Atoms and Nuclei — Page 16 | JEE Physics

Q151

The waves emitted when a metal target is bombarded with high energy electrons are

A Infrared rays

B Radio Waves

C Microwaves

D X-rays

Correct Answer

Option D

Solution

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.

Q152

For a nucleus

{ }_{\mathrm{A}}^{\mathrm{A}} \mathrm{X}

having mass number

\mathrm{A}

and atomic number

\mathrm{Z}

A. The surface energy per nucleon

\left(b_{\mathrm{s}}\right)=-a_{1} A^{2 / 3}

. B. The Coulomb contribution to the binding energy

\mathrm{b}_{\mathrm{c}}=-a_{2} \dfrac{Z(Z-1)}{A^{4 / 3}}

C. The volume energy

\mathrm{b}_{\mathrm{v}}=a_{3} A

D. Decrease in the binding energy is proportional to surface area. E. While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons. (

a_{1}, a_{2}

and

a_{3}

are constants) Choose the most appropriate answer from the options given below:

A C, D only

B B, C, E only

C B, C only

D A, B, C, D only

Correct Answer

Option A

Solution

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 $\dfrac{Z(Z-1)}{A^{1/3}}$ .

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 $b_{c}=-a_{2} \dfrac{Z(Z-1)}{A^{4 / 3}}$ is incorrect.

The correct form should be proportional to $\dfrac{Z(Z-1)}{A^{1/3}}$ .

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.

Q153

A small particle of mass

m

moves in such a way that its potential energy

U=\dfrac{1}{2} m ~\omega^{2} r^{2}

where

\omega

is constant and

r

is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of

n^{\text{th }}

orbit will be proportional to,

A

\sqrt{n}

B

n^{2}

C

\dfrac{1}{n}

D

n

Correct Answer

Option A

Solution

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}

.

Q154

From the statements given below : (A) The angular momentum of an electron in

n^{\text{th }}

orbit is an integral multiple of

\hbar

. (B) Nuclear forces do not obey inverse square law. (C) Nuclear forces are spin dependent. (D) Nuclear forces are central and charge independent. (E) Stability of nucleus is inversely proportional to the value of packing fraction. Choose the correct answer from the options given below :

A (B), (C), (D), (E) only

B (A), (C), (D), (E) only

C (A), (B), (C), (E) only

D (A), (B), (C), (D) only

Correct Answer

Option C

Solution

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:

L = n\hbar

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:

\text{Option C : (A), (B), (C), (D) only}

Q155

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :

A

13.6 \mathrm{eV}

B

1.5 \mathrm{eV}

C

12.1 \mathrm{eV}

D

1.9 \mathrm{eV}

Correct Answer

Option C

Solution

The energy $E_n$ of an electron in the $n$ th energy level of a hydrogen atom is given by the formula: $E_n = -\dfrac{13.6 \, \text{eV}}{n^2}$ 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.

Ground State Energy ( $E_1$ ): The energy of the electron in the ground state ( $n=1$ ) of hydrogen is: $E_1 = -\dfrac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV}$ This represents the electron's energy level when it's closest to the nucleus.

Energy for the $n=3$ Level ( $E_3$ ): For an electron in the $n=3$ level, its energy is: $E_3 = -\dfrac{13.6 \, \text{eV}}{3^2} = -1.51 \, \text{eV}$ 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 $n=3$ .

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.

Q156

The radius of third stationary orbit of electron for Bohr's atom is R. The radius of fourth stationary orbit will be:

A

\dfrac{4}{3} \mathrm{R}

B

\dfrac{16}{9} R

C

\dfrac{3}{4} R

D

\dfrac{9}{16} \mathrm{R}

Correct Answer

Option B

Solution

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 $\dfrac{16}{9}$ times the radius of the third stationary orbit, $R$ .

Therefore,

r_4 = \frac{16}{9} R

.

Q157

The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192 is :

A 32

B 24

C 20

D 40

Correct Answer

Option B

Solution

\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}

Q158

The explosive in a Hydrogen bomb is a mixture of

{ }_1 \mathrm{H}^2,{ }_1 \mathrm{H}^3

and

{ }_3 \mathrm{Li}^6

in some condensed form. The chain reaction is given by

\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 \end{aligned}

During the explosion the energy released is approximately [Given ;

\mathrm{M}(\mathrm{Li})=6.01690 \mathrm{~amu}, \mathrm{M}\left({ }_1 \mathrm{H}^2\right)=2.01471 \mathrm{~amu}, \mathrm{M}\left({ }_2 \mathrm{He}^4\right)=4.00388

\mathrm{amu}

, and

1 \mathrm{~amu}=931.5 \mathrm{~MeV}]

A 22.22 MeV

B 28.12 MeV

C 16.48 MeV

D 12.64 MeV

Correct Answer

Option A

Solution

\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 \\ & \\{ }_3 \mathrm{Li}^6+{ }_1 \mathrm{H}^2 \rightarrow 2\left({ }_2 \mathrm{He}^4\right) \\ & \\ \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}

Q159

Given below are two statements: Statement I : Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model. Statement II : An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model. In the light of the above statements, choose the most appropriate from the options given below

A Both Statement I and Statement II are true

B Statement I is true but Statement II is false

C Statement I is false but Statement II is true

D Both statement I and statement II are false

Correct Answer

Option B

Solution

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.

Q160

An electron revolving in

n^{\text{th }}

Bohr orbit has magnetic moment

\mu_n

. If

\mu_n \propto n^x

, the value of

x

is

A 2

B 0

C 3

D 1

Correct Answer

Option D

Solution

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)