Atoms and Nuclei

Total energy remains conserved. Energy is shared by antineutrino, proton and electron. Kinetic energy of electron has continuous spectrum and it is maximum when antineutrino does not share any kinetic energy.

So total energy is shared with proton and electron only.

$\therefore$ K $\le$ 0.8 $\times$ 10⁶ eV

and kinetic energy of electron will be minimum or zero when total energy is shared by proton and antineutrino.

$\therefore$ 0 $\le$ K $\le$ 0.8 $\times$ 10⁶ eV

${1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$

For singly-ionized helium atom, Z = 2. For hydrogen atom Z = 1.

For Balmer series n₁ = 2.

For first spectral line of hydrogen :

${1 \over {6561}} = {R_H} \times {(1)^2}\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right] = {{5R} \over {36}}$

$ \Rightarrow {R_H} = {{36} \over {5 \times 6561}}$

For second spectral line of helium,

${1 \over \lambda } = {R_H} \times {(2)^2}\left[ {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right] = {{3{R_H}} \over 4}$

$ = {3 \over 4} \times {{36} \over {5 \times 6561}}$

$ \Rightarrow \lambda = 1215\,\mathop A\limits^o $

To find the quantized rotational energy of a diatomic molecule, we can use Bohr's quantization condition for angular momentum. Bohr's quantization condition states that the angular momentum $ L $ of an electron in a hydrogen atom is quantized and given by:

$ L = n \hbar $

where $ n $ is a positive integer (known as the quantum number) and $ \hbar $ is the reduced Planck's constant, given by $ \hbar = {\hbar \over {2\pi}} $.

For a diatomic molecule, we extend this concept to its rotational motion. A rigid diatomic molecule can be approximated as a rigid rotor with a moment of inertia $ I $. The angular momentum $ L $ for the molecule in the nth level is given by:

$ L = n \hbar $

where $ n $ is the quantum number (note that $ n = 0 $ is not allowed).

The rotational kinetic energy for a rotating body with moment of inertia $ I $ is given by:

$ E = {1 \over 2} I \omega^2 $

where $ \omega $ is the angular velocity. The angular momentum $ L $ is related to the angular velocity $ \omega $ by:

$ L = I \omega $

Solving for $ \omega $, we get:

$ \omega = {L \over I} $

Substituting this back into the expression for energy, we get:

$ E = {1 \over 2} I \left( {L \over I} \right)^2 = {L^2 \over 2I} $

Using Bohr's quantization condition $ L = n \hbar $, the energy expression becomes:

$ E = {{(n \hbar)^2} \over {2I}} $

Substituting $ \hbar = {\hbar \over {2\pi}} $, we obtain:

$ E = {n^2 \over {2I}} \left( {{\hbar^2} \over {4\pi^2}} \right) = n^2 \left( {{\hbar^2} \over {8\pi^2 I}} \right) $

Therefore, the quantized rotational energy of the diatomic molecule in the nth level is:

$ E_n = n^2 \left( {{\hbar^2} \over {8\pi^2 I}} \right) $

So, the correct option is:

Option D

$ n^2 \left( {{{h^2} \over {8\pi^2 I}}} \right) $

The energy of photon is equal to the energy difference between the ground level and first excited level.

$hv = {E_2} - {E_1}$

$hv = {{(4 - 1){h^2}} \over {8{\pi ^2}I}} \Rightarrow I = {{3h} \over {8{\pi ^2}v}}$

$I = {{3 \times 2\pi \times {{10}^{ - 34}}} \over {[(8{\pi ^2}4)/\pi ] \times {{10}^{11}}}} = {3 \over {16}} \times {10^{ - 45}}$ kg-m² = 1.87 $\times$ 10^$-$46 kg-m²

The moment of inertia of CO molecule is

I = $\mu$r² ....... (i)

where, $\mu$ = reduced mass of the CO molecule, r = distance between C and O or bond length

The reduced mass $\mu$ of the CO molecule is

$\mu = {{{m_1}{m_2}} \over {{m_1} + {m_2}}} = \left[ {{{(12)(16)} \over {12 + 16}}} \right] \times {5 \over 3} \times {10^{ - 27}}$ kg

But $I = 1.87 \times {10^{ - 46}}$ kg m² (From the above question)

From equation (i), we get

${r^2} = {I \over \mu }$

Substituting the values of I and $\mu$ in above equation, we get

${r^2} = {{1.87 \times {{10}^{ - 46}}} \over {\left[ {{{12 \times 16} \over {28}} \times {5 \over 3} \times {{10}^{ - 27}}} \right]}}$

or ${r^2} = {{1.87 \times {{10}^{ - 46}} \times 28 \times 3} \over {12 \times 16 \times 5 \times {{10}^{ - 27}}}}$ or $r = 1.3 \times {10^{ - 10}}$ m

To find the relationship between the speed of the particle and the quantum number $n$, we need to follow a few steps using the given information and fundamental quantum mechanical principles.

Firstly, for a particle moving in a one-dimensional box of length $a$, the allowed wavelengths $\lambda$ of standing waves can be given by:

$ \lambda = \frac{2a}{n} $

for $ n = 1, 2, 3, \ldots $.

The de Broglie relation links the wavelength $\lambda$ of a particle to its momentum $p$:

$ \lambda = \frac{h}{p} $

From this, we can express the momentum $p$ in terms of the quantum number $n$:

$ \frac{h}{p} = \frac{2a}{n} $

Solving for $p$, we get:

$ p = \frac{nh}{2a} $

The kinetic energy $E$ of the particle is related to its momentum $p$ by the equation:

$ E = \frac{p^2}{2m} $

Substituting $ p = \frac{nh}{2a} $ into the energy expression, we get:

$ E = \frac{(nh/2a)^2}{2m} = \frac{n^2 h^2}{8ma^2} $

The kinetic energy can also be written in terms of the speed $v$ of the particle:

$ E = \frac{1}{2} m v^2 $

Equating the two expressions for energy, we have:

$ \frac{1}{2} m v^2 = \frac{n^2 h^2}{8ma^2} $

Solving for the speed $v$, we get:

$ v^2 = \frac{n^2 h^2}{4m^2 a^2} $

Therefore, the speed $v$ of the particle is:

$ v = \frac{nh}{2ma} $

From the above expression, we see that the speed $v$ of the particle is directly proportional to $n$:

$ v \propto n $

Therefore, the speed of the particle is proportional to:

Option D: $ n $

In a nuclear fusion reactor, the gas becomes plasma primarily because of the high temperature maintained inside the reactor core. At such high temperatures, typically in the range of millions of degrees Kelvin, the thermal energy is sufficient to ionize the atoms, meaning electrons are stripped from the nuclei, resulting in a collection of positively charged deuteron nuclei and negatively charged electrons. This ionized state of matter is known as plasma.

The strong nuclear force and Coulomb force (between deuterons and between deuteron-electron pairs) play roles in the fusion process and the behavior of particles. However, the ionization into plasma is primarily due to the extremely high temperatures.

Therefore, the correct option is:

Option D: the high temperature maintained inside the reactor core.

To determine the minimum temperature required for two deuteron nuclei to reach a separation of $4 \times 10^{-15}$ m, we need to equate the initial kinetic energy of the deuterons with the potential energy at the specified separation distance.

The initial kinetic energy of each deuteron is given as :

$ KE = 1.5 \, k \, T $

Since there are two deuterons, the total kinetic energy is :

$ KE_{\text{total}} = 2 \times 1.5 \, k \, T = 3 \, k \, T $

The Coulomb potential energy (PE) between two deuterons at a separation distance $r = 4 \times 10^{-15}$ m is given by :

$ PE = \frac{e^2}{4 \pi \epsilon_0 r} $

Using the given value:

$ \frac{e^2}{4 \pi \epsilon_0} = 1.44 \times 10^9 \, \text{eV} \cdot \text{m} $

So, the potential energy becomes :

$ PE = \frac{1.44 \times 10^9 \, \text{eV} \cdot \text{m}}{4 \times 10^{-15} \, \text{m}} $

Simplifying :

$ PE = \frac{1.44 \times 10^9}{4 \times 10^{-15}} \, \text{eV} $

$ PE = 0.36 \times 10^{24} \, \text{eV} $

Now, we equate the total kinetic energy to the potential energy to find the temperature $ T $ :

$ 3 \, k \, T = 0.36 \times 10^{24} \, \text{eV} $

Solving for $ T $ :

$ T = \frac{0.36 \times 10^{24}}{3 \times 8.6 \times 10^{-5}} \, \text{K} $

$ T = \frac{0.36 \times 10^{24}}{2.58 \times 10^{-4}} \, \text{K} $

$ T = 1.395 \times 10^{9} \, \text{K} $

This temperature is in the range :

Option A: $1.0 \times {10^9}K < T < 2.0 < {10^9}K$

So, the correct answer is :

Option A : $1.0 \times {10^9}K < T < 2.0 < {10^9}K$

To determine the most promising design based on the Lawson criterion, we need to calculate the Lawson number ($nt_0$) for each design and check if it exceeds $5 \times 10^{14}$ s/cm$^3$. We will do this for each of the given options:

Option A:

Deuteron density, $n = 2.0 \times 10^{12}~\mathrm{cm^{-3}}$

Confinement time, $t_0 = 5.0 \times 10^{-3}~\mathrm{s}$

Lawson number, $nt_0 = n \cdot t_0 = (2.0 \times 10^{12}) \cdot (5.0 \times 10^{-3}) = 1.0 \times 10^{10}~\mathrm{s/cm^3}$

Option B:

Deuteron density, $n = 8.0 \times 10^{14}~\mathrm{cm^{-3}}$

Confinement time, $t_0 = 9.0 \times 10^{-1}~\mathrm{s}$

Lawson number, $nt_0 = n \cdot t_0 = (8.0 \times 10^{14}) \cdot (9.0 \times 10^{-1}) = 7.2 \times 10^{14}~\mathrm{s/cm^3}$

Option C:

Deuteron density, $n = 4.0 \times 10^{23}~\mathrm{cm^{-3}}$

Confinement time, $t_0 = 1.0 \times 10^{-11}~\mathrm{s}$

Lawson number, $nt_0 = n \cdot t_0 = (4.0 \times 10^{23}) \cdot (1.0 \times 10^{-11}) = 4.0 \times 10^{12}~\mathrm{s/cm^3}$

Option D:

Deuteron density, $n = 1.0 \times 10^{24}~\mathrm{cm^{-3}}$

Confinement time, $t_0 = 4.0 \times 10^{-12}~\mathrm{s}$

Lawson number, $nt_0 = n \cdot t_0 = (1.0 \times 10^{24}) \cdot (4.0 \times 10^{-12}) = 4.0 \times 10^{12}~\mathrm{s/cm^3}$

Based on the calculations, the Lawson numbers for each option are:

• Option A: $1.0 \times 10^{10}~\mathrm{s/cm^3}$
• Option B: $7.2 \times 10^{14}~\mathrm{s/cm^3}$
• Option C: $4.0 \times 10^{12}~\mathrm{s/cm^3}$
• Option D: $4.0 \times 10^{12}~\mathrm{s/cm^3}$

The most promising design based on the Lawson criterion, which requires the Lawson number to be greater than $5 \times 10^{14}~\mathrm{s/cm^3}$, is Option B because its Lawson number exceeds the threshold.

To solve this problem, let's begin with some fundamental concepts of radioactive decay.

The activity of a radioactive sample, which is its decay rate, is given by the equation:

$A = \lambda N$

where:

• $A$ is the activity.
• $\lambda$ is the decay constant.
• $N$ is the number of radioactive nuclei present.

Given that sample S₁ has an activity of 5 $\mu$Ci and sample S₂ has an activity of 10 $\mu$Ci, and that sample S₁ has twice the number of nuclei as sample S₂, we have:

$A_1 = \lambda_1 N_1$

$A_2 = \lambda_2 N_2$

We are also given that:

$N_1 = 2N_2$

Substitute $N_1 = 2N_2$ into the first equation:

$5 \, \mu\text{Ci} = \lambda_1 (2N_2)$

And for the second equation:

$10 \, \mu\text{Ci} = \lambda_2 N_2$

Dividing the first equation by the second equation to eliminate $N_2$, we get:

$\frac{5 \, \mu\text{Ci}}{10 \, \mu\text{Ci}} = \frac{\lambda_1 (2N_2)}{\lambda_2 N_2}$

Simplifying this, we obtain:

$\frac{1}{2} = \frac{2 \lambda_1}{\lambda_2}$

Therefore:

$\lambda_2 = 4 \lambda_1$

Next, we can relate the decay constant to the half-life using the equation:

$\lambda = \frac{\ln(2)}{T_{1/2}}$

Using the relationship between the decay constants, we have:

$\frac{\lambda_2}{\lambda_1} = \frac{T_{1/2,1}}{T_{1/2,2}}$

Substituting $\lambda_2 = 4 \lambda_1$, we get:

$4 = \frac{T_{1/2,1}}{T_{1/2,2}}$

Therefore:

$T_{1/2,1} = 4 T_{1/2,2}$

Given the options, the half-lives that satisfy this relationship are:

Option A: 20 years and 5 years, respectively.

In this case:

$\frac{20 \text{ years}}{5 \text{ years}} = 4$

Thus, option A is correct. Therefore, the half-lives of S₁ and S₂ are 20 years and 5 years, respectively.

For H atom,

${E_1} = - {{13.6} \over {{1^2}}} = - 13.6\,eV$

${E_2} = - {{13.6} \over {{2^2}}} = - 3.4\,eV$

Energy released by H atom = E$_2$ $-$ E$_1$

$ = - 3.4 - ( - 13.6) = 10.2\,eV$

This energy will be absorbed by He atom. Thus, for He atom

$10.2 = - 13.6 \times {2^2}\left( {{1 \over {{2^2}}} - {1 \over {{n^2}}}} \right)$

$0.1875 = {1 \over 4} - {1 \over {{n^2}}}$

${n^2} = 16$

$n = 4$

${{hc} \over \lambda } = {E_4} - {E_3}$

${{hc} \over \lambda } = - 13.68 \times {2^2}\left( {{1 \over {{4^2}}} - {1 \over {{3^2}}}} \right)$

${{hc} \over \lambda } = 13.6 \times {2^2}\left( {{1 \over 9} - {1 \over {16}}} \right)$

${{4.1356 \times {{15}^{ - 15}}\,eV.s \times 3 \times {{10}^8}\,m{s^{ - 1}}} \over 1} = 2.644$

$ \Rightarrow \lambda \cong 4.8\, * \,{10^{ - 7}}\,m$

To determine the ratio of the kinetic energy of the $n=2$ electron for the H atom to that of the He$^+$ ion, we need to consider the Bohr model of the atom. According to the Bohr model, the kinetic energy (KE) of an electron in a hydrogen-like ion in the nth orbit can be given by:

$ KE_n = \frac{1}{2} m v_n^2 = \frac{Z^2 e^4 m}{8 \epsilon_0^2 h^2 n^2} $

where:

• $ m $ is the mass of the electron
• $ Z $ is the atomic number of the nucleus
• $ e $ is the charge of the electron
• $ \epsilon_0 $ is the permittivity of free space
• $ h $ is Planck's constant
• $ n $ is the principle quantum number, which is 2 in this case

For a hydrogen atom (H) ($ Z = 1 $) in the $ n = 2 $ state:

$ KE_{H (n=2)} = \frac{1^2 \cdot e^4 m}{8 \epsilon_0^2 h^2 \cdot 2^2} = \frac{e^4 m}{32 \epsilon_0^2 h^2} $

For a singly ionized helium ion (He$^+$) ($ Z = 2 $) in the $ n = 2 $ state:

$ KE_{He^+ (n=2)} = \frac{2^2 \cdot e^4 m}{8 \epsilon_0^2 h^2 \cdot 2^2} = \frac{4 e^4 m}{32 \epsilon_0^2 h^2} = \frac{e^4 m}{8 \epsilon_0^2 h^2} $

To find the ratio of the kinetic energy of the $ n=2 $ electron for the H atom to that of the He$^+$ ion, we divide the kinetic energies:

$ \text{Ratio} = \frac{KE_{H (n=2)}}{KE_{He^+ (n=2)}} = \frac{\frac{e^4 m}{32 \epsilon_0^2 h^2}}{\frac{e^4 m}{8 \epsilon_0^2 h^2}} = \frac{1}{4} $

Therefore, the ratio of the kinetic energy of the $ n=2 $ electron for the H atom to that of the He$^+$ ion is:

Option A: $\frac{1}{4}$

Iodine and Yttrium are medium sized nuclei and therefore, have more binding energy per nucleon as compared to uranium which has a big nuclei and less binding energy per nucleon. So that Iodine and Yttrium are more stable and therefore possess less energy and less restmass. When uranium nuclei explodes, it will convert into I and Y nuclei having kinetic energies.

For largest $\lambda$ in UV $\to$ 122 nm

For $\Delta$E$_{\mathrm{min}}$,

${1 \over {{\lambda _{lyman}}}} = {R_c}\left( {1 - {1 \over 4}} \right)$

[$\because$ ${1 \over \lambda } = {R_c}\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$]

${1 \over {{\lambda _{lyman}}}} = {{3{R_c}} \over 4}$ ..... (i)

R$_c$ = Rydber constant = 1.097 $\times$ 10$^7$ m$^{-1}$

For min $\lambda$ in IR region, n = D $\to$ n = 3

${1 \over {{\lambda _{IR}}}} = {{{R_c}} \over 9}$ ...... (ii)

On dividing eq. (i) by (ii)

${1 \over {{\lambda _{lyman}}}} \times {\lambda _{IR}} = {1 \over {{{3{R_c}} \over 4}}} \times {9 \over {{R_c}}} = {{3{R_c}} \over 4} \times {9 \over {{R_c}}} = {{27} \over 4}$

$\therefore$ ${\lambda _{IR}} = {{27} \over 4} \times 122$ nm = 823.5 nm

In a nuclear fusion reaction, two or more lighter nuclei combine to give a comparatively heavier nucleus and some matter is converted into energy.

In a nuclear fission reaction, a heavy nuclear breaks into two or more lighter nuclei and some matter is converted into energy.

$\beta$-decay essentially proceeds by weak nuclear forces and converts some matter into energy. Exothermic nuclear reaction is possible for both type of nuclei (with low and high atomic number) and releases energy.

(A) Any static charge produces electric field $\forall$ invariant with time. Uniformly charged dielectric does the same.

(B) Rotating charged behaves like a current. hence, magnetic field is produced around the ring and also if there is current in rotating ring, ring has magnetic moment.

(C) Since, net charge is zero there will be no time independent electric field. The current produces magnetic field and magnetic moment.

(D) A changing magnetic field will be produced. This will create a induced electric field.

Also a changing magnetic moment will be produced.

(A) Using

$R=R_{0} A^{1 / 3}$

$\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$

$\frac{\mathrm{R}}{\mathrm{R}_{\mathrm{He}}}=\left(\frac{\mathrm{A}}{4}\right)^{1 / 3}$

$(14)^{1 / 3}=\left(\frac{\mathrm{A}}{4}\right)^{1 / 3}$

$A=14 \times 4=56$

So, $Z=(56-30)=26$

$=\text { atomic number }$

(B) Frequency of $K_{\alpha}$ line $=f$

$\therefore f=\mathrm{R} c(z-b)^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right], b=1$ for $\mathrm{K} \alpha, z=26$

$f=\left(1.1 \times 10^{7}\right)\left(3 \times 10^{8}\right)(26-1)\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$

$f=\frac{3.3 \times 10^{15} \times 25 \times 25 \times 3}{4}$

$f=1.547 \times 10^{18} \mathrm{~Hz}$

$\therefore$ Frequency of $\mathrm{K} \alpha$ line of $\mathrm{X}$-rays

$=1.547 \times 10^{18} \mathrm{~Hz}=1.55 \times 10^{18} \mathrm{~Hz} \text {. }$

The central force will provide necessary centripetal force

$\Rightarrow \mathrm{kr}=\frac{\mathrm{mv}^2}{\mathrm{r}}$

or, $\mathrm{kr}^2=\mathrm{mv}^2\quad \text{... (1)}$

By quantisation rule $\mathrm{n} \hbar=\mathrm{mvr}$

or, $\frac{\mathrm{n} \hbar}{\mathrm{r}}=\mathrm{mv}\quad \text{... (2)}$

$\begin{aligned} & \frac{(1)}{(2)^2} \Rightarrow \frac{\mathrm{kr}^2}{\frac{\mathrm{n}^2 \hbar^2}{\mathrm{r}^2}}=\frac{\mathrm{mv}^2}{\mathrm{~m}^2 \mathrm{v}^2} \\ & \Rightarrow \frac{\mathrm{k}}{\mathrm{n}^2 \hbar^2} \mathrm{r}^4=\frac{1}{\mathrm{~m}} \\ & \Rightarrow \mathrm{r}=\left(\frac{\mathrm{n}^2 \hbar^2}{\mathrm{~km}}\right)^{\frac{1}{4}} \Rightarrow \mathrm{r}^2=\frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}} \end{aligned}$

(B) Using (1), $\mathrm{K} \cdot \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}=\mathrm{mv}^2$

$\Rightarrow \mathrm{v}^2=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}^3}}$

(C) $\frac{\mathrm{L}}{\mathrm{mr}^2}=\frac{\mathrm{mvr}}{\mathrm{mr}^2}=\frac{\mathrm{v}}{\mathrm{r}}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ from (1)

(D) $\mathrm{E}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{kr}^2=\frac{\mathrm{n} \hbar}{2} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}+\frac{1}{2} \mathrm{k} \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}$

$\mathrm{E}=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

Total binding energy (without considering repulsions),

${E_b} = [Z{m_p} + (A - Z){m_n} - {m_x}]{c^2}$

Where, $_Z^AX$ is the nuclei under consideration.

Now, considering repulsion :

Number of proton pairs $ = {}^z{C_2}$

$\Rightarrow$ This repulsion energy $ \propto \,{{Z(Z - 1)} \over 2} \times {1 \over {4\pi { \in _0}}}{{{e^2}} \over R}$

Where R is the radius of the nucleus

$ \Rightarrow E_b^p - E_b^n \propto Z(Z - 1)$ $\therefore$ there will be no repulsion term for neutrons.

Also, since $R = {R_0}{A^{1/3}}$

$ \Rightarrow E_b^p - E_b^n \propto {A^{ - 1/3}}$

Because of repulsion among protons,

$E_b^p < E_b^n$

Since in $\beta^+$ decay, number of protons decrease $\Rightarrow$ repulsion would decrease

$ \Rightarrow E_b^p$ increases

(p_total) = 0 $\Rightarrow$ (p_total)_f = 0

So, momentum of one nucleus is p in forward direction, then momentum of the other nucleus will be p in backwards direction.

Energy released = ($\Delta$m)c² = $\delta$c² = KE_P + KE_Q

$K{E_P} = {{{p^2}} \over {2{m_P}}},K{E_Q} = {{{p^2}} \over {2{m_Q}}}$

$ \Rightarrow K{E_P}:K{E_Q} = {1 \over {{m_P}}}:{1 \over {{m_Q}}} = {m_Q}:{m_P}$

$K{E_P} = \left( {{{{m_Q}} \over {{m_P} + {m_Q}}}} \right)\delta {c^2}$

$K{E_Q} = \left( {{{{m_P}} \over {{m_P} + {m_Q}}}} \right)\delta {c^2}$

$K{E_P} + K{E_Q} = \delta {c^2}$

${{{p^2}} \over {2{m_P}}} + {{{p^2}} \over {2{m_Q}}} = \delta {c^2} \Rightarrow p = c\sqrt {2\left( {{{{m_P}{m_Q}} \over {{m_P} + {m_Q}}}} \right)\delta } $

For Balmer series

$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{n^2}\right) n>2 $

$\lambda$ is minimum for transition from $n_1 \rightarrow \infty$ to $n_2=2$.

$ \frac{1}{\lambda_{\min }}=\frac{R}{4} $

$\lambda$ is maximum for transition from $n_1=3$ to $n_2=2$

$ \frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 \mathrm{R}}{36} $

Dividing the above two equations, we get

$ \frac{\lambda_{\max }}{\lambda_{\min }}=\frac{R}{4} \times \frac{36}{5 R}=\frac{9}{5} $

Option (A) is correct.

For Lyman series,

$\lambda_{\min }=91.18 \mathrm{~nm}$

$\lambda_{\max }=121.57 \mathrm{~nm}$

For Balmer series,

$\lambda_{\min }=364.60 \mathrm{~nm}$

$\lambda_{\max }=656.30 \mathrm{~nm}$

For Pashen series,

$\lambda_{\min }=820.40 \mathrm{~nm}$

$\lambda_{\max }=1875 \mathrm{~nm}$

We can see from the above that there is no overlap between the wavelength ranges of Balmer and Paschen series. so, option (B) is incorrect.

Similarly wavelength ranges of Lyman and Balmer series also do not overlap. Hence, option (D) is correct.

Shortest wavelength of Lyman series

$ \begin{aligned} \frac{1}{\lambda_0} & =R\left(1-\frac{1}{\infty}\right) \\\\ \lambda_0 & =\frac{1}{R} \end{aligned} $

For Lyman series, $\frac{1}{\lambda}=\mathrm{R}\left(1-\frac{1}{n^2}\right)$

$ \begin{aligned} & =\frac{1}{\lambda_0}\left(1-\frac{1}{n^2}\right) \\\\ \lambda & =\frac{\lambda_0}{1-\frac{1}{n^2}}=\frac{\lambda_0 n^2}{n^2-1} \\\\ & =\frac{\lambda_0\left(n^2-1\right)+1}{n^2-1} \\\\ & =\lambda_0\left(1+\frac{1}{n^2-1}\right) \end{aligned} $

If $n^2-1=m^2$ ( $m$ is an integer)

then this is rarely true as $\left(n^2-1\right)$ cannot always be the square of an integer. Hence, option (C) is incorrect.

$\lambda \propto {1 \over V}$

So, when V double, then cut-off wavelength become half. Characteristic does not depend on voltage. When I is halved, then intensity will decrease.

$|Force| = \left| { - {{dV} \over {dr}}} \right| \Rightarrow F = {{m{v^2}} \over r}$ .....(i)

$mvr = {{nh} \over {2\pi }}$ .....(ii)

From Eqs. (i) and (ii), we get

$ \therefore $ $F = {m \over r}{\left( {{{nh} \over {2\pi mr}}} \right)^2}$ $ \because $ $\left( {v = {{nh} \over {2\pi mr}}} \right)$

or $F = {{m{n^2}{h^2}} \over {4{\pi ^2}{m^2}{r^3}}}$

or $F = {{{n^2}{h^2}} \over {4{\pi ^2}m{r^3}}}$

${r^3} \propto \,{n^2}$ (as F = constant)

$r \propto \,{n^{2/3}}$

$v = {{nh} \over {2\pi mr}}$

$v \propto {n \over {{n^{2/3}}}}$

or $v \propto {n^{1/3}}$

$ \therefore $ $KE = {1 \over 2}m{v^2} = {{Fr} \over 2}$

$ \because $ $TE = PE + KE = Fr + {{Fr} \over 2}$

$ \Rightarrow {{3Fr} \over 2} = TE$

or $TE = E = {{3F} \over 2} \times {\left( {{{{n^2}{h^2}} \over {4{\pi ^2}rnF}}} \right)^{1/3}}$

$ = {3 \over 2}{\left( {{{{n^2}{h^2}{F^2}} \over {4{\pi ^2}m}}} \right)^{1/3}}$

A hydrogen atom in its ground state (n = 1) absorbs a photon with wavelength $\lambda_a$ and gets excited to the n = 4 state. Subsequently, the electron transitions to the n = m state, emitting a photon with wavelength $\lambda_e$. Given that ${{{\lambda _a}} \over {{\lambda _e}}} = {1 \over 5}$, let's analyze the options systematically.

Calculation for the Wavelengths:

The energy difference between levels in a hydrogen atom can be given by the Rydberg formula for hydrogen:

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

The energy of the absorbed photon that excites the electron from n = 1 to n = 4 could be expressed as:

$ E_a = E_4 - E_1 = - \frac{{13.6 \ eV}}{{4^2}} - (- \frac{{13.6 \ eV}}{{1^2}}) = - \frac{{13.6 \ eV}}{{16}} + 13.6 \ eV = 13.6 \ eV (1 - \frac{1}{16}) = 13.6 \ eV \cdot \frac{15}{16} = 12.75 \ eV$

Wavelength $\lambda_a$ associated with this energy is:

$ \lambda_a = \frac{{hc}}{{E_a}} = \frac{{1242 \ nm \cdot eV}}{{12.75 \ eV}} = 97.4 \ nm$

Given ${{{\lambda _a}} \over {{\lambda _e}}} = {1 \over 5}$, we get:

$ \lambda_e = 5 \lambda_a = 5 \cdot 97.4 \ nm = 487 \ nm$

Option A: The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is ${1 \over 4}$

The kinetic energy of an electron in a hydrogen atom in the n^th state is proportional to $- \frac{{1}}{{n^2}}$. Therefore, the ratio of kinetic energy in state m to that in state n can be written as:

$ \frac{{E_m}}{{E_1}} = \frac{{- \frac{{13.6}}{{m^2}} eV}}{{- 13.6 \ eV}} = \frac{1}{{m^2}} $

According to the question, the ratio is $\frac{1}{4}$:

$ \frac{1}{{m^2}} = \frac{1}{4} \Rightarrow m = 2 $

Therefore, Option A is correct if m = 2.

Option B: m = 2

Based on the energy calculations above, we find m = 2. Therefore, Option B is correct.

Option C: ${{\Delta {p_a}} \over {\Delta {p_e}}} = {1 \over 2}$

The change in momentum for each photon is given by:

$ \Delta p = \frac{h}{\lambda} $

So, the ratio of the momentum changes is:

$ \frac{{\Delta p_a}}{{\Delta p_e}} = \frac{{\frac{h}{\lambda_a}}}{{\frac{h}{\lambda_e}}} = \frac{\lambda_e}{\lambda_a} = 5 $

This shows the given option C (i.e., $\frac{1}{2}$) is incorrect. The correct ratio is 5.

Option D: $\lambda_e = 418 \ nm$

From our previous calculations, we found $\lambda_e = 487 \ nm$, which does not match 418 nm. Option D is incorrect.

Hence, the correct options are A and B.

Given,

${}_{90}^{232}Th\buildrel {} \over \longrightarrow {}_{82}^{212}pb$

where N_$\alpha$ = number of $\alpha$ particles, N_$\beta$ = number of $\beta$ particles

We know that $\alpha$-decay is given as

${}_Z^AX\buildrel {} \over \longrightarrow {}_{Z - 2}^{A - 4}Y + {\alpha ^ - }$ particle

and $\beta$-decay is given as

${}_Z^AX\buildrel {} \over \longrightarrow {}_{Z + 1}^AX + {\beta ^ - }$ particle

In above decay process change in mass number A is 20 and change in atomic number Z is 8.

Therefore, change in mass number A $\Rightarrow$ number of $\alpha$-particles emitted are

${N_\alpha } = {{20} \over 4} = 5$

Now, emission of 5 $\alpha$ particles implies atomic number reduces by 10 but change in atomic number is 8, therefore, number of $\beta$-particles emitted is 2.

Therefore, N_$\alpha$ = 5 and N_$\beta$ = 2

So, option (A) and option (C) are correct.

The radius of $n^{\text {th }}$ orbital for a hydrogen-like atom of atomic number $Z$ is given by $r_n={{{n^2}{a_0}} \over Z}$, where $a_0=0.53 $$\mathop A\limits^o $ is the Bohr's radius.

$ \therefore $ $ \text { Radius of orbit, } r \propto \frac{n^2}{Z} $

The relative change in the radii of two consecutive orbitals is

$ \frac{r_{n+1}-r_n}{r_n}=\frac{\frac{(n+1)^2}{Z}-\frac{n^2}{Z}}{\frac{n^2}{Z}}=\frac{2 n+1}{n^2} \approx \frac{2}{n} \quad(\because n>>1) $

The energy of the n^th orbital is given by

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

$ \therefore $ $ E_n \propto \frac{-Z^2}{n^2} $

The relative change in the energy of two consecutive orbitals is

$ \begin{aligned} & \frac{E_{n+1}-E_n}{E_n}=\frac{\frac{-Z^2}{(n+1)^2}+\frac{Z^2}{n^2}}{\frac{-Z^2}{n^2}}=-\frac{(2 n+1)}{(n+1)^2} \\\\ & \approx-\frac{2 n}{n^2}=-\frac{2}{n} \quad(n>>1) \end{aligned} $

The angular momentum of the $n^{\text {th }}$ orbital is given by $L_n={{nh} \over {2\pi }}$. The relative change in the angular momentum of two consecutive orbitals is

$ \frac{L_{n+1}-L_n}{L_n}=\frac{(n+1) {{h} \over {2\pi }}-{{nh} \over {2\pi }}}{{nh} \over {2\pi }}=\frac{1}{n} $

${}_{92}^{236}U \to {}_{54}^{140}Xe + {}_{38}^{94}Sr + x + y$

K_x = 2 MeV, K_y = 2 MeV, K_Xe = ?, K_Sr = ?

By conservation of charge number and mass number, x $\equiv$ y $\equiv$ n

B.E. per nucleon of ${}_{92}^{236}U = 7.5$ MeV

B.E. per nucleon of ${}_{54}^{140}Xe$ or ${}_{38}^{94}Sr = 8.5$ MeV

Q value of reaction,

Q = Net kinetic energy gained in the process

$ = {K_{Xe}} + {K_{Sr}} + 2 + 2 - 0 = {K_{Xe}} + {K_{Sr}} + 4$ ..... (1)

As number of nucleons is conserved in a reaction, so Q = Difference of binding energies of the nuclei

$ = 140 \times 8.5 + 94 \times 8.5 - 236 \times 7.5 = 219$ MeV ...... (2)

From eqns. (i) and (ii)

${K_{Xe}} + {K_{Sr}} = 219 - 4 = 215$ MeV ....... (3)

The linear momentum of a particle of mass m and kinetic energy K is given by $p = \sqrt {2mK} $. Since the masses and kinetic energies of x and y are very small in comparison to that of ${}_{54}^{140}Xe$ and ${}_{38}^{94}Sr$, we can neglect the linear momentum of these particles. Initially, the linear momentum of ${}_{92}^{236}U$ is zero (at rest). Finally, the products ${}_{54}^{140}Xe$ and ${}_{38}^{94}Sr$ will move in opposite direction with equal linear momentum (by conservation of linear momentum). Thus,

$\sqrt {2{M_{Xe}}{K_{Xe}}} = \sqrt {2{M_{Sr}}{K_{Sr}}} $, i.e.,

$140{K_{Xe}} = 94{K_{Sr}}$. ...... (4)

Solve equations (3) and (4) to get K_Xe = 86 MeV and K_Sr = 129 MeV.

The orbital angular momentum is

$L = {{nh} \over {2\pi }}$

${{3h} \over {2\pi }} = {{nh} \over {2\pi }} \Rightarrow n = 3$

The radius of the orbit is

$4.5({a_0}) = {a_0}\left( {{{{n^2}} \over Z}} \right)$

$Z = {{{n^2}} \over {4.5}} = {{{3^2}} \over {4.5}} = {9 \over {4.5}} = 2$

Thus, the possible transitions are 3$\to$2, 3$\to$1 and 2$\to$1. For the transition 3$\to$2, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {{1 \over 4} - {1 \over 9}} \right] = 4R\left[ {{{9 - 4} \over {36}}} \right] = {{5R} \over 9}$

$ \Rightarrow \lambda = {9 \over {5R}}$

For the transition 3$\to$1, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {1 - {1 \over 9}} \right] = 4R\left( {{8 \over 9}} \right) = {{32R} \over 9}$

$ \Rightarrow \lambda = {9 \over {32R}}$

For the transition 2$\to$1, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {1 - {1 \over 4}} \right] = 4R\left( {{3 \over 4}} \right) = 3R$

$ \Rightarrow \lambda = {1 \over {3R}}$

When binding energy per nucleon increases for a nuclear process, energy is released.

(a) For 1 < A < 50, on fusion, mass number of the resulting nucleus will be less than 100. No energy will be released.

(b) For 51 < A < 100, on fusion, mass number of the resulting nucleus will be between 100 and 200. As BE increases, energy will be released.

(c) On fission for 100 < A < 200, the mass number for fission nuclei will be between 50 and 100. As BE decreases, no energy will be released.

(d) On fission for 200 < A < 260, the mass number for the fission nuclei will be between 100 and 130. Since BE will increase, energy will be released.