Atoms and Nuclei — Page 18 | JEE Physics

Q171

Binding energy of a certain nucleus is

18 \times 10^8 \mathrm{~J}

. How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus:

A 20

\mu

g

B 2

\mu

g

C 10

\mu

g

D 0.2

\mu

g

Correct Answer

Option A

Solution

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.

Q172

An electron rotates in a circle around a nucleus having positive charge

\mathrm{Ze}

. Correct relation between total energy (E) of electron to its potential energy (U) is :

A

2 \mathrm{E}=3 \mathrm{U}

B

\mathrm{E}=\mathrm{U}

C

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

D

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

Correct Answer

Option C

Solution

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}

Q173

The angular momentum of an electron in a hydrogen atom is proportional to : (Where

\mathrm{r}

is the radius of orbit of electron)

A

\dfrac{1}{\mathrm{r}}

B

\dfrac{1}{\sqrt{\mathrm{r}}}

C

\sqrt{\mathrm{r}}

D r

Correct Answer

Option C

Solution

\begin{aligned} & L=m v r \propto n \\ & \therefore \quad m v r \propto \sqrt{r} \end{aligned}

Q174

The longest wavelength associated with Paschen series is : (Given

\mathrm{R}_{\mathrm{H}}=1.097 \times 10^7 \mathrm{SI}

unit)

A

2.973 \times 10^{-6} \mathrm{~m}

B

1.876 \times 10^{-6} \mathrm{~m}

C

1.094 \times 10^{-6} \mathrm{~m}

D

3.646 \times 10^{-6} \mathrm{~m}

Correct Answer

Option B

Solution

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: $\dfrac{1}{\lambda} = R_{H} \left( \dfrac{1}{3^2} - \dfrac{1}{n^2} \right)$ 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 $n=3$ ): $\dfrac{1}{\lambda} = 1.097 \times 10^7 \left( \dfrac{1}{3^2} - \dfrac{1}{4^2} \right)$ $\dfrac{1}{\lambda} = 1.097 \times 10^7 \left( \dfrac{1}{9} - \dfrac{1}{16} \right)$ $\dfrac{1}{\lambda} = 1.097 \times 10^7 \left( \dfrac{16 - 9}{144} \right)$ $\dfrac{1}{\lambda} = 1.097 \times 10^7 \times \dfrac{7}{144}$ $\dfrac{1}{\lambda} = 1.097 \times 10^7 \times \dfrac{7}{144}$ $\dfrac{1}{\lambda} = 1.097 \times 10^7 \times \dfrac{7}{144}$ $\dfrac{1}{\lambda} = 1.097 \times 10^7 \times \dfrac{7}{144} = 76403.47\, \text{m}^{-1}$ Finally, we calculate $\lambda$ by taking the reciprocal of this value: $\lambda = \dfrac{1}{76403.47} \approx 1.308 \times 10^{-5} \, \text{m} = 1.308 \times 10^{-5} \times 10^{2} \, \text{cm} = 1.308 \times 10^{-3} \, \text{cm}$ It seems there was a mistake in my calculation.

Revising the calculation properly: $\lambda = \dfrac{1}{1.097 \times 10^7 \times \dfrac{7}{144}}$ Correctly evaluating this, we should directly compute: $\lambda \approx \dfrac{144}{7 \times 1.097 \times 10^7} = \dfrac{144}{7.679 \times 10^7} \approx 1.876 \times 10^{-6} \, \text{m}$ Therefore, the correct option that matches our calculation for the longest wavelength in the Paschen series is: Option B: $1.876 \times 10^{-6} \, \text{m}$ .

Q175

Choose the correct nuclear process from the below options [ p : proton, n : neutron,

\mathrm{e}^{-}

: electron,

\mathrm{e}^{+}

: positron,

v:

neutrino,

\bar{v}:

antineutrino]

A

n \rightarrow p+e^{+}+{v}

B

\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{-}+\bar{v}

C

\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{+}+\bar{v}

D

\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{-}+v

Correct Answer

Option B

Solution

We know, theoretical equation for $\beta^-$ decay, Hence, option 2 is correct.

Q176

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

A

1: 2

B

1: 4

C

2: 1

D

4: 1

Correct Answer

Option D

Solution

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

.

Q177

A radioactive nucleus

\mathrm{n}_2

has 3 times the decay constant as compared to the decay constant of another radioactive nucleus

n_1

. If initial number of both nuclei are the same, what is the ratio of number of nuclei of

n_2

to the number of nuclei of

n_1

, after one half-life of

n_1

?

A 1/4

B 1/8

C 4

D 8

Correct Answer

Option A

Solution

Let's work through the problem step by step. The decay of a radioactive nucleus is given by the formula:

N(t) = N_0 \, e^{-\lambda t},

where: $N(t)$ is the number of nuclei remaining at time $t$ , $N_0$ is the initial number of nuclei, $\lambda$ is the decay constant.

For nucleus $n_1$ with decay constant $\lambda_1$ , its half-life $t_{1/2}$ is:

t_{1/2} = \frac{\ln 2}{\lambda_1}.

After one half-life of $n_1$ , the number of $n_1$ nuclei remaining is:

N_1 = N_0 \, e^{-\lambda_1 \, t_{1/2}} = N_0 \, e^{-\lambda_1 \, \frac{\ln2}{\lambda_1}} = N_0 \, e^{-\ln2} = \frac{N_0}{2}.

Nucleus $n_2$ has a decay constant that is 3 times that of $n_1$ :

\lambda_2 = 3\lambda_1.

For nucleus $n_2$ , the number of nuclei remaining after one half-life of $n_1$ (which is $t = \dfrac{\ln2}{\lambda_1}$ ) is: $N_2 = N_0 \, e^{-\lambda_2 \, t} = N_0\, e^{-3\lambda_1 \, \dfrac{\ln2}{\lambda_1}} = N_0 \, e^{-3\ln2} = N_0 \cdot \dfrac{1}{2^3} = \dfrac{N_0}{8}.$ Now, we find the ratio of $n_2$ nuclei to $n_1$ nuclei after this time: $\dfrac{N_2}{N_1} = \dfrac{\dfrac{N_0}{8}}{\dfrac{N_0}{2}} = \dfrac{1/8}{1/2} = \dfrac{1}{4}.$ Therefore, the correct ratio is $\dfrac{1}{4}$ , which corresponds to Option A.

Q178

Given below are two statements: one is labelled as

\mathbf{A s s e r t i o n} \mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A : The Bohr model is applicable to hydrogen and hydrogen-like atoms only. Reason

\mathbf{R}

: The formulation of Bohr model does not include repulsive force between electrons. In the light of the above statements, choose the correct answer from the options given below

A

\mathbf{A}

is true but

\mathbf{R}

is false

B Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

C

\mathbf{A}

is false but

\mathbf{R}

is true

D Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

Correct Answer

Option B

Solution

Option B Both statements are true, and the absence of electron–electron repulsion in Bohr’s formulation is exactly why it works only for one‐electron systems (hydrogen and hydrogen‐like ions).

• Assertion A is true: Bohr’s model applies only to atoms with a single electron.

• Reason R is true: Bohr treated only the Coulomb attraction between nucleus and electron, omitting any repulsive force between multiple electrons.

• Since ignoring e–e repulsion prevents extension to multi–electron atoms, R correctly explains A.

Q179

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The binding energy per nucleon is found to be practically independent of the atomic number A , for nuclei with mass numbers between 30 and 170 . Reason (R) : Nuclear force is long range. In the light of the above statements, choose the correct answer from the options given below :

A (A) is true but (R) is false

B Both (A) and (R) are true and (R) is the correct explanation of (A)

C (A) is false but (R) is true

D Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

Correct Answer

Option A

Solution

Let's analyze both statements: Assertion (A): “The binding energy per nucleon is found to be practically independent of the atomic number A, for nuclei with mass numbers between 30 and 170.”

In fact, for nuclei in this mass range, the binding energy per nucleon is nearly constant (around 7–8 MeV per nucleon).

This is because the attractive nuclear force saturates, leading to a relatively uniform binding energy per nucleon in medium to heavy nuclei.

Therefore, Assertion (A) is true.

Reason (R): “Nuclear force is long range.”

The nuclear force (or strong force) that holds the nucleons (protons and neutrons) together is actually very short-ranged.

Its effective range is approximately 1 to 3 femtometers (fm), after which the force rapidly decreases.

Hence, Reason (R) is false.

Since the assertion is correct and the reason is incorrect, the correct answer is: Option A: (A) is true but (R) is false.

Q180

During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is

2000 \mathop A\limits^o

and it becomes

6000 \mathop A\limits^o

when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is

A

4000 \mathop A\limits^o

B

6000 \mathop A\limits^o

C

2000 \mathop A\limits^o

D

3000 \mathop A\limits^o

Correct Answer

Option D

Solution

\frac{1}{\lambda_{AC}} = \frac{E_A - E_C}{hc}, \quad \frac{1}{\lambda_{BC}} = \frac{E_B - E_C}{hc}

Since the electron transitions from state A to state C in two steps (A to B and then B to C), the energy difference for the transition from A to B is given by

E_A - E_B = (E_A - E_C) - (E_B - E_C)

Expressing these energy differences in terms of wavelengths,

\frac{hc}{\lambda_{AB}} = \frac{hc}{\lambda_{AC}} - \frac{hc}{\lambda_{BC}}

Cancelling out $hc$ , we obtain

\frac{1}{\lambda_{AB}} = \frac{1}{\lambda_{AC}} - \frac{1}{\lambda_{BC}}

Substitute the given values:

\frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000}

Finding a common denominator:

\frac{1}{\lambda_{AB}} = \frac{3 - 1}{6000} = \frac{2}{6000} = \frac{1}{3000}

Thus, the wavelength for the A to B transition is

\lambda_{AB} = 3000 \, \mathring{A}.

This matches Option D.