Atoms and Nuclei

$t=30$ 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} $

$ \begin{aligned} &\text { From the given condition, }\\ &E_C-E_A=\left(E_C-E_B\right)+\left(E_B-E_A\right) \end{aligned} $

In a nuclear reactor, the fuel consumption rate is given as $1 \times 10^{-3} \, \text{g/s}$.

The speed of light is $ c = 3 \times 10^8 \, \text{m/s} $.

For a nuclear reaction, the power generated $ P $ is given by the formula:

$ P = \frac{E}{t} = \frac{m c^2}{t} $

Substituting the given values, we have:

$ P = \frac{1 \times 10^{-3} \text{ g} \times (3 \times 10^8 \, \text{m/s})^2}{1 \, \text{s}} $

Converting grams to kilograms (since $1 \text{ g} = 10^{-3} \text{ kg}$):

$ P = 1 \times 10^{-6} \times 3 \times 10^8 \times 3 \times 10^8 $

This simplifies to:

$ P = 9 \times 10^{10} \, \text{W} $

Converting watts to kilowatts (since $1 \text{ W} = 10^{-3} \text{ kW}$):

$ P = 9 \times 10^7 \, \text{kW} $

Given:

$\lambda_1 = 1026 \, \mathring{A}$

$\lambda_2 = 304 \, \mathring{A}$

$R = 1.097 \times 10^7 \, \mathrm{m}^{-1}$

The excited state of $\mathrm{He}^{+}$ is denoted as $n_2$. The electron transitions from the $n_2$ state to the ground state via an intermediate state $n_1$, emitting two photons successively.

For the first transition:

$ \frac{1}{\lambda_1} = R Z^2 \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right] $

For the second transition:

$ \frac{1}{\lambda_2} = R Z^2 \left[\frac{1}{1} - \frac{1}{n_1^2}\right] $

Adding the above two equations:

$ \left[\frac{1}{\lambda_1} + \frac{1}{\lambda_2}\right] = R Z^2 \left[1 - \frac{1}{n_2^2}\right] $

Substitute the given values:

$ \left[\frac{1}{1026 \times 10^{-10}} + \frac{1}{304 \times 10^{-10}}\right] = 1.097 \times 10^7 \times 4 \left[1 - \frac{1}{n_2^2}\right] $

Calculate:

$ \left[0.974 \times 10^7 + 3.289 \times 10^7\right] = 1.097 \times 4 \times 10^7 \left[1 - \frac{1}{n_2^2}\right] $

$ \frac{4.263}{4.388} = 1 - \frac{1}{n_2^2} $

From the above,

$ \frac{1}{n_2^2} = 0.0284 \quad \Rightarrow \quad n_2^2 = 35.21 $

Thus,

$ n_2 = \sqrt{35.21} \approx 6 $

According to the calculations, the principal quantum number $n_2 = 6$.

The physical quantity measured in barns is the scattering cross-section.

In nuclear physics, a barn (symbol : b) is a unit of area used to quantity the likelihood of interaction between particles, such as the collision between a neutron and a nucleus. One barn is equal to $10^{-28}$ square metres.

To determine the ratio of the minimum wavelength in the Balmer series to the maximum wavelength in the Brackett series of the hydrogen spectrum, we start with the following calculations:

Maximum Wavelength of the Brackett Series:

The formula for the maximum wavelength in the Brackett series is given by:

$ \frac{1}{\lambda_{\max}} = R \left[\frac{1}{4^2} - \frac{1}{5^2}\right] $

Simplifying this, we have:

$ \frac{1}{\lambda_{\max}} = R \left[\frac{1}{16} - \frac{1}{25}\right] = \frac{9R}{25 \times 16} $

Minimum Wavelength of the Balmer Series:

For the Balmer series, the formula for the minimum wavelength is:

$ \frac{1}{\lambda_{\min}} = R \left(\frac{1}{2^2} - \frac{1}{\infty}\right) = \frac{R}{4} $

Ratio of Wavelengths:

We can now find the ratio of the minimum wavelength in the Balmer series to the maximum wavelength in the Brackett series:

$ \frac{\text{Minimum wavelength of Balmer series}}{\text{Maximum wavelength of Brackett series}} = \frac{9R}{25 \times 16} \times \frac{4}{R} = \frac{9}{100} $

Therefore, the ratio is:

$ 9:100 $

To find the age of the sample where the radioactive element $ A $ and the stable element $ B $ are present in a 1:15 ratio, we start by using the information given:

The ratio of $ A $ (denoted as $ N_X $) to $ B $ (denoted as $ N_Y $) is $ \frac{1}{15} $.

From this ratio:

$ \frac{N_X}{N_Y} = \frac{1}{15} $

The total presence of elements $ A $ and $ B $ can be written as:

$ N_X + N_Y = 16N_X $

We deduce that:

$ \frac{N_X}{N_X + N_Y} = \frac{1}{16} $

Rearranging, we have:

$ N_X = \frac{1}{16} (N_X + N_Y) $

Which implies:

$ N_X = \frac{1}{2^4} (N_X + N_Y) $

This result indicates that only $\frac{1}{16}$ of the original $ A $ remains. It is now a problem of calculating how many half-lives this corresponds to. Using the relationship that each half-life divides the remaining sample by 2:

After 1 half-life, the remaining amount is $ \frac{1}{2} $.

After 2 half-lives, the remaining amount is $ \frac{1}{4} $.

After 3 half-lives, the remaining amount is $ \frac{1}{8} $.

After 4 half-lives, the remaining amount is $ \frac{1}{16} $.

Thus, 4 half-lives have passed. With a half-life period of 62 years for element $ A $:

The age of the sample is:

$ 4 \times 62 = 248 \text{ years} $

Given:

Potential energy, $ \text{PE} = -6.8 \text{ eV} $

According to Bohr's postulate, the relationship between total energy $ E $ and kinetic energy $ K $ is:

$ E = K + \text{PE} $

We also know from Bohr's postulate:

$ E = \frac{\text{PE}}{2} $

Substituting the given potential energy:

$ E = \frac{-6.8 \text{ eV}}{2} = -3.4 \text{ eV} $

Using Bohr's relationship between kinetic and total energy:

$ -E = K $

Thus:

$ K = 3.4 \text{ eV} $

Given in equation form, where $ m_e $ is the mass of the electron and $ v_e $ is the speed of the electron:

$ K = \frac{1}{2} m_e v_e^2 $

Converting to joules:

$ 3.4 \times 1.6 \times 10^{-19} = \frac{1}{2} \times 9.1 \times 10^{-31} \times v_e^2 $

Solving for $ v_e $:

$ v_e = 1.093 \times 10^6 \text{ m/s} $

Expressing this speed in terms of the speed of light, $ c $:

$ v_e = \frac{c}{274} $

After calculating:

$ \frac{c}{274} = \frac{3 \times 10^8}{274} = 1.093 \times 10^6 \text{ m/s} $

Given:

The surface areas of two nuclei are in the ratio $9:25$.

This can be expressed mathematically as:

$ \frac{A_1}{A_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{9}{25} $

Simplifying, we get:

$ \frac{r_1^2}{r_2^2} = \frac{9}{25} $

Taking the square root of both sides, we find the ratio of the radii:

$ \frac{r_1}{r_2} = \frac{3}{5} \tag{i} $

Assuming the density of the nuclei is $\rho$, the ratio of their masses can be calculated using their volumes:

$ \frac{m_1}{m_2} = \frac{V_1 \cdot \rho}{V_2 \cdot \rho} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} $

Substituting the radius ratio:

$ \frac{m_1}{m_2} = \left(\frac{r_1}{r_2}\right)^3 = \left(\frac{3}{5}\right)^3 = \frac{27}{125} $

Thus, the mass number ratio of the nuclei is:

$ m_1 : m_2 = 27:125 $

Given that the ground state energy of a hydrogen atom is $-13.6 \text{ eV}$, this represents the total energy of the atom.

The kinetic energy (KE) of the electron in this state is equal to the negative of the total energy:

$ \mathrm{KE} = -E = -(-13.6) = 13.6 \text{ eV} $

The potential energy (PE) is calculated as the negative of two times the kinetic energy:

$ \begin{aligned} \mathrm{PE} & = -2 \times \mathrm{KE} \\ & = -2 \times 13.6 \\ & = -27.2 \text{ eV} \end{aligned} $

To determine the energy released in the fission of 0.1 kg of $\text{U}_{92}^{235}$, we start by calculating the number of atoms in this amount. The number of atoms in 235 g of uranium is $6.023 \times 10^{23}$.

Calculate the number of atoms in 100 g of uranium:

$ \text{Number of atoms} = \frac{6.023 \times 10^{23}}{235} \times 100 $

Given that the energy released per fission is 200 MeV, the total energy released is:

$ E_T = \frac{6.023 \times 10^{23} \times 200 \times 100}{235} \text{ MeV} $

Convert energy from MeV to Joules (since $1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}$):

$ E_T = \frac{6.023 \times 10^{23} \times 200 \times 1.6 \times 10^{-13} \times 100}{235} \text{ J} $

Convert energy from Joules to kilowatt-hours (since $1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$):

$ E_T = \frac{6.023 \times 10^{23} \times 200 \times 1.6 \times 10^{-13} \times 100}{235 \times 3.6 \times 10^6} \text{ kWh} $

Simplify the calculation:

$ E_T = 0.02278 \times 10^6 \times 10^2 $

The final energy released in kilowatt-hours is:

$ E_T = 22.8 \times 10^5 \text{ kWh} $

Ultraviolet rays are produced when inner shell electrons in atoms move from one energy level to a lower level.

To calculate the speed of the electron in a hydrogen atom when it is in the $ n=3 $ energy level, we use the formula:

$ v_n = \frac{e^2}{2 \varepsilon_0 h n} $

where:

$ e = 1.6 \times 10^{-19} \mathrm{C} $ is the charge of the electron,

$ \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m} $ is the permittivity of free space,

$ h = 6.64 \times 10^{-34} \mathrm{Js} $ is Planck's constant, and

$ n = 3 $ is the principal quantum number for the energy level.

Substituting these values into the formula, we calculate:

$ v_3 = \frac{(1.6 \times 10^{-19})^2}{2 \times 8.85 \times 10^{-12} \times 6.64 \times 10^{-34} \times 3} $

Simplifying this, we find:

$ v_3 \approx 0.0073 \times 10^8 \mathrm{~m/s} $

Thus, the speed of the electron in the $ n=3 $ level is:

$ v_3 = 7.3 \times 10^5 \mathrm{~m/s} $

To determine the decay constant for one mole of radium, we start with its given activity:

Activity (A) equation:

$ A = \frac{1}{3.7} \times 10^3 \text{ curie} $

Key constants:

1 Curie = $ 3.7 \times 10^{10} $ decays per second

Avogadro's Number, $ N_A = 6 \times 10^{23} \text{ atoms/mol} $

Calculate Decay Constant ($\lambda$):

The decay constant $\lambda$ can be found using the formula:

$ \lambda = \frac{A}{N_A} $

Substitute the values:

$ A = \frac{1}{3.7} \times 10^3 \times 3.7 \times 10^{10} \text{ decays/s} $

Therefore:

$ \lambda = \frac{\frac{1}{3.7} \times 10^3 \times 3.7 \times 10^{10}}{6 \times 10^{23}} $

Solving this gives:

$ \lambda = \frac{1}{6} \times 10^{-10} \text{ s}^{-1} $

Hence, the decay constant of radium is calculated to be $\frac{1}{6} \times 10^{-10} \text{ s}^{-1}$.

The Bohr radius $ r_n $ for the $ n $-th orbit in a hydrogen atom is defined as

$ r_n = n^2 \frac{\hbar^2}{m_e e^2} $

For the first Bohr orbit ( $ n=1 $ ), the formula simplifies to:

$ r_1 = \frac{\hbar^2}{m_e e^2} $

Here, $ \hbar $ is the reduced Planck's constant.

Now, consider the Bohr radius for the $\mu$-meson system:

$ r_n^{\prime} = n^2 \frac{\hbar^2}{m_\mu Z e^2} $

Given that $ m_\mu=208 m_e $ and $ Z=3 $, we want to find the value of $ n $ such that the $\mu$-meson's orbit $ r_n^{\prime} $ is approximately equal to the radius of the first Bohr orbit $ r_1 $ of hydrogen:

$ \begin{align*} r_n^{\prime} &= r_1 \\ n^2 \frac{\hbar^2}{208 m_e 3 e^2} &= \frac{\hbar^2}{m_e e^2} \\ n^2 \frac{1}{208 \times 3} &= 1 \\ n^2 &= 208 \times 3 = 624 \\ n &= \sqrt{624} \approx 25 \end{align*} $

Hence, the quantum state $ n $ is approximately 25.

An alpha particle has an atomic mass number of 4. Let's assume the original nucleus has an atomic mass number $ A $. After losing 2 alpha particles, the new nucleus will have an atomic mass number $ A' = A - 2 \times 4 = A - 8 $.

The volume of a nucleus is proportional to its atomic mass number $ A $, i.e., $ V \propto A $.

Given that the volume of the new nucleus is 60 times that of an alpha particle, we write:

$ V_{\text{new}} = 60 \times V_{\text{alpha}} $

Thus,

$ A' = 60 \times 4 = 240 $

Since,

$ A' = A - 8 $

We equate:

$ A - 8 = 240 $

Solving for $ A $:

$ A = 240 + 8 = 248 $

Therefore, the atomic mass number $ A $ of the original nucleus is 248.

To determine the initial energy level $ n $ of a hydrogen atom that releases 12.75 eV when transitioning to the first energy level, follow these steps:

The change in energy ($ \Delta E $) for an electron falling from the $ n $-th level to the first level is given by:

$ \Delta E = -\frac{13.6}{n^2} - \left(-\frac{13.6}{1^2}\right) = 12.75 $

Simplify the equation:

$ -\frac{13.6}{n^2} + 13.6 = 12.75 $

Isolate the energy term:

$ -\frac{13.6}{n^2} = 12.75 - 13.6 $

Calculate the energy difference:

$ -\frac{13.6}{n^2} = -0.85 $

Rearrange to solve for $ n^2 $:

$ n^2 = \frac{13.6}{0.85} $

Calculate $ n^2 $:

$ n^2 = 16 $

Finally, find $ n $:

$ n = 4 $

Therefore, the $ n $-th orbit from which the electron falls is $ n = 4 $.

To determine the daily decrease in uranium mass in a reactor operating at a power of 12 MW, we start by calculating the total energy produced in one day.

Total Energy Per Day:

The power output of the reactor is 12 MW. Since power is energy per unit time, the energy output over one day is calculated as follows:

$ \text{Total energy per day} = 12 \times 10^6 \, \text{J/s} \times 24 \times 3600 \, \text{s} $

Energy Released Per Fission:

For each fission event of ${ }_{92}\mathrm{U}^{235}$, approximately 200 MeV of energy is released. To convert this energy to joules:

$ 200 \, \text{MeV} = 200 \times 1.6 \times 10^{-13} \, \text{J} = 3.2 \times 10^{-11} \, \text{J} $

Calculating Fissions Per Day:

Using the relationship between power, energy per fission, and the number of fissions, we get:

$ \text{Power} = \frac{\text{Number of Fissions} \times \text{Energy per Fission}}{\text{Time}} $

Solving for the number of fissions ($n$):

$ 12 \times 10^6 = \frac{n \times 3.2 \times 10^{-11}}{24 \times 3600} $

$ n = \frac{12 \times 10^6 \times 24 \times 3600}{3.2 \times 10^{-11}} = 3.2 \times 10^{22} $

Calculating Mass of Uranium Consumed:

To find how much uranium is consumed, we first determine the number of moles:

$ \text{Number of moles} = \frac{3.2 \times 10^{22}}{6.023 \times 10^{23}} = 0.053 $

The mass of uranium consumed is then:

$ \text{Mass of uranium} = 0.0531 \times 235 \approx 12.64 \, \text{g} $

Thus, the daily decrease in uranium mass in the reactor is approximately 12.64 grams.

Given:

The binding energy for a hydrogen atom is 13.6 eV.

We know:

The energy of an electron in the $ n $-th state is given by:

$ E_n = -13.6 \frac{Z^2}{n^2} $

For $\mathrm{Li}^{2+}$, the atomic number $ Z = 3 $.

To find the energy required to remove the electron from the first excited state ($ n = 2 $):

Substitute into the equation:

$ \begin{aligned} E_2 &= -13.6 \times \frac{3^2}{2^2} = -13.6 \times \frac{9}{4} \\ E_2 &= -30.6 \, \text{eV} \end{aligned} $

Since $ E_2 + E = 0 $, we have:

$ -30.6 + E = 0 $

Thus, solving for $ E $:

$ E = +30.6 \, \text{eV} $

Therefore, the energy required to remove the electron from the first excited state of $\mathrm{Li}^{2+}$ is 30.6 eV.

Given:

Initial amount of $ A_1 = 40 \, \text{g} $

Initial amount of $ A_2 = 160 \, \text{g} $

Half-life of $ A_1, T_{A_1} = 20 \, \text{s} $

Half-life of $ A_2, T_{A_2} = 10 \, \text{s} $

To determine when the amounts of $ A_1 $ and $ A_2 $ will be equal, we use the decay formula:

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

For $ A_1 $, the remaining amount is:

$ N_{A_1} = 40 \left(\frac{1}{2}\right)^{t/20} $

For $ A_2 $, the remaining amount is:

$ N_{A_2} = 160 \left(\frac{1}{2}\right)^{t/10} $

To find the time when these amounts are equal, set $ N_{A_1} = N_{A_2} $:

$ \begin{aligned} 40 \left(\frac{1}{2}\right)^{t/20} &= 160 \left(\frac{1}{2}\right)^{t/10} \\ 2^{t/10} &= 4 \cdot 2^{t/20} \Rightarrow 2^{\frac{t}{10}-\frac{t}{20}} = 4 \\ 2^{t/20} &= 2^2 \Rightarrow \frac{t}{20} = 2 \\ t &= 40 \, \text{s} \end{aligned} $

Therefore, the amounts of $ A_1 $ and $ A_2 $ in the mixture will become equal after 40 seconds.

Change in angular momentum of the electron in H-atom when it is excited from 2nd orbit to 4th orbit is given as

$\Delta L=L_2-L_I$

Where, $L_2$ is angular momentum of 4th excited state $(n=5)$ and $L_1$ is angular moment of 2nd excited state $(n=3)$.

$\begin{aligned} \therefore \Delta L & =\frac{n_2 h}{2 \pi}-\frac{n_1 h}{2 \pi} \\ & =\frac{h}{2 \pi}\left[n_2-n_1\right]=\frac{6.64 \times 10^{-34}}{2 \times 314}[5-4] \\ & =2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s} \end{aligned}$

We know that, the radius $(R)$ of nucleus is given as

$R=R_0 A^{1 / 3} \text { or } R \propto A^{1 / 3}$

Where,

$\begin{aligned} R_0 & =\text { constant } \\ A & =\text { mass number } \end{aligned}$

$\therefore$ Nuclear density, $\rho=\frac{\text { mass }}{\text { volume }}=\frac{m A}{\frac{4}{3} \pi R^3}$

(here, $m$ is mass of each nucleon)

$\begin{aligned} \Rightarrow \quad \rho & =\frac{m A}{\frac{4}{3} \pi\left(R_0 A^{1 / 3}\right)^3}=\frac{m A}{\frac{4}{3} \pi R_0^3 A} \\ \rho & =\frac{3 m}{4 \pi R_0^3} \end{aligned}$

Hence, it is clear from above formula, nuclear density is independent of mass number $A$.

The binding energy, $E=\Delta m c^2$

i. e $E \propto \Delta m$

Where, $\Delta m=$ mass defect

Energy is released when heavy nuclei undergo transmutation into light nuclei.

For a carbon sample $(\mathrm{C}^{14})$,

$T_{1 / 2}=5730 \text { year }$

Decay constant,

$\begin{aligned} & K=\frac{0.693}{T_{1 / 2}}\\ & =\frac{0.693}{5730} \\ & =1.209 \times 10^{-4} / \text { year } \end{aligned}$

The rate of counts is proportional to the number of $\mathrm{C}^{14}$ atom in the sample

$N_0=100, N=75$

The age of the sample is given as,

$\begin{aligned} t & =\frac{2.303}{K} \log \frac{N_0}{N}=\frac{2.303}{K} \log \frac{1}{0.75} \\ & =\frac{2303}{1.209 \times 10^{-4}} \log \frac{100}{75} \\ & =2456 \times 10^3 \text { years } \\ & =2456 \text { years } \end{aligned}$

Energy of hydrogen atom in ground state,

$E_0=-13.6 \mathrm{~eV}$

Energy of electron in excited state $(n=4)$,

$\begin{aligned} E^{\prime}=\frac{-13.6}{n^2} & \\ & =\frac{-13.6}{4^2}=\frac{-13.6}{16}=-0.85 \mathrm{eV} \end{aligned}$

Potential energy of the electron in the excited state $(n=4)$,

$\begin{aligned} U & =2 E^{\prime} \\ & =2 \times(-0.85 \mathrm{eV}) \\ & =-1 \cdot 7 \mathrm{eV} \end{aligned}$

We know that, radius (R) of atomic nucleus and mass number (A) are related as

$\begin{array}{ll} & R=R_0 A^{1 / 3} \\ \Rightarrow & R \propto A^{1 / 3} \\ \Rightarrow & \frac{R_2}{R_1}=\left(\frac{A_2}{A_1}\right)^{1 / 3} \\ \Rightarrow & \frac{A_2}{A_1}=\left(\frac{R_2}{R_1}\right)^3 \end{array}$

Given, $A_1=64, R_1=4.8$ fermi, $R_2=6$ fermi

Putting these values in above equation, we get

$\begin{aligned} & \frac{A_2}{64}=\left(\frac{6}{4.8}\right)^3=\left(\frac{5}{4}\right)^3 \\ & \Rightarrow \quad \frac{A_2}{64}=\frac{125}{64} \\ & \Rightarrow \quad A_2^{\prime}=125 \\ & \end{aligned}$

Energy of electron in $n$th orbit of hydrogen atom,

$E=\frac{-13.6}{n^2} \mathrm{~eV}$

energy of electron in ground state ($n=1$),

$E_0=\frac{-13.6}{1^2}=-13.6 \mathrm{~eV}$

for second excited state, $n=3$

$\therefore$ energy of electron in H -atom in second excited state

$E^{\prime}=\frac{-13.6}{3^2}=-1.51 \mathrm{~eV}$

Energy required to excite the electron in hydrogen atom to second excited state

$=E^{\prime}-E_0$

$\begin{aligned} & =-1.51-(-13.6)=1209 \mathrm{eV} \\ & \simeq 12 \mathrm{~eV} \end{aligned}$

Energy of ionised electron of $(n=\infty) \mathrm{H}$ - atom.

$E^{\prime \prime}=\frac{-13.6}{\infty}=0 \mathrm{~eV}$

energy required to ionise the electron

$=E^{\prime \prime}-E_0=0-(-13.6)=13.6 \mathrm{~eV}$

Radius of nucleus (R) having mass number (A) is given as,

$R=R_0 A^{1 / 3}$

where, $R_0$ is constant.

Taking $\log$ both side, we have

$\begin{aligned} & \ln R=\ln R_0 A^{1 / 3} \\ & \Rightarrow \ln R=\ln R_0+\ln A^{1 / 3} \\ & \Rightarrow \ln R=\ln R_0+\frac{1}{3} \ln A \\ & \Rightarrow \ln R=\frac{1}{3} \ln A+\ln R_0 \end{aligned}$

this is in the form of equation of straight line

$y=m x+c$

where, $y=\ln R$

$m=\frac{1}{3}, x=\ln A$

and $\quad c=\ln R_0$

As we know that,

By using de-Broglie wavelength,

$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m e V}}$

where, $\lambda$ is wavelength of $X$-ray, $h$ is Planck's constant i.e. $6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}, p$ is, momentum, $m$ is mass, $E$ is energy, $e$ is charge and $V$ is electric potential.

$\therefore \quad \lambda \propto \frac{1}{\sqrt{V}} $

By using Rydberg's formula

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

For Lyman series,

$\begin{array}{rlrl} n_1 & =1 \text { and } n_2=2 \\ \therefore \quad \quad \frac{1}{\lambda} & =R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R\left(\frac{1}{1}-\frac{1}{4}\right)=R\left(\frac{3}{4}\right) \\ \Rightarrow \quad \lambda & =4 / 3 R=1215 \times 10^{-10} \mathrm{~m} \\ & =\frac{4}{3 \times 1.097 \times 10^7}=1215 \mathop A\limits^o\end{array}$

As we know that,

In case of $(-\beta)$ - decay we get $+e_1^0$ positron, so atomic number of reactant increases get by 1

and in case of $\gamma$-decay we get only gamma-ray and no change in atomic number as well as mass.

$\therefore \mathrm{Np}_{93}^{239} \longrightarrow \mathrm{Pu}_{94}^{239}+\beta^{-}+\gamma^{-}$ is possible whereas (a), (b) and (d) are not possible.

As we know that, angular momentum of electron, $L=nh/2\pi$.

$\therefore$ L is integral multiple of $h/2\pi$.

As we know that, nuclear density of any atom $=2\times10^{17}$ kg/m$^3$

$\therefore$ Option (d) is correct.