Atoms and Nuclei — Page 15 | JEE Physics

Q141

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A : The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range 30 to 170 . Reason R : Nuclear force is short ranged. In the light of the above statements, choose the correct answer from the options given below

A

\mathrm{A}

is false but

\mathbf{R}

is true

B

\mathrm{A}

is true but

\mathbf{R}

is false

C Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

D Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

Correct Answer

Option C

Solution

The statement about the binding energy per nucleon is true, and is known as the semi-empirical mass formula.

According to this formula, the binding energy per nucleon for nuclei in the range of mass numbers 30 to 170 is nearly constant, with a maximum value around mass number 60.

The statement about nuclear force being short ranged is also true.

The strong nuclear force that binds nucleons together is a short-range force that acts only over distances of a few femtometers.

Therefore, both Assertion A and Reason R are true, and Reason R provides a valid explanation for Assertion A.

The correct answer is: So, Both Assertion A and Reason R are true, and Reason R is the correct explanation for Assertion A.

Q142

_{92}^{238}A \to _{90}^{234}B + _2^4D + Q

In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of

{ }_{92}^{238} \mathrm{~A}=238.05079 \times 931.5 ~\mathrm{MeV} / \mathrm{c}^{2},

mass of

{ }_{90}^{234} B=234 \cdot 04363 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2},

mass of

\left.{ }_{2}^{4} D=4 \cdot 00260 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2}\right]

A 2.12 MeV

B 4.25 MeV

C 3.82 MeV

D 5.9 MeV

Correct Answer

Option B

Solution

The energy released in a nuclear reaction can be determined by the mass difference between the reactants and the products, multiplied by the speed of light squared, as per Einstein's mass-energy equivalence relation,

E=mc^2

. In the given nuclear reaction, the energy released is:

Q = \left( m_{A} - m_{B} - m_{D} \right) \times 931.5 \ \text{MeV/c}^2

We are given the masses of A, B, and D as follows:

m_{A} = 238.05079 \times 931.5 \ \text{MeV/c}^2

m_{B} = 234.04363 \times 931.5 \ \text{MeV/c}^2

m_{D} = 4.00260 \times 931.5 \ \text{MeV/c}^2

Now, we can substitute these values into the equation for Q:

Q = \left( (238.05079 - 234.04363 - 4.00260) \times 931.5 \right) \ \text{MeV}

Q = \left( (0.00456) \times 931.5 \right) \ \text{MeV}

Calculating the energy released:

Q \approx 4.25 \ \text{MeV}

So, the approximate amount of energy released in the given nuclear reaction is 4.25 MeV.

Q143

A

12.5 \mathrm{~eV}

electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

A 2

B 4

C 3

D 1

Correct Answer

Option C

Solution

The energy of an electron in an excited state of hydrogen is given by the equation: Code snippet

E = -13.6 \frac{1}{n^2} \mathrm{~eV}

where n is the principal quantum number of the state.

The energy of the electron beam is 12.5 eV, which is enough to excite the electron to the n = 3 state.

The possible transitions from n = 3 to lower energy states are: n = 3 to n = 2, with a wavelength of 656.33 nm (H-alpha) n = 3 to n = 1, with a wavelength of 102.57 nm (Lyman-alpha) n = 2 to n = 1, with a wavelength of 121.57 nm (Lyman-beta) Therefore, there are 3 possible spectral lines that can be emitted.

Q144

The energy of

\mathrm{He}^{+}

ion in its first excited state is, (The ground state energy for the Hydrogen atom is

-13.6 ~\mathrm{eV})

:

A

-13.6 ~\mathrm{eV}

B

-27.2 ~\mathrm{eV}

C

-3.4 ~\mathrm{eV}

D

-54.4 ~\mathrm{eV}

Correct Answer

Option A

Solution

The energy levels of a one-electron ion can be described by the formula:

E_n = -\frac{Z^2}{n^2} \times E_0

where $E_n$ is the energy of the nth level, Z is the atomic number (number of protons), n is the principal quantum number, and $E_0$ is the ground state energy of the hydrogen atom (-13.6 eV).

For the

\mathrm{He}^{+}

ion, the atomic number Z is 2 (since helium has 2 protons).

We are looking for the energy of the first excited state, which corresponds to n = 2.

Plugging these values into the formula, we get:

E_2 = -\frac{2^2}{2^2} \times (-13.6 ~\mathrm{eV}) = -13.6 ~\mathrm{eV}

So, the energy of the

\mathrm{He}^{+}

ion in its first excited state is

-13.6 ~\mathrm{eV}

.

Q145

Consider the nuclear fission Ne20

\to

2He4 + C12 Given that the binding energy/ nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement -

A 8.3 MeV energy will be released

B energy of 11.9 MeV has to be supplied

C energy of 12.4 MeV will be supplied

D energy of 3.6 MeV will be released

Correct Answer

Option B

Solution

Ne20

~ \to

2He4 + C12 Q – value, EB = (BE)react $-$ (BE)product = (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12) = 9.72 MeV

Q146

If

M_0

is the mass of isotope

{ }_5^{12} B, M_p

and

M_n

are the masses of proton and neutron, then nuclear binding energy of isotope is:

A

(5 M_p+7 M_n-M_o) C^2

B

(M_o-5 M_p-7 M_n) C^2

C

(M_o-5 M_p) C^2

D

(M_0-12 M_n) C^2

Correct Answer

Option A

Solution

To determine the nuclear binding energy of the isotope

_5^{12}B

, we need to consider the mass defect concept.

The mass defect is the difference between the sum of the individual masses of nucleons (protons and neutrons) and the actual mass of the nucleus.

Let's calculate the mass defect first.

The isotope

_5^{12}B

has 5 protons and 7 neutrons (since the total number of nucleons is 12). Therefore, the mass defect can be written as:

\Delta M = (5 M_p + 7 M_n) - M_0

Once we have the mass defect, the binding energy can be found using Einstein's mass-energy equivalence principle, which is given by:

E = \Delta M \cdot c^2

Substituting the mass defect into this equation, we get:

E = ((5 M_p + 7 M_n) - M_0) \cdot c^2

Therefore, the correct answer is Option A:

(5 M_p + 7 M_n - M_0) \cdot c^2

Q147

Two radioactive elements A and B initially have same number of atoms. The half life of A is same as the average life of B. If

\lambda_{A}

and

\lambda_{B}

are decay constants of A and B respectively, then choose the correct relation from the given options.

A

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

B

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

C

\lambda_{\mathrm{A}}=2 \lambda_{\mathrm{B}}

D

\lambda_{\mathrm{A}}=\lambda_{\mathrm{B}}

Correct Answer

Option A

Solution

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

Q148

The half life of a radioactive substance is T. The time taken, for disintegrating

\dfrac{7}{8}

th part of its original mass will be:

A 8T

B 3T

C T

D 2T

Correct Answer

Option B

Solution

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.

Q149

The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be

A L

B

\dfrac{L}{2}

C zero

D 2 L

Correct Answer

Option A

Solution

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 $\dfrac{h}{2\pi}$ 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 $L$ .

Q150

A radio active material is reduced to

1 / 8

of its original amount in 3 days. If

8 \times 10^{-3} \mathrm{~kg}

of the material is left after 5 days the initial amount of the material is

A 64 g

B 256 g

C 32 g

D 40 g

Correct Answer

Option B

Solution

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}