Atoms and Nuclei

To determine the relationship between the magnetic moment and the principal quantum number $ n $, we need to understand the formula for the magnetic moment of an electron in a Bohr orbit.

The magnetic moment ($ \mu_n $) of an electron in the nth Bohr orbit is given by:

$ \mu_n = \frac{n \cdot e}{2m} \cdot \frac{e \cdot Z \cdot h}{2 \pi m} $

Where:

• $ e $ is the charge of the electron
• $ m $ is the mass of the electron
• $ Z $ is the atomic number (for a hydrogen atom, $ Z = 1 $)
• $ h $ is Planck's constant

Since the Bohr's magneton ($ \mu_B $) is defined as:

$ \mu_B = \frac{e \hbar}{2m} $

The magnetic moment for the nth orbit can be written as:

$ \mu_n = n \cdot \mu_B $

From the above formula, it is clear that the magnetic moment $ \mu_n $ is directly proportional to the principal quantum number $ n $. Mathematically, this relationship can be expressed as:

$ \mu_n \propto n^1 $

Therefore, the value of $ x $ is 1.

Answer: Option D (1)

$\begin{aligned} & \begin{aligned} & (\mathrm{X}) \rightarrow(\mathrm{Y})+(\mathrm{Z})+(\mathrm{P}) \\ & \begin{array}{llll} \mathrm{M} & \mathrm{M} / 3 & \mathrm{M} / 3 & \mathrm{M} / 3 \end{array} \\ \end{aligned} \\ & \Delta \mathrm{Mc}^2=\frac{1}{2} \frac{\mathrm{M}}{3} \mathrm{~V}^2+\frac{1}{2} \frac{\mathrm{M}}{3} \mathrm{~V}^2+\frac{1}{2} \frac{\mathrm{M}}{3} \mathrm{~V}^2 \\ & \mathrm{~V}=\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{M}}{\mathrm{M}}} \end{aligned}$

$\frac{1}{2}|P E|=K E$ for each value of $\mathrm{n}$ (orbit)

$\therefore \frac{K E}{|P E|}=\frac{1}{2}$

According to Rutherford atomic model, most of mass of atom and all its positive charge is concentrated in tiny nucleus & electron revolve around it.

According to Thomson atomic model, atom is spherical cloud of positive charge with electron embedded in it.

Hence, Statement I is true but statement II false.

$\begin{aligned} & { }_3 \mathrm{Li}^6+{ }_0 \mathrm{n}^1 \rightarrow{ }_2 \mathrm{He}^4+{ }_1 \mathrm{H}^3 \\ & { }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^3 \rightarrow{ }_2 \mathrm{He}^4+{ }_0 \mathrm{n}^1 \\ & \hline{ }_3 \mathrm{Li}^6+{ }_1 \mathrm{H}^2 \rightarrow 2\left({ }_2 \mathrm{He}^4\right) \\ & \hline \end{aligned}$

Energy released in process

$\begin{aligned} & \mathrm{Q}=\Delta \mathrm{mc}^2 \\ & \mathrm{Q}=\left[\mathrm{M}(\mathrm{Li})+\mathrm{M}\left(\mathrm{H}^2\right)-2 \times \mathrm{M}\left({ }_2 \mathrm{He}^4\right)\right] \times 931.5 \mathrm{~MeV} \\ & \mathrm{Q}=[6.01690+2.01471-2 \times 4.00388] \times 931.5 \mathrm{~MeV} \\ & \mathrm{Q}=22.216 \mathrm{~MeV} \\ & \mathrm{Q}=22.22 \mathrm{~MeV} \end{aligned}$

$\begin{aligned} & { }_6 \mathrm{C}^{13}+\text { Energy } \rightarrow{ }_6 \mathrm{C}^{12}+{ }_0 \mathrm{n}^1 \\ & \Delta \mathrm{m}=(12.000000+1.008665)-13.003354 \\ & =-0.00531 \mathrm{u} \\ & \therefore \text { Energy required }=0.00531 \times 931.5 \mathrm{~MeV} \\ & =4.95 \mathrm{~MeV} \end{aligned}$

The radius of the nth stationary orbit in the Bohr model of an atom is directly proportional to the square of its principal quantum number n and inversely proportional to the atomic number Z. This relationship is represented by the formula $r_n = \frac{n^2h^2}{4\pi^2kme^2Z},$ where :

• $n$ is the principal quantum number,

• $h$ is Planck's constant,

• $k$ is the Coulomb constant,

• $m$ is the mass of the electron,

• $e$ is the charge of the electron, and

• $Z$ is the atomic number of the nucleus (for hydrogen, Z=1).

To find the radius of the fourth orbit ($r_4$), in comparison to the third orbit ($r_3$), we apply the formula with $n=4$ for the fourth orbit and $n=3$ for the third orbit, and simplify as follows:

$\begin{aligned} & \frac{r_4}{r_3} = \frac{(4^2)h^2}{4\pi^2kme^2Z} \div \frac{(3^2)h^2}{4\pi^2kme^2Z} = \frac{(4^2)}{(3^2)} \\\\ & = \frac{16}{9} \end{aligned}$

Thus, the radius of the fourth stationary orbit is $\frac{16}{9}$ times the radius of the third stationary orbit, $R$. Therefore, $r_4 = \frac{16}{9} R$.

We are given that the half-life of A is the same as the average life of B. The relationship between half-life ($T_{1/2}$) and the decay constant ($\lambda$) is:

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

For the average life ($\tau$), the relationship with the decay constant is:

$\tau = \frac{1}{\lambda}$

According to the given information, the half-life of A is equal to the average life of B:

$T_{1/2(A)} = \tau_{B}$

Now, we can substitute the relationships for half-life and average life:

$\frac{\ln 2}{\lambda_{A}} = \frac{1}{\lambda_{B}}$

To find the correct relationship between $\lambda_{A}$ and $\lambda_{B}$, we can rearrange the equation:

$\lambda_{A} = \lambda_{B} \ln 2$

Let's use the formula for the remaining mass of a radioactive substance after a certain time:

$N(t) = N_0(1/2)^{t/T}$

where N(t) is the mass at time t, N₀ is the initial mass, T is the half-life, and t is the time elapsed.

We are given that $\frac{7}{8}$th of the original mass has disintegrated. Therefore, the remaining mass is $\frac{1}{8}$th of the original mass:

$\frac{N(t)}{N_0} = \frac{1}{8}$

Using the formula, we have:

$\frac{1}{8} = (1/2)^{t/T}$

Taking the logarithm of both sides:

$\log_{1/2}\frac{1}{8} = \frac{t}{T}$

$3 = \frac{t}{T}$

Now, solving for t:

$t = 3T$

So, the time taken for disintegrating $\frac{7}{8}$th part of the original mass is 3T.

According to Bohr's model of the hydrogen atom, the angular momentum of an electron in an orbit is an integral multiple of Planck's constant divided by $2\pi$ (or $h/2\pi$, where $h$ is the Planck's constant). This can be expressed as:

$ L = n \frac{h}{2\pi} $

where $n$ is the principal quantum number or the orbit number.

So, for the first orbit ($n=1$), the angular momentum $L_1$ is:

$ L_1 = 1 \times \frac{h}{2\pi} = \frac{h}{2\pi} $

And for the second orbit ($n=2$), the angular momentum $L_2$ is:

$ L_2 = 2 \times \frac{h}{2\pi} = \frac{h}{\pi} $

The change in angular momentum when moving from the first to the second orbit is the difference between $L_2$ and $L_1$:

$ \Delta L = L_2 - L_1 = \frac{h}{\pi} - \frac{h}{2\pi} = \frac{h}{2\pi} $

Since $\frac{h}{2\pi}$ is equal to the initial angular momentum $L_1$, the change in angular momentum when the electron moves to the second orbit is $L$.

Therefore, the correct answer is $L$.

The decay of a radioactive material follows an exponential decay law, which can be expressed as:

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

where:

• $N$ is the final amount of the material,
• $N_0$ is the initial amount of the material,
• $t$ is the elapsed time,
• $T$ is the half-life of the material.

From the problem, we know that the material is reduced to $1/8$ of its original amount in 3 days. Therefore, the half-life of the material can be calculated as follows:

$ \frac{1}{8} = \left(\frac{1}{2}\right)^{\frac{3}{T}} $

This simplifies to $2^{-3} = 2^{-\frac{3}{T}} $, which gives $ T = 1 \, \text{day} $.

Knowing the half-life, we can now find the initial amount of the material. We know that after 5 days, $8 \times 10^{-3} \, \text{kg}$ of the material is left. Therefore, we can write:

$ 8 \times 10^{-3} \, \text{kg} = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{5}{1}} $

Solving this equation for $N_0$ gives:

$ N_0 = 8 \times 10^{-3} \, \text{kg} \cdot 2^{5} = 256 \times 10^{-3} \, \text{kg} = 0.256 \, \text{kg} = 256 \, \text{g} $

When a metal target is bombarded with high-energy electrons, the phenomenon known as X-ray emission occurs. This is due to the excitation of the inner-shell electrons in the metal atoms by the high-energy electrons. When these inner-shell electrons drop back to their original energy levels, they emit energy in the form of X-ray photons.

In the semi-empirical mass formula, the terms have the following forms:

• The volume term (binding energy) is proportional to the volume of the nucleus, and therefore to the number of nucleons $A$.


• The surface term takes into account that the nucleons at the surface of the nucleus have fewer nearest neighbors than those in the interior, and so the binding energy is less for surface nucleons. This term is usually proportional to $A^{2/3}$.


• The Coulomb term arises because the protons in the nucleus repel each other. This term is usually proportional to $\frac{Z(Z-1)}{A^{1/3}}$.


• The symmetry term arises because of the Pauli exclusion principle, but it's not mentioned here.
• The pairing term arises because of the tendency of protons and neutrons to pair up, but it's not mentioned here.

Now, let's check the statements:

• A: The surface energy per nucleon $b_{s}=-a_{1} A^{2 / 3}$ is incorrect. The correct form should be proportional to $A^{2/3}$, but it should not be per nucleon.


• B: The Coulomb contribution to the binding energy $b_{c}=-a_{2} \frac{Z(Z-1)}{A^{4 / 3}}$ is incorrect. The correct form should be proportional to $\frac{Z(Z-1)}{A^{1/3}}$.


• C: The volume energy $b_{v}=a_{3} A$ is correct.


• D: Decrease in the binding energy is proportional to surface area is correct. This represents the surface term in the semi-empirical mass formula.


• E: While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons is incorrect. While the exact number can vary depending on the model used, it's generally agreed that each nucleon interacts with a few nearest neighbors, not necessarily 12.

So, the correct answer is C, D only.

According to Bohr's quantization of angular momentum, the angular momentum $L$ of a particle in a circular orbit is given by:

$L = n\hbar$

Where $n$ is an integer and $\hbar$ is the reduced Planck's constant. The angular momentum $L$ can also be expressed as:

$L = mvr$

Where $m$ is the mass of the particle, $v$ is its linear velocity, and $r$ is the radius of the orbit.

Now, we are given the potential energy $U = \frac{1}{2} m\omega^2r^2$. Since the particle is in a circular orbit, its centripetal force is provided by the gradient of the potential energy:

$m\frac{v^2}{r} = -\frac{\mathrm{d}U}{\mathrm{d}r} = -m\omega^2r$

We can simplify this equation to get the relation between $v$ and $r$:

$v^2 = \omega^2r^2$

Now, let's combine the equations for angular momentum and the relation between $v$ and $r$:

$n\hbar = mvr = m\sqrt{\omega^2r^2}r = m\omega r^2$

We can now solve for the radius $r$ in terms of $n$:

$r^2 = \frac{n\hbar}{m\omega}$

Taking the square root of both sides, we get:

$r \propto \sqrt{n}$

So, the radius of the $n^{\text{th}}$ orbit is proportional to $\sqrt{n}$.

$R = {R_0}{A^{{1 \over 3}}}$, using this

$\rho = {M \over {{4 \over 3}\pi {R^3}}} = {{A{m_P}} \over {{4 \over 3}\pi R_0^3A}} = {{{m_P}} \over {{4 \over 3}\pi R_0^3}}$

$\rho$ is independent of mass number.

$\therefore$ A is false

$v\,\alpha {z \over n}$

${{{v_1}} \over {{v_2}}} = \left( {{{{n_2}} \over {{n_1}}}} \right)$

$ \Rightarrow {{3.6 \times {{10}^6}} \over {{v_2}}} = {3 \over 7}$

$ \Rightarrow {v_2} = {7 \over 3} \times 3.6 \times {10^6}$ m/s

$ = 8.4 \times {10^6}$ m/s

${n_A} = 20$ moles

${n_B} = 10$ moles

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

${N_A} = (20\,N){e^{ - \left( {{{\ln 2} \over 1} \times 2} \right)}}$

$ = {{20\,N} \over 4} = 5\,N$ (N = Avogadro's Number)

${N_B} = 10\,N\,{e^{ - 4\ln 2}}$

$ = \left( {{{10\,N} \over {16}}} \right)$

${N_A} + {N_B} = 5\,N + {{10\,N} \over {16}} = \left( {{{90\,N} \over {16}}} \right) = 3.38 \times {10^{24}}$

We know,

Radius of nucleus, $r = {r_0}{A^{{1 \over 3}}}$

where, A = mass number

$\therefore$ Volume $(v) = {4 \over 3}\pi {r^3}$

$ = {4 \over 3}\pi r_0^3\,.\,A$

$\therefore$ $v \propto A$

$\therefore$ Volume is directly proportional to mass number.

We know,

density $(d) = {m \over v}$

$ = {{z{m_p} + (A - z){m_N}} \over {{4 \over 3}\pi r_0^3{{({A^{1/3}})}^3}}}$

$ = {{z{m_p} + A{m_p} - z{m_p}} \over {{4 \over 3}\pi r_0^3A}}$ [$\because$ ${m_p} \simeq {m_N}$]

$ = {{A{m_p}} \over {{4 \over 3}\pi r_0^3A}}$

$ = {{{m_p}} \over {{4 \over 3}\pi r_0^3}}$

$\therefore$ Density of nucleus is independent of mass number.

${E_1} = {E_0}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right) = {E_0} \times {3 \over 4}$

${E_2} = {E_0}$

$\therefore$ ${{{E_1}} \over {{E_2}}} = {3 \over 4}$

$N = {{{N_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}$

$ \Rightarrow {2^{{t \over {{T_{1/2}}}}}} = {{{N_0}} \over N} = {{{N_0}} \over {\left( {{{{N_0}} \over 8}} \right)}} = 8$

$ \Rightarrow {t \over {{T_{1/2}}}} = 3$

$ \Rightarrow {T_{1/2}} = {{15} \over 3} = 5$ min

$\because$ $N = {{{N_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}$

$ \Rightarrow {2^{{t \over {{T_{1/2}}}}}} = {{{N_0}} \over N} = {{{N_0}} \over {\left( {{{{N_0}} \over 8}} \right)}}$

$ \Rightarrow {2^{{t \over {{T_{1/2}}}}}} = 8 = {2^3}$

$ \Rightarrow t = 3 \times {T_{1/2}} = 3 \times 60$

$ = 180$ days

$\because$ $A = {{{A_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}$

$ \Rightarrow {2^{t/5}} = {{6.4 \times {{10}^{ - 4}}} \over {5 \times {{10}^{ - 6}}}} = 128 = {2^7}$

$ \Rightarrow {t \over 5} = 7$

$ \Rightarrow t = 35$ days

$\because$ $A = {{{A_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}$

$ \Rightarrow {2^{{t \over {{T_{1/2}}}}}} = {{{A_0}} \over A} = 16$

$ \Rightarrow {t \over {{T_{1/2}}}} = 4$

$ \Rightarrow {{30} \over {{T_{1/2}}}} = 4$

$ \Rightarrow {T_{1/2}} = {{30} \over 4}$

$ = 7.5$ years

Linear momentum is conserved

so, ${p_{M'/3}} = {p_{2M'/3}}$

so, ${{{\lambda _{M'/3}}} \over {{\lambda _{2M'/3}}}} = {1 \over 1}$

$\therefore$ $R = {R_0}{A^{{1 \over 3}}}$

$ \Rightarrow {{{R_1}} \over {{R_2}}} = {\left( {{{{A_1}} \over {{A_2}}}} \right)^{{1 \over 3}}} = {\left( {{4 \over 3}} \right)^{{1 \over 3}}}$

$\therefore$ Density ratio, ${{{\rho _1}} \over {{\rho _2}}} = {{{A_1}/{V_1}} \over {{A_2}/{V_2}}}$

$ = \left( {{{{A_1}} \over {{A_2}}}} \right) \times {\left( {{{{R_2}} \over {{R_1}}}} \right)^3}$

$ = 1:1$

${A_0} = 4250$

$A = 2250 = {A_0}{e^{ - \lambda t}}$

$ \Rightarrow {{2250} \over {4250}} = {e^{ - \lambda t}}$

$ \Rightarrow \lambda (10) = \ln \left( {{{4250} \over {2250}}} \right)$

$\lambda (10) = 0.636$

$\lambda = 0.063$

$\because$ ${1 \over \lambda } = R\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right)$

$ \Rightarrow {1 \over {\lambda R}} = 1 - {1 \over {{n^2}}}$

$ \Rightarrow {1 \over {{n^2}}} = 1 - {1 \over {\lambda R}} = {{\lambda R - 1} \over {\lambda R}}$

$ \Rightarrow n = \sqrt {{{\lambda R} \over {\lambda R - 1}}} $

$\because$ $mvr = {{nh} \over {2\pi }}$

$ \Rightarrow mv = {{nh} \over {2\pi r}}$

$\because$ $\overrightarrow \mu = {{q\overrightarrow L } \over {2m}}$

$ \Rightarrow \overrightarrow \mu = {{ - e\overrightarrow L } \over {2m}}$

Equivalent reaction can be written as

$D\buildrel {} \over \longrightarrow {D_4} + 2\alpha + {\beta ^ - } + \gamma $

$\Rightarrow$ Mass number of D₄ = Mass number of D $-$ 2 $\times$ 4

= 182 $-$ 8 = 174

$\Rightarrow$ Atomic number of D₄

= Atomic number of D $-$ 2 $\times$ 2 + 1

= 74 $-$ 4 + 1 = 71

Radioactive decay is a random and spontaneous process it depends on unbalancing of nucleus.

$N = {N_0}{e^{ - \lambda t}}$ ..... (B)

$\ln N = - \lambda t + \ln {N_0}$

So, slope $ = - \lambda $ ..... (C)

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

So ${t_{1/2}} \times \lambda = \ln 2 = $ Constant

K_p > 0

If Q is released $\Rightarrow$ Q > 0

$\Rightarrow$ K_p + Q > 0

Even the particle has to be given kinetic energy greater than magnitude of Q to maintain momentum conservation.

$\Rightarrow$ K + Q > 0

Radiation is not emitted but absorbed when an electron jumps from low energy to high energy.

Also, E₂ $-$ E₁ is the energy of photon

$\Rightarrow$ E₂ $-$ E₁ = hf

$ \Rightarrow f = {{{E_2} - {E_1}} \over h}$