Atoms and Nuclei — Page 17 | JEE Physics

Q161

In a nuclear fission reaction of an isotope of mass

M

, three similar daughter nuclei of same mass are formed. The speed of a daughter nuclei in terms of mass defect

\Delta M

will be :

A

c \sqrt{\dfrac{3 \Delta M}{M}}

B

\dfrac{\Delta M c^2}{3}

C

c \sqrt{\dfrac{2 \Delta M}{M}}

D

\sqrt{\dfrac{2 c \Delta M}{M}}

Correct Answer

Option C

Solution

\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}

Q162

The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5th excited state of a hydrogen atom is :

A 4

B 1

C

\dfrac{1}{2}

D

\dfrac{1}{4}

Correct Answer

Option C

Solution

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

for each value of

\mathrm{n}

(orbit)

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

Q163

A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of

2: 1

. After disintegration they will move :

A in opposite directions with speed in the ratio of

1: 2

respectively.

B in the same direction with same speed.

C in opposite directions with speed in the ratio of

2: 1

respectively.

D in opposite directions with the same speed.

Correct Answer

Option A

Solution

In nuclear disintegration, the conservation of momentum plays a crucial role in determining the motion of the resulting fragments.

Since the original nucleus is at rest, its total initial momentum is zero.

After the disintegration, the total momentum of the system must still be zero to conserve momentum.

Given the mass ratio of the resulting two smaller nuclei is $2:1$ , let's denote the masses of the two nuclei as $2m$ and $m$ , respectively.

Applying Conservation of Momentum For the nucleus with mass $2m$ and velocity $v_1$ and for the nucleus with mass $m$ and velocity $v_2$ , the conservation of momentum equation is: $2m \cdot v_1 + m \cdot v_2 = 0$ Given that momentum is a vector quantity, and the total initial momentum was zero, the nuclei must move in opposite directions for their momenta to cancel each other out.

Therefore, we rearrange the equation: $2m \cdot v_1 = -m \cdot v_2$ $2 \cdot v_1 = -v_2$ $v_2 = -2 \cdot v_1$ The negative sign indicates that $v_1$ and $v_2$ are in opposite directions.

Taking magnitudes and considering the ratio: $\left| v_2 \right| = 2 \cdot \left| v_1 \right|$ This equation shows that the velocity of the lighter fragment (mass $m$ ) is twice the velocity of the heavier fragment (mass $2m$ ).

Since they must move in opposite directions for momentum conservation, this confirms: Correct Answer: Option A - in opposite directions with speed in the ratio of $1:2$ respectively.

Q164

The energy E and momentum p of a moving body of mass m are related by some equation. Given that c represents the speed of light, identify the correct equation

A

\mathrm{E}^2=\mathrm{pc}^2+\mathrm{m}^2 \mathrm{c}^4

B

\mathrm{E}^2=\mathrm{pc}^2+\mathrm{m}^2 \mathrm{c}^2

C

\mathrm{E}^2=\mathrm{p}^2 \mathrm{c}^2+\mathrm{m}^2 \mathrm{c}^4

D

\mathrm{E}^2=\mathrm{p}^2 \mathrm{c}^2+\mathrm{m}^2 \mathrm{c}^2

Correct Answer

Option C

Solution

\begin{aligned} & {[E]=M^1 L^2 T^{-2}} \\ & {[P c]=M^1 L^1 T^{-1} \cdot L^1 T^{-1}=M^1 L^2 T^{-2}} \\ & {\left[\mathrm{mc}^2\right]=\mathrm{M}^1 L^2 T^{-2}} \\ & E^2=\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2} \\ & E^2=\mathrm{P}^2 \mathrm{c}^2+\mathrm{m}^2 \mathrm{c}^4 \end{aligned}

Q165

The energy released in the fusion of

2 \mathrm{~kg}

of hydrogen deep in the sun is

E_H

and the energy released in the fission of

2 \mathrm{~kg}

of

{ }^{235} \mathrm{U}

is

E_U

. The ratio

\dfrac{E_H}{E_U}

is approximately: (Consider the fusion reaction as

4_1^1H+2 \mathrm{e}^{-} \rightarrow{ }_2^4 \mathrm{He}+2 v+6 \gamma+26.7 \mathrm{~MeV}

, energy released in the fission reaction of

{ }^{235} \mathrm{U}

is

200 \mathrm{~MeV}

per fission nucleus and

\mathrm{N}_{\mathrm{A}}= 6.023 \times 10^{23})

A 7.62

B 25.6

C 9.13

D 15.04

Correct Answer

Option A

Solution

To determine the ratio

\frac{E_H}{E_U}

, let's first consider the energy released in the fusion of hydrogen and the energy released in the fission of

^{235}U

. For the fusion reaction: The given reaction is:

4_1^1H + 2 \mathrm{e}^{-} \rightarrow {}_2^4 \mathrm{He} + 2 \mathrm{\nu} + 6 \gamma + 26.7 \, \mathrm{MeV}

This reaction shows that 4 hydrogen atoms and 2 electrons produce 1 helium atom, 2 neutrinos, and 6 gamma photons while releasing 26.7 MeV of energy.

The mass of 1 mole of

H_1^1

is approximately 1 gram, so 2 kg of hydrogen is equal to:

2 \times 10^3 \, \mathrm{g}

The number of moles of hydrogen in 2 kg is:

\mathrm{N}_\text{moles} = \frac{2 \times 10^3}{1} = 2 \times 10^3 \, \mathrm{moles}

The number of hydrogen atoms in 2 kg is:

\mathrm{N}_\text{atoms} = N_A \times 2 \times 10^3 = 6.023 \times 10^{23} \times 2 \times 10^3 = 1.2046 \times 10^{27} \, \mathrm{atoms}

Since 4 hydrogen atoms release 26.7 MeV, the total energy released (

E_H

) is:

E_H = \frac{1.2046 \times 10^{27}}{4} \times 26.7 \, \mathrm{MeV}

E_H = 3.0115 \times 10^{26} \times 26.7 \, \mathrm{MeV}

E_H \approx 8.04 \times 10^{27} \, \mathrm{MeV}

For the fission reaction: The energy released per fission of one

^{235}U

nucleus is 200 MeV. The mass of 1 mole of

^{235}U

is approximately 235 grams, so 2 kg of

{}^{235}U

is equal to:

2 \times 10^3 \, \mathrm{g}

The number of moles of

^{235}U

in 2 kg is:

\mathrm{N}_\text{moles} = \frac{2 \times 10^3}{235} \approx 8.51 \, \mathrm{moles}

The number of

^{235}U

nuclei is:

\mathrm{N}_\text{nuclei} = N_A \times 8.51 \approx 6.023 \times 10^{23} \times 8.51 \approx 5.12 \times 10^{24} \, \mathrm{nuclei}

The total energy released (

E_U

) is:

E_U = 200 \, \mathrm{MeV} \times 5.12 \times 10^{24} \approx 1.024 \times 10^{27} \, \mathrm{MeV}

Finally, the ratio

\frac{E_H}{E_U}

is:

\frac{E_H}{E_U} \approx \frac{8.04 \times 10^{27}}{1.024 \times 10^{27}} \approx 7.85

Therefore, the ratio is closest to option A: 7.62.

Q166

A hydrogen atom in ground state is given an energy of

10.2 \mathrm{~eV}

. How many spectral lines will be emitted due to transition of electrons?

A 3

B 6

C 10

D 1

Correct Answer

Option D

Solution

To determine how many spectral lines will be emitted due to transitions of electrons in a hydrogen atom when it is given an energy of $10.2 \, \text{eV}$ , we first need to ascertain which energy level the electron will reach with this energy and then count the possible transitions (spectral lines) as it returns to the ground state.

Energy Levels of Hydrogen Atom : The energy levels $E_n$ of a hydrogen atom can be calculated using the formula: $E_n = -\dfrac{13.6 \, \text{eV}}{n^2}$ where $n$ is the principal quantum number.

Ground State Energy : The ground state (n=1) energy is $E_1 = -13.6 \, \text{eV}$ .

Determine the Excited State : If the ground state electron is given $10.2 \, \text{eV}$ , its total energy becomes: $E_{\text{total}} = E_1 + 10.2 \, \text{eV} = -13.6 \, \text{eV} + 10.2 \, \text{eV} = -3.4 \, \text{eV}$ Now, we find the principal quantum number $n$ for which the energy is closest to $-3.4 \, \text{eV}$ : For $n = 2$ : $E_2 = -\dfrac{13.6}{2^2} = -3.4 \, \text{eV}$ For $n = 3$ : $E_3 = -\dfrac{13.6}{3^2} = -1.51 \, \text{eV}$ Since $-3.4 \, \text{eV}$ matches exactly with $E_2$ , the electron reaches the second energy level ( $n = 2$ ).

Counting Spectral Lines : When the electron falls back to the ground state from $n = 2$ , it can do so in a single transition: $n = 2$ to $n = 1$ Thus, only one spectral line will be emitted during this transition.

Conclusion : The number of spectral lines emitted when a hydrogen atom in the ground state is given $10.2 \, \text{eV}$ and the electron transitions back to the ground state from $n = 2$ is just one.

Therefore, the correct answer is: Option D: 1.

Q167

The energy equivalent of

1 \mathrm{~g}

of substance is :

A

5.6 \times 10^{26} \mathrm{~MeV}

B

5.6 \times 10^{12} \mathrm{~MeV}

C

5.6 \mathrm{~eV}

D

11.2 \times 10^{24} \mathrm{~MeV}

Correct Answer

Option A

Solution

To determine the energy equivalent of a mass, we use Einstein's mass-energy equivalence principle given by the equation:

E = mc^2

where: -

E

is the energy -

m

is the mass -

c

is the speed of light in a vacuum, which is approximately

3 \times 10^8 \mathrm{~m/s}

Given the mass

m = 1 \mathrm{~g} = 1 \times 10^{-3} \mathrm{~kg}

, we can substitute these values into the equation:

E = (1 \times 10^{-3} \mathrm{~kg}) \times (3 \times 10^8 \mathrm{~m/s})^2

Calculating this, we get:

E = 1 \times 10^{-3} \times 9 \times 10^{16}

E = 9 \times 10^{13} \mathrm{~J}

Next, to convert this energy into electron volts (

\mathrm{eV}

), we use the conversion factor:

1 \mathrm{~J} = 6.242 \times 10^{12} \mathrm{~MeV}

. Therefore:

E = 9 \times 10^{13} \mathrm{~J} \times 6.242 \times 10^{12} \mathrm{~MeV/J}

Calculating this, we get:

E = 5.6178 \times 10^{26} \mathrm{~MeV}

Therefore, the energy equivalent of

1 \mathrm{~g}

of a substance is: Option A:

5.6 \times 10^{26} \mathrm{~MeV}

Q168

Which of the following nuclear fragments corresponding to nuclear fission between neutron

\left({ }_0^1 \mathrm{n}\right)

and uranium isotope

\left({ }_{92}^{235} \mathrm{U}\right)

is correct :

A

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

B

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

C

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

D

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

Correct Answer

Option D

Solution

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}

.

Q169

According to Bohr's theory, the moment of momentum of an electron revolving in

4^{\text{th }}

orbit of hydrogen atom is:

A

2 \dfrac{h}{\pi}

B

\dfrac{h}{2 \pi}

C

\dfrac{h}{\pi}

D

8 \dfrac{h}{\pi}

Correct Answer

Option A

Solution

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 $\dfrac{h}{2\pi}$ 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: $2 \dfrac{h}{\pi}$

Q170

In a hypothetical fission reaction

{ }_{92} X^{236} \rightarrow{ }_{56} \mathrm{Y}^{141}+{ }_{36} Z^{92}+3 R

The identity of emitted particles (R) is :

A Proton

B Neutron

C Electron

D

\gamma

-radiations

Correct Answer

Option B

Solution

In the given hypothetical fission reaction:

^{236}_{92} X \rightarrow \, ^{141}_{56} Y + \, ^{92}_{36} Z + 3 R

We need to determine the identity of particles denoted by $R$ .

Let's use the conservation of charge and mass number (nucleon number) to identify $R$ .

First, for the conservation of nucleon number (mass number), we have:

236 = 141 + 92 + 3 \times A_R

Where $A_R$ is the mass number of $R$ . This simplifies to:

236 = 233 + 3A_R

3A_R = 236 - 233

3A_R = 3

A_R = 1

Next, we use the conservation of charge (atomic number), we have:

92 = 56 + 36 + 3Z_R

Where $Z_R$ is the atomic number of $R$ . This simplifies to:

92 = 92 + 3Z_R

3Z_R = 92 - 92

3Z_R = 0

Z_R = 0

Since the particle $R$ has a mass number of 1 and atomic number of 0, it must be a neutron.

So, the identity of the emitted particles $R$ is: Option B: Neutron