\left(\frac{27 - 20}{144}\right) \\ &= RC\left(\frac{7}{144}\right) \end{aligned} $

Relationship between differences:

$ \begin{aligned} \frac{f'}{f} &= \frac{\left(\frac{7}{144}\right)}{\left(\frac{5}{36}\right)} = \frac{7}{144} \times \frac{36}{5} = \frac{252}{720} = \frac{7}{20} \\ f' &= \frac{7}{20} f \end{aligned} $

Average energy of a neutron emitted in ${ }_{92}^{235} \mathrm{U}$ fission is 2 MeV

$ \begin{aligned} \therefore \quad \varepsilon & =2 \mathrm{MeV} \\ & =2 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J} \\ & =3.2 \times 10^{-13} \mathrm{~J}=320 \times 10^{-15} \mathrm{~J} \end{aligned} $

We use this formula to find how much of the substance is left after a certain time:

$ N = N_0 \left(\frac{1}{2}\right)^{t / T_{1/2}} $

It is said that $96.875\%$ of the substance has decayed after $10$ days.

This means that only $3.125\%$ of the substance is left, because $100\% - 96.875\% = 3.125\%$.

We can also write $3.125\%$ as a fraction: $3.125\% = \frac{1}{32}$.

So, $\frac{N}{N_0} = \frac{1}{32}$. We set up the formula:
$ \left(\frac{1}{2}\right)^{t / T_{1/2}} = \frac{1}{32} $
Now, $\frac{1}{32} = \left(\frac{1}{2}\right)^5$, so $t / T_{1/2} = 5$.

So, $t = 5 \times T_{1/2}$.
We know $t = 10$ days. So,
$ T_{1/2} = \frac{10}{5} = 2\ \text{days} $
The half-life of the substance is $2$ days.

$t=30$ years, $T_{1 / 2}=10$ years

$\therefore$ number of half lives, $n=\frac{t}{T_{1 / 2}}=\frac{30}{10{}}=3$

$ \therefore N=N_0\left(\frac{1}{2}\right)^n \Rightarrow N=N_0\left(\frac{1}{2}\right)^3 \Rightarrow N=\frac{N_0}{8} $

Percentage of material decayed

$ \begin{aligned} & =\frac{N_0-N}{N_0} \times 100 \% \\ & =\frac{N_0-\frac{N_0}{8}}{N_0} \times 100 \% \\ & =\frac{7}{8} \times 100 \%=87.5 \% \end{aligned} $

$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For the shortest wavelength of Bracket series

$ \begin{aligned} n_1 & =4 \text { and } n_2=\infty \\ \therefore \frac{1}{\lambda^{\prime}} & =R\left(\frac{1}{4^2}-\frac{1}{\infty}\right)=\frac{R}{16} \\ \therefore \lambda^{\prime} & =\frac{16}{R} \end{aligned} $

For the shortest wavelength of Balmer series

$ n_1=2 \text { and } n_2=\infty $

$ \begin{array}{lll} \therefore & \frac{1}{\lambda^{\prime \prime}} & =R\left(\frac{1}{2^2}-\frac{1}{\infty}\right)=\frac{R}{4} \\ \therefore & \lambda^{\prime \prime} & =\frac{4}{R} \\ & \frac{\lambda^{\prime}}{\lambda^{\prime \prime}} & =\frac{\frac{16}{R}}{\frac{4}{R}}=4 \Rightarrow \lambda^{\prime}: \lambda^{\prime \prime}=4: 1 \end{array} $

Total binding energy of one deutron

$ =2 \times 1.15=2.3 \mathrm{MeV} $

For two deutron,

$ (B E)_{\text {total }}=2 \times 23=4.6 \mathrm{MeV} $

Total binding energy of $\alpha$-particle

$ =4 \times 7.1=28.4 \mathrm{MeV} $

∴ Energy released

$ Q=28.4-4.6=23.8 \mathrm{MeV} $

∴ Energy released per nucleon

$ =\frac{Q}{4}=\frac{23.8}{4}=5.95 \mathrm{MeV} $

$T \propto n^3$

$ \frac{T_2}{T_3}=\left(\frac{n_2}{n_3}\right)^3 $

For the second excited state, $n_2=3$

For the third excited state, $n_3=4$

$ \begin{array}{ll} \therefore & \frac{T_2}{T_3}=\left(\frac{3}{4}\right)^3=\frac{27}{64} \\ \Rightarrow & T_2: T_3=27: 64 \end{array} $

$ \begin{aligned} & \text { } \frac{a_1}{a_2}=\frac{9}{49} \\ & \Rightarrow \frac{4 \pi R_1^2}{4 \pi R_2^2}=\frac{9}{49} \Rightarrow\left(\frac{R_1}{R_2}\right)^2=\left(\frac{3}{7}\right)^2 \\ & \Rightarrow \frac{R_1}{R_2}=\frac{3}{7} \end{aligned} $

Since, $R \propto A^{\frac{1}{3}}$

$ \begin{aligned} & \frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{\frac{1}{3}} \\ & \left(\frac{R_1}{R_2}\right)^3=\frac{A_1}{A_2} \\ \Rightarrow & \frac{A_1}{A_2}=\left(\frac{R_1}{R_2}\right)^3=\left(\frac{3}{7}\right)^3=\frac{27}{343} \\ \therefore & A_1: A_2=27: 343 \end{aligned} $

Using, $E_n=\frac{-13.6}{n^2} \mathrm{eV}$;

Energy of photon in transition $n=2$ to $n=1$

$ \begin{aligned} & =\Delta E_1=E_2-E_1 \\ & =(-3.4)-(-13.6)=10.2 \mathrm{eV} \end{aligned} $

Energy of photon in transition $n=5$ to $n=2$

$ \begin{aligned} =\Delta E_2=E_5-E_2= & -0.544-(-3.4) \\ & =2.856 \mathrm{eV} \end{aligned} $

$ \text { Ratio } \frac{\Delta E_1}{\Delta E_2}=\frac{10.2}{2.8}=\frac{102}{28}=\frac{25.5}{7} \approx \frac{25}{7} $

Decay occurs are

$ \begin{aligned} N_0 \xrightarrow{T_{1 / 2}=10 \mathrm{~min}} \frac{N_0}{2} & \xrightarrow{T_{1 / 2}=10 \mathrm{~min}} \frac{N_0}{4} \xrightarrow{T_{1 / 2}=10 \mathrm{~min}} \frac{N_0}{8} \end{aligned} $

$\therefore$ Number of atoms decayed in 20 min

$ n_1=N_0-\frac{N_0}{4}=\frac{3}{4} N_0 . $

Number of atoms decayed in 30 min

$ n_2=N_0-\frac{N_0}{8}=\frac{7}{8} N_0 $

Ratio $\frac{n_1}{n_2}=\frac{3}{4} \times \frac{8}{7}=\frac{6}{7}$.

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\ & \qquad \begin{aligned} \frac{1}{\lambda_{3 \rightarrow 2}} & =R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ \frac{1}{\lambda_{3 \rightarrow 2}} & =\frac{5 R}{36} \Rightarrow \lambda_{3 \rightarrow 2}=\frac{36}{5 R} \\ \text { and } \frac{1}{\lambda_{2 \rightarrow 1}} & =R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3 R}{4} \\ \Rightarrow \quad \lambda_{2 \rightarrow 1} & =\frac{4}{3 R} \\ \therefore \quad \frac{\lambda_{3 \rightarrow 2}}{\lambda_{2 \rightarrow 1}} & =\frac{\frac{36}{5 R}}{\frac{4}{3 R}} \\ & =\frac{36 \times 3 R}{5 R \times 4}=\frac{27}{5} \end{aligned} \end{aligned} \end{aligned} $

Step 1: Typical nuclear parameters

A nucleus has:

• Radius $ R = R_0 A^{1/3} $,

  where $ R_0 \approx 1.2 \times 10^{-15} \, \mathrm{m} $

• Each nucleon (proton or neutron) has a mass $ m_n \approx 1.67 \times 10^{-27} \, \mathrm{kg} $

Step 2: Estimate the number of nucleons per unit volume

For nuclear matter, the volume per nucleon is roughly the volume of a sphere of radius $ R_0 $:

$ V_{\text{per nucleon}} \approx \frac{4}{3}\pi R_0^3 $

$ V_{\text{per nucleon}} \approx \frac{4}{3}\pi (1.2 \times 10^{-15})^3 \approx 7.2 \times 10^{-45} \, \mathrm{m^3} $

Hence, number density of nucleons:

$ n = \frac{1}{V_{\text{per nucleon}}} \approx \frac{1}{7.2 \times 10^{-45}} \approx 1.4 \times 10^{44} \, \mathrm{m^{-3}} $

Step 3: Compute the mass density

$ \rho = n \times m_n = (1.4 \times 10^{44})(1.67 \times 10^{-27}) \approx 2.3 \times 10^{17} \, \mathrm{kg\,m^{-3}} $

✅ Answer:

$ \boxed{\text{Density of nuclear matter } \sim 10^{17} \, \mathrm{kg\,m^{-3}}} $

Correct option: B ($10^{17}$).

Bohr's atomic model is applicable to only hydrogenic atoms.

Step 1. Typical energy scale of atomic levels

Atomic energy levels (like in the hydrogen atom) are of the order of electron volts (eV).

$ E_{\text{atomic}} \sim 1\,\text{eV} $

Step 2. Typical energy scale of nuclear levels

Nuclear energy level spacings are typically in the range of MeV (mega electron volts):

$ E_{\text{nuclear}} \sim 1\,\text{MeV} = 10^6\,\text{eV} $

Step 3. Ratio of orders

$ \frac{E_{\text{nuclear}}}{E_{\text{atomic}}} \sim \frac{10^6\,\text{eV}}{1\,\text{eV}} = 10^6 $

✅ Final Answer:

$ \boxed{10^6} $

Correct Option: (B) $10^6$

The formula for wavelength is:

$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

First Line of Lyman Series:

For Lyman series, $n_1=1$. The first line means $n_2=2$.

$\frac{1}{\lambda_L} = R \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R \left(1 - \frac{1}{4}\right) = R \left(\frac{3}{4}\right) = \frac{3R}{4}$

So, $\lambda_L = \frac{4}{3R}$

Second Line of Balmer Series:

For Balmer series, $n_1=2$. The second line means $n_2=4$.

$\frac{1}{\lambda_B} = R \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R \left(\frac{1}{4} - \frac{1}{16}\right) = R \left(\frac{3}{16}\right) = \frac{3R}{16}$

So, $\lambda_B = \frac{16}{3R}$

Find the Ratio:

Now, find $\frac{\lambda_L}{\lambda_B}$:

$\frac{\lambda_L}{\lambda_B} = \frac{\frac{4}{3R}}{\frac{16}{3R}} = \frac{4}{16} = \frac{1}{4}$

This means $\lambda_L : \lambda_B = 1:4$

Energy per fission $=200 \mathrm{MeV}$

$ \begin{aligned} & =200 \times 1.6 \times 10^{-13} \text { Joule } \\ E & =3.2 \times 10^{11} \mathrm{~J} \end{aligned} $

Power of reactor

$ P=1 \mathrm{MW} $

So, Number of fissions in time $t$

$ N=\frac{P t}{E} $

So, number of fissions per second

$ \begin{aligned} \frac{N}{t} & =\frac{P}{E}=\frac{10^6}{3.2 \times 10^{11}} \\ & \simeq 3.125 \times 10^{16} \end{aligned} $

Rate of fission $=3.125 \times 10^6$ per second

Frequency of a spectral line in the hydrogen atom,

$ v=R c\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) $

For the first line of Paschen series,

$ \begin{aligned} & & n_1 & =3 \text { and } n_2=4 \\ \therefore & & v_1 & =R c\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \\ \Rightarrow & & v_1 & =\frac{7 R c}{144} \end{aligned} $

For the second line of Paschen series,

$ \begin{gathered} n_1=3 \text { and } n_2=5 \\ \therefore \quad v_2=R c\left(\frac{1}{3^2}-\frac{1}{5^2}\right) \Rightarrow v_2=\frac{16 R c}{225} \\ \because \quad v_2-v_1=\frac{16 R c}{225}-\frac{7 R c}{144} \\ =\frac{81 R c}{3600}=\frac{9 R c}{400} \end{gathered} $

When $8 \%$ of nuclei decays, then remaining nuclei are $92 \%$ from

As, $N=N_0 e^{-\lambda t}$

Here, $N=0.92 N_0$

So, $0.92 N_0=N_0 e^{-\lambda t_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

When, $77 \%$ nuclei decay remaining number of nuclei are $23 \%$.

So, $\quad N=0.23 N_0$

$ 0.23 N_0=N_0 e^{-\lambda t_2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

By dividing in equation (i) by equation (ii)

$ \frac{0.92}{0.23}=\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}=e^{\lambda\left(t_2-t_1\right)} $

So, $e^{\lambda\left(t_2-t_1\right)}=4$

By taking natural $\log$ of both sides

$ \begin{aligned} \lambda\left(t_2-t_1\right) & =\ln 4 \\ \text { and } t_2-t_1 & =12 \text { minutes } \\ 12 \lambda & =2 \ln 2 \\ \lambda & =\frac{\ln 2}{6} \end{aligned} $

and relation between half life and decay constant is

$ \lambda=\frac{\ln 2}{t_{1 / 2}} $

So, $\frac{\ln 2}{6}=\frac{\ln 2}{t_{1 / 2}}$

and $t_{1 / 2}=6$ minutes

Given data

• Wavelength emitted by the star (rest wavelength):

  $ \lambda_0 = 6600~\text{Å} $

• Observed redshift (increase in wavelength):

  $ \Delta \lambda = 22~\text{Å} $

The redshift implies that the star is moving away from the Earth.

Formula (for non-relativistic Doppler shift)

For light at non-relativistic speeds:

$ \frac{\Delta \lambda}{\lambda_0} = \frac{v}{c} $

where

$v$ = radial speed of the source relative to the observer,

$c = 3 \times 10^8~\text{m s}^{-1}$.

Substituting values

$ \frac{22}{6600} = \frac{v}{3\times10^8} $

$ v = 3\times10^8 \times \frac{22}{6600} $

$ v = 3\times10^8 \times 3.33\times10^{-3} $

$ v = 1.0\times10^6~\text{m s}^{-1} $

$ v = 10^6~\text{m s}^{-1} = 10 \times 10^5~\text{m s}^{-1} $

Direction

Since the wavelength is red shifted, the star is receding from Earth.

✅ Answer:

Option B

$ \boxed{\text{Receding away from Earth with a speed of } 10 \times 10^5~\mathrm{m/s}} $

Frequency of spectral line of hydrogen atom

$ v=R c\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) $

For the first line of Lyman series

$ \begin{aligned} & n_1=1 \text { and } n_2=2 \\ & \therefore v_1=R c\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3 R c}{4} \end{aligned} $

For the second line of Lyman series,

$ \begin{aligned} & n_1=1 \text { and } n_2=3 \\ & v_2=R c\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=\frac{8}{9} R c \end{aligned} $

Thus, $v_2-v_1=\frac{8}{9} R c-\frac{3}{4} R c=\frac{5}{36} R c$

$ \begin{aligned} &\text { } T_{1 / 2}=12.5 \text { hours }\\ &\begin{aligned} & \therefore N_0=256 \mathrm{~g}, N=1 \mathrm{~g} \\ & \text { Since, } N=N_0\left(\frac{1}{2}\right)^n \end{aligned} \end{aligned} $

$ \begin{aligned} 1 & =256\left(\frac{1}{2}\right)^n \\ \Rightarrow\left(\frac{1}{2}\right)^n & =\frac{1}{256} \\ \Rightarrow \quad\left(\frac{1}{2}\right)^n & =\left(\frac{1}{2}\right)^8 \\ \Rightarrow \quad n & =8 \\ \quad \Rightarrow \quad \frac{t}{T_{1 / 2}} & =8 \\ \Rightarrow \quad t & =8 \times T_{1 / 2}=8 \times 12.5 \\ & =100 \text { hours } \end{aligned} $

$ \text { } \begin{aligned} & \frac{N_t}{N_0}=\left(\frac{1}{2}\right)^n \\ \Rightarrow \quad & \frac{1}{8}=\left(\frac{1}{2}\right)^n \Rightarrow n=3 \\ & \frac{t}{T_{1 / 2}}=3 \end{aligned} $

$ \begin{aligned} \Rightarrow \quad t & =3 \times T_{1 / 2}=3 \times 1.4 \times 10^9 \\ & =4.2 \times 10^9 \text { years } \end{aligned} $

Kinetic energy of ejected electron $=$ Energy of incident photon - Energy required to ionize the electron from $n$th orbit (all in eV )

$ \begin{array}{ll} \therefore & 10.4=\frac{1242}{90}-\left|E_n\right| \\ & 10.4=13.8-\frac{13.6}{n^2} \\ \Rightarrow & \frac{13.6}{n^2}=3.4 \\ & \left(\text { as } E_n \propto \frac{1}{n^2} \text { and } E_1=-13.6 \mathrm{eV}\right) \end{array} $

Solving this equation, we get

               $ n=2 $

The nucleus of ${ }_2 \mathrm{He}^4$ contains 2 neutrons and 2 protons.

Mass of 2 protons $=2 \times 1.0073=2.0146 u$

Mass of 2 neutrons $=2 \times 1.0087=2.0174 u$

Total mass of 2 protons and 2 neutrons

$ =(2.0146+2.0174)=4.032 u $

Mass of helium nucleus $=4.0015 u$

Thus, mass defect is lacking of mass in forming the helium nucleus from 2 protons and 2 neutrons.

$ \begin{aligned} \therefore \Delta m & =\text { Mass defect } \\ & =(4.032-4.0015) \mathrm{u}=0.0305 \mathrm{u} \\ & =0.0305 \mathrm{amu} \end{aligned} $

but $1 \mathrm{amu}=931 \mathrm{MeV}$

Hence, binding energy,

$ \begin{aligned} \Delta E & =(\Delta m) \times 931 \\ & =0.0305 \times 931=28.4 \mathrm{MeV} \end{aligned} $

In nuclear disintegration, the conservation of momentum plays a crucial role in determining the motion of the resulting fragments. Since the original nucleus is at rest, its total initial momentum is zero. After the disintegration, the total momentum of the system must still be zero to conserve momentum.

Given the mass ratio of the resulting two smaller nuclei is $2:1$, let's denote the masses of the two nuclei as $2m$ and $m$, respectively.

Applying Conservation of Momentum

For the nucleus with mass $2m$ and velocity $v_1$ and for the nucleus with mass $m$ and velocity $v_2$, the conservation of momentum equation is:

$ 2m \cdot v_1 + m \cdot v_2 = 0 $

Given that momentum is a vector quantity, and the total initial momentum was zero, the nuclei must move in opposite directions for their momenta to cancel each other out. Therefore, we rearrange the equation:

$ 2m \cdot v_1 = -m \cdot v_2 $

$ 2 \cdot v_1 = -v_2 $

$ v_2 = -2 \cdot v_1 $

The negative sign indicates that $v_1$ and $v_2$ are in opposite directions. Taking magnitudes and considering the ratio:

$ \left| v_2 \right| = 2 \cdot \left| v_1 \right| $

This equation shows that the velocity of the lighter fragment (mass $m$) is twice the velocity of the heavier fragment (mass $2m$). Since they must move in opposite directions for momentum conservation, this confirms:

Correct Answer:

Option A - in opposite directions with speed in the ratio of $1:2$ respectively.

To determine the ratio $\frac{E_H}{E_U}$, let's first consider the energy released in the fusion of hydrogen and the energy released in the fission of $^{235}U$.

For the fusion reaction:

The given reaction is:

$4_1^1H + 2 \mathrm{e}^{-} \rightarrow {}_2^4 \mathrm{He} + 2 \mathrm{\nu} + 6 \gamma + 26.7 \, \mathrm{MeV}$

This reaction shows that 4 hydrogen atoms and 2 electrons produce 1 helium atom, 2 neutrinos, and 6 gamma photons while releasing 26.7 MeV of energy.

The mass of 1 mole of $H_1^1$ is approximately 1 gram, so 2 kg of hydrogen is equal to:

$2 \times 10^3 \, \mathrm{g}$

The number of moles of hydrogen in 2 kg is:

$\mathrm{N}_\text{moles} = \frac{2 \times 10^3}{1} = 2 \times 10^3 \, \mathrm{moles}$

The number of hydrogen atoms in 2 kg is:

$\mathrm{N}_\text{atoms} = N_A \times 2 \times 10^3 = 6.023 \times 10^{23} \times 2 \times 10^3 = 1.2046 \times 10^{27} \, \mathrm{atoms}$

Since 4 hydrogen atoms release 26.7 MeV, the total energy released ($E_H$) is:

$E_H = \frac{1.2046 \times 10^{27}}{4} \times 26.7 \, \mathrm{MeV}$

$E_H = 3.0115 \times 10^{26} \times 26.7 \, \mathrm{MeV}$

$E_H \approx 8.04 \times 10^{27} \, \mathrm{MeV}$

For the fission reaction:

The energy released per fission of one $^{235}U$ nucleus is 200 MeV.

The mass of 1 mole of $^{235}U$ is approximately 235 grams, so 2 kg of ${}^{235}U$ is equal to:

$2 \times 10^3 \, \mathrm{g}$

The number of moles of $^{235}U$ in 2 kg is:

$\mathrm{N}_\text{moles} = \frac{2 \times 10^3}{235} \approx 8.51 \, \mathrm{moles}$

The number of $^{235}U$ nuclei is:

$\mathrm{N}_\text{nuclei} = N_A \times 8.51 \approx 6.023 \times 10^{23} \times 8.51 \approx 5.12 \times 10^{24} \, \mathrm{nuclei}$

The total energy released ($E_U$) is:

$E_U = 200 \, \mathrm{MeV} \times 5.12 \times 10^{24} \approx 1.024 \times 10^{27} \, \mathrm{MeV}$

Finally, the ratio $\frac{E_H}{E_U}$ is:

$\frac{E_H}{E_U} \approx \frac{8.04 \times 10^{27}}{1.024 \times 10^{27}} \approx 7.85$

Therefore, the ratio is closest to option A: 7.62.

To determine how many spectral lines will be emitted due to transitions of electrons in a hydrogen atom when it is given an energy of $10.2 \, \text{eV}$, we first need to ascertain which energy level the electron will reach with this energy and then count the possible transitions (spectral lines) as it returns to the ground state.

Energy Levels of Hydrogen Atom :

The energy levels $ E_n $ of a hydrogen atom can be calculated using the formula:

$ E_n = -\frac{13.6 \, \text{eV}}{n^2} $

where $ n $ is the principal quantum number.

Ground State Energy :

The ground state (n=1) energy is $ E_1 = -13.6 \, \text{eV} $.

Determine the Excited State :

If the ground state electron is given $10.2 \, \text{eV}$, its total energy becomes:

$ E_{\text{total}} = E_1 + 10.2 \, \text{eV} = -13.6 \, \text{eV} + 10.2 \, \text{eV} = -3.4 \, \text{eV} $

Now, we find the principal quantum number $ n $ for which the energy is closest to $-3.4 \, \text{eV}$:

• For $ n = 2 $: $ E_2 = -\frac{13.6}{2^2} = -3.4 \, \text{eV} $


• For $ n = 3 $: $ E_3 = -\frac{13.6}{3^2} = -1.51 \, \text{eV} $

Since $-3.4 \, \text{eV}$ matches exactly with $ E_2 $, the electron reaches the second energy level ($ n = 2 $).

Counting Spectral Lines :

When the electron falls back to the ground state from $ n = 2 $, it can do so in a single transition:

• $ n = 2 $ to $ n = 1 $

Thus, only one spectral line will be emitted during this transition.

Conclusion :

The number of spectral lines emitted when a hydrogen atom in the ground state is given $10.2 \, \text{eV}$ and the electron transitions back to the ground state from $ n = 2 $ is just one.

Therefore, the correct answer is:

To determine the energy equivalent of a mass, we use Einstein's mass-energy equivalence principle given by the equation:

$E = mc^2$

where:

- $E$ is the energy

- $m$ is the mass

- $c$ is the speed of light in a vacuum, which is approximately $3 \times 10^8 \mathrm{~m/s}$

Given the mass $m = 1 \mathrm{~g} = 1 \times 10^{-3} \mathrm{~kg}$, we can substitute these values into the equation:

$E = (1 \times 10^{-3} \mathrm{~kg}) \times (3 \times 10^8 \mathrm{~m/s})^2$

Calculating this, we get:

$E = 1 \times 10^{-3} \times 9 \times 10^{16}$

$E = 9 \times 10^{13} \mathrm{~J}$

Next, to convert this energy into electron volts ($\mathrm{eV}$), we use the conversion factor: $1 \mathrm{~J} = 6.242 \times 10^{12} \mathrm{~MeV}$.

Therefore:

$E = 9 \times 10^{13} \mathrm{~J} \times 6.242 \times 10^{12} \mathrm{~MeV/J}$

Calculating this, we get:

$E = 5.6178 \times 10^{26} \mathrm{~MeV}$

Therefore, the energy equivalent of $1 \mathrm{~g}$ of a substance is:

Option A: $5.6 \times 10^{26} \mathrm{~MeV}$

To determine the nuclear binding energy of the isotope $_5^{12}B$, we need to consider the mass defect concept. The mass defect is the difference between the sum of the individual masses of nucleons (protons and neutrons) and the actual mass of the nucleus.

Let's calculate the mass defect first. The isotope $_5^{12}B$ has 5 protons and 7 neutrons (since the total number of nucleons is 12). Therefore, the mass defect can be written as:

$\Delta M = (5 M_p + 7 M_n) - M_0$

Once we have the mass defect, the binding energy can be found using Einstein's mass-energy equivalence principle, which is given by:

$E = \Delta M \cdot c^2$

Substituting the mass defect into this equation, we get:

$E = ((5 M_p + 7 M_n) - M_0) \cdot c^2$

Therefore, the correct answer is Option A:

$ (5 M_p + 7 M_n - M_0) \cdot c^2 $

In the given hypothetical fission reaction:

$^{236}_{92} X \rightarrow \, ^{141}_{56} Y + \, ^{92}_{36} Z + 3 R$

We need to determine the identity of particles denoted by $ R $. Let's use the conservation of charge and mass number (nucleon number) to identify $ R $.

First, for the conservation of nucleon number (mass number), we have:

$236 = 141 + 92 + 3 \times A_R$

Where $ A_R $ is the mass number of $ R $. This simplifies to:

$236 = 233 + 3A_R$

$3A_R = 236 - 233$

$3A_R = 3$

$A_R = 1$

Next, we use the conservation of charge (atomic number), we have:

$92 = 56 + 36 + 3Z_R$

Where $ Z_R $ is the atomic number of $ R $. This simplifies to:

$92 = 92 + 3Z_R$

$3Z_R = 92 - 92$

$3Z_R = 0$

$Z_R = 0$

Since the particle $ R $ has a mass number of 1 and atomic number of 0, it must be a neutron.

So, the identity of the emitted particles $ R $ is:

Option B: Neutron

To determine the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus, we need to use the concept of mass-energy equivalence provided by Einstein's famous equation:

$ E= \Delta m \cdot c^2 $

where:

• E is the binding energy.
• $ \Delta m $ is the mass defect (difference in mass).
• $ c $ is the speed of light in vacuum, which is approximately $3 \times 10^8 \, \text{m/s}$.

Given:

• Binding energy, $ E = 18 \times 10^8 \, \text{J} $.

We need to find $ \Delta m $, so rearranging the equation:

$ \Delta m = \frac{E}{c^2} $

Substitute the given values:

$ \Delta m = \frac{18 \times 10^8}{(3 \times 10^8)^2} $

Calculate the value:

$ \Delta m = \frac{18 \times 10^8}{9 \times 10^{16}} $

$ \Delta m = \frac{18}{9} \times 10^{-8} $

$ \Delta m = 2 \times 10^{-8} \, \text{kg} $

To convert the mass from kilograms to micrograms ($ \mu g $), we use the conversion factor $ 1 \, \text{kg} = 10^9 \, \mu g $:

$ \Delta m = 2 \times 10^{-8} \, \text{kg} \times 10^9 \, \frac{\mu g}{\text{kg}} $

$ \Delta m = 2 \times 10^{1} \, \mu g $

$ \Delta m = 20 \, \mu g $

Therefore, the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus is 20 $ \mu g $.

Option A (20 $ \mu g $) is the correct answer.

To determine the longest wavelength associated with the Paschen series, we need to understand what the Paschen series is and how its wavelength can be calculated. The Paschen series pertains to the spectral line emissions of the hydrogen atom as an electron transitions from higher energy levels (n > 3) down to n = 3. The formula used to calculate the wavelength ($\lambda$) of the emitted photon during such a transition in the hydrogen atom is given by the Rydberg formula:

$\frac{1}{\lambda} = R_{H} \left( \frac{1}{3^2} - \frac{1}{n^2} \right)$

where,

• $R_{H}$ is the Rydberg constant for hydrogen, $R_{H} = 1.097 \times 10^7 \, \text{m}^{-1}$.
• $n$ is the principal quantum number of the higher energy level (For the longest wavelength, we use the smallest possible value of $n$ that is larger than 3, which is $n=4$, since for longer wavelength transitions, the difference in energy levels should be the smallest).

To find the longest wavelength, we plug in $n=4$ into the Rydberg equation, as this corresponds to the smallest energy transition within the Paschen series ($n=4$ to $n=3$):

$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$

$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{16} \right)$

$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{16 - 9}{144} \right)$

$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144}$

$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144}$

$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144}$

$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144} = 76403.47\, \text{m}^{-1}$

Finally, we calculate $\lambda$ by taking the reciprocal of this value:

$\lambda = \frac{1}{76403.47} \approx 1.308 \times 10^{-5} \, \text{m} = 1.308 \times 10^{-5} \times 10^{2} \, \text{cm} = 1.308 \times 10^{-3} \, \text{cm}$

It seems there was a mistake in my calculation. Revising the calculation properly:

$\lambda = \frac{1}{1.097 \times 10^7 \times \frac{7}{144}}$

Correctly evaluating this, we should directly compute:

$\lambda \approx \frac{144}{7 \times 1.097 \times 10^7} = \frac{144}{7.679 \times 10^7} \approx 1.876 \times 10^{-6} \, \text{m}$

Therefore, the correct option that matches our calculation for the longest wavelength in the Paschen series is:

Option B: $1.876 \times 10^{-6} \, \text{m}$.

The wavelength of light emitted when an electron transitions between energy levels in a hydrogen atom is given by the Rydberg formula:

$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

where:

• $\lambda$ is the wavelength of the emitted light,
• $R$ is the Rydberg constant,
• $n_1$ is the lower energy level, and
• $n_2$ is the higher energy level.

The Balmer series corresponds to electron transitions where the lower energy level, $n_1$, is 2, and the Lyman series corresponds to transitions where $n_1$ is 1. The shortest wavelength in each series occurs for transitions from the highest possible energy level ($n_2 = \infty$) to the specified lower level ($n_1$).

For the Balmer series (shortest wavelength):

$n_1 = 2, n_2 = \infty$,

Putting these values into the Rydberg formula gives:

$\frac{1}{\lambda_{B}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$

$\frac{1}{\lambda_{B}} = R \left( \frac{1}{4} \right)$

$\lambda_{B} = \frac{1}{R \cdot \frac{1}{4}} = \frac{4}{R}$

For the Lyman series (shortest wavelength):

$n_1 = 1, n_2 = \infty$,

Putting these values into the Rydberg formula gives:

$\frac{1}{\lambda_{L}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)$

$\frac{1}{\lambda_{L}} = R \cdot 1$

$\lambda_{L} = \frac{1}{R}$

The ratio of the shortest wavelength of Balmer series ($\lambda_{B}$) to the shortest wavelength of Lyman series ($\lambda_{L}$) is:

$\frac{\lambda_{B}}{\lambda_{L}} = \frac{\frac{4}{R}}{\frac{1}{R}} = \frac{4}{1} = 4:1$

Therefore, the correct answer is Option D: $4: 1$.

$\begin{aligned} & L=m v r \propto n \\ & \therefore \quad m v r \propto \sqrt{r} \end{aligned}$

In the context of an electron orbiting around a nucleus with a positive charge of $\mathrm{Ze}$, we are dealing with classical physics approximations and the electrostatic force between the electron and the nucleus. In such a setup, the electron's potential energy (U) is due to electrostatic interaction, and it is given by Coulomb's law:

$U = -\frac{kZe^2}{r}$

Where:

• $U$ is the potential energy of the electron,
• $k$ is Coulomb's constant,
• $Z$ is the atomic number (number of protons in the nucleus),
• $e$ is the charge of an electron, and
• $r$ is the radius of the orbit of the electron around the nucleus.

The negative sign indicates that the potential energy is negative because the electron and nucleus attract each other.

The total energy (E) of the electron in orbit is the sum of its kinetic energy (K) and its potential energy (U). Since the electron is in a stable orbit, its kinetic energy can be shown to be exactly half the magnitude of its potential energy but positive:

$K = -\frac{1}{2}U$

Therefore,

$E = K + U = -\frac{1}{2}U + U = \frac{1}{2}U$

To find a relation between total energy (E) and potential energy (U), we rearrange the equation as follows:

$2E = U$

This is to say, the total energy (E) is half the magnitude of potential energy (U) but negative, and the correct relationship between them, when looking for a positive proportionality, yields to $2E = U$. Hence, the correct option is:

Option C: $2 \mathrm{E} = \mathrm{U}$