Atoms and Nuclei

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

According to Bohr's theory, one of the postulates specifies that the angular momentum of an electron in orbit around a nucleus is quantized. This quantization can be expressed by the formula:

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

Where:

• $L$ is the angular momentum of the electron,
• $n$ is the principal quantum number (or the orbit number in simpler terms), which can take positive integer values (1, 2, 3, ...),
• $h$ is Planck's constant ($6.62607015 \times 10^{-34} m^2 kg / s$), and
• $\frac{h}{2\pi}$ is often denoted as $\hbar$ (h-bar), known as the reduced Planck's constant.

For an electron in the 4th orbit ($n = 4$) of a hydrogen atom, we substitute $n = 4$ into the equation:

$ L = 4\frac{h}{2\pi} $

Therefore, the moment of momentum (or angular momentum) of an electron in the $4^{\text{th}}$ orbit of a hydrogen atom is:

$ L = 4\frac{h}{2\pi} = 2\frac{2h}{2\pi} = 2\frac{h}{\pi} $

Hence, the correct option is:

Option A: $2 \frac{h}{\pi}$

In order to identify the correct nuclear fragments resulting from the fission of uranium-235 by a neutron, $\left({ }_0^1 \mathrm{n}\right) + \left({ }_{92}^{235} \mathrm{U}\right)$, we need to apply the conservation of mass number and atomic number. These conservation laws tell us that the sum of mass numbers (top numbers, representing the total count of protons and neutrons) and the sum of atomic numbers (bottom numbers, representing the total count of protons) before and after the fission must be equal. Let's apply these rules to each option:

For the uranium-235 fission reaction, before the reaction, the total mass number is $235 + 1 = 236$ and the total atomic number is $92$ (since a neutron doesn't contribute to the atomic number).

Option A: ${ }_{56}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+3{ }_0^1 \mathrm{n}$

• Total mass number after = $140 + 94 + 3(1) = 234 + 3 = 237$
• Total atomic number after = $56 + 38 = 94$

Option B: ${ }_{51}^{153} \mathrm{Sb}+{ }_{41}^{99} \mathrm{Nb}+3{ }_0^1 \mathrm{n}$

• Total mass number after = $153 + 99 + 3(1) = 252 + 3 = 255$
• Total atomic number after = $51 + 41 = 92$

Option C: ${ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+4{ }_0^1 \mathrm{n}$

• Total mass number after = $144 + 89 + 4(1) = 233 + 4 = 237$
• Total atomic number after = $56 + 36 = 92$

Option D: ${ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+3{ }_0^1 \mathrm{n}$

• Total mass number after = $144 + 89 + 3(1) = 233 + 3 = 236$
• Total atomic number after = $56 + 36 = 92$

By examining the conservation of mass numbers and atomic numbers, Option D is identified as the correct option because both the total mass number and atomic number after the reaction match exactly with the total mass number and atomic number before the reaction. Therefore, the nuclear fission fragments and the neutron count for the reaction are accurately represented by ${ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+3{ }_0^1 \mathrm{n}$.

Let's analyze each of the given statements to determine which ones are correct:

(A) The angular momentum of an electron in $n^{\text{th}}$ orbit is an integral multiple of $\hbar$.

This statement reflects Bohr's quantization rule for angular momentum in the Bohr model of the hydrogen atom. According to this rule, the angular momentum of an electron in a stationary orbit is quantized and given by:

where $n$ is a principal quantum number (which can be any positive integer), and $\hbar$ is the reduced Planck's constant. Therefore, this statement (A) is correct.

(B) Nuclear forces do not obey inverse square law.

Nuclear forces, specifically strong nuclear forces, do act over short ranges within the nucleus but do not obey the inverse square law, which is characteristic of the electromagnetic and gravitational forces. Thus, statement (B) is correct.

(C) Nuclear forces are spin dependent.

The strength of the nuclear force can depend on the spin alignment of the nucleons. This is why some isotopes are more stable than others depending on spin-related factors. Thus, statement (C) is correct.

(D) Nuclear forces are central and charge independent.

The strong nuclear force is indeed charge independent, meaning it is the same regardless of the types of nucleons involved (neutrons or protons). However, nuclear forces are not always central; they can be tensor forces too, which involve more complex interactions that are not purely central. Therefore, the entirety of statement (D) is not correct; the statement should specify that nuclear forces are charge independent but may not always be central.

(E) Stability of nucleus is inversely proportional to the value of packing fraction.

This statement (E) is correct.

Based on the above explanations, the correct statements are (A), (B), (C) and (E). Hence, the correct answer is:

The energy $E_n$ of an electron in the $n$th energy level of a hydrogen atom is given by the formula:

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

where:

• $E_n$ is the energy in electron volts (eV),


• $13.6 \, \text{eV}$ is the ionization energy of hydrogen (the energy required to remove the electron from the ground state, $n=1$),


• $n$ is the principal quantum number (an integer starting from 1).

The Balmer Series

The Balmer series is a specific set of spectral lines of hydrogen that are visible to the naked eye. It results from the electron making transitions between energy levels, specifically from levels with $n \geq 3$ (i.e., from higher energy states) down to $n = 2$. These transitions release energy in the form of electromagnetic radiation, which we observe as the Balmer series.

1. Ground State Energy ($E_1$): The energy of the electron in the ground state ($n=1$) of hydrogen is:

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

This represents the electron's energy level when it's closest to the nucleus.

1. Energy for the $n=3$ Level ($E_3$): For an electron in the $n=3$ level, its energy is:

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

1. Minimum Energy for Balmer Series Transition: The minimum energy transition within the Balmer series is from $n=3$ to $n=2$, but to understand the total energy required for an electron to emit radiation in the Balmer series, starting from the ground state, we consider the energy needed to first excite the electron from $n=1$ to $n=3$.


2. Energy Difference ($\Delta E$): The energy emitted as radiation for the electron to transition from the ground state to a state where it can participate in the Balmer series is the difference between the ground state energy and the energy of the $n=3$ state:

$ \Delta E = E_3 - E_1 = -1.51 \, \text{eV} - (-13.6 \, \text{eV}) = 12.09 \, \text{eV} $

Note

This calculation shows that the minimum energy required by a hydrogen atom in the ground state to emit radiation in the Balmer series is approximately 12.1 eV. The energy difference essentially represents the energy that must be supplied to an electron to excite it from the ground state ($n=1$) to an excited state ($n=3$) from which it can then transition to $n=2$, emitting radiation observable as part of the Balmer series.

$\begin{aligned} & \mathrm{R}_1=\frac{\mathrm{R}_2}{2} \\ & \mathrm{R}_0\left(\mathrm{~A}_1\right)^{1 / 3}=\frac{\mathrm{R}_0}{2}\left(\mathrm{~A}_2\right)^{1 / 3} \\ & \mathrm{~A}_1=\frac{1}{8} \mathrm{~A}_2 \\ & \mathrm{~A}_1=\frac{192}{8}=24 \end{aligned}$

$\begin{aligned} & \frac{1}{\lambda}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right] ...... \text{(i)}\\ & \frac{1}{\lambda^{\prime}}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{3^2}\right] ...... \text{(ii)} \end{aligned}$

On dividing (i) & (ii)

$\lambda^{\prime}=\frac{27}{32} \lambda$

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 $\sqrt{n}$.