Physics M.C.Q (1 Marks)

2. $\Big(\text{ln}\frac{2}{\lambda}\Big)$ and $\frac{1}{\lambda}$

Explanation:

The half-life of a radioactive sample $\Big(\text{t}_{\frac{1}2{}}\Big)$ is defined as the time elapsed before half the active nuclei decays.

Let the initial number of the active nuclei present in the sample be N₀.

$\frac{\text{N}_{0}}{2}=\text{N}_{\text{0}}\text{e}^{-\lambda\text{t}_{\frac{1}2{}}}$

$\Rightarrow\text{t}_{\frac{1}{2}}=\frac{\text{In}2}{\lambda}$

Average life of the nuclei, $\text{t}_{\text{av}}=\frac{\text{S}}{\text{N}_{0}}=\frac{1}{\lambda}$

Here, S is the sum of all the lives of all the N nuclei that were active at t = 0 and $\lambda$ is the decay constant of the sample.

4. $\gamma-\text{rays}$

Explanation:

Magnetic force acts on a charged particle, due to which it deflects from its path. The magnitude of this force is measured as $\Big|\overrightarrow{\text{F}}\Big|=\Big|\text{q}\Big(\overrightarrow{\text{v}}\times\overrightarrow{\text{B}}\Big)\Big|.$

Here, q is the charge on the particle that is moving with speed v in a uniform magnetic field B.

Since alpha, beta-plus and beta-minus are charged particles, they suffer deflection due to the field applied. On the other hand, gamma rays are photons and due to zero charge, they do not suffer any deflection.

4. $\gamma-\text{decay}$

Explanation:

In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number Z by 4 and neutron number N by 2 such that the element gets changed.

$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}-4}_{\text{Z}-2}\text{Y}+\text{ }^4_2\text{He}$

During $\beta^--\text{decay},$ a neutron is converted to a proton​, an electron and an antineutrino, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed.

$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}+1}\text{Y}+\text{e}+\bar{\text{v}}$

During $\beta^+-\text{decay},$ a proton in the nucleus is converted to a neutron​, a positron and a neutrino in order to maintain the stability of the nucleus, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed.

$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}-1}\text{Y}+\beta^++\text{v}$

When a nucleus is in higher excited state or has excess of energy, it comes to the ground state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. Hence, the element in gamma decay doesn't change.

2. Proton.

Explanation:

If an alpha particle is bombarded on a nitrogen (N-14) nucleus, an oxygen (O-17) nucleus and a proton are released.

According to the conservation of mass and charge,

$^4_2\text{He}+\text{ }^{14}_7\text{N}\rightarrow\text{ }^{17}_6\text{O}+\text{ }^1_1\text{p}$

So, the emitted particle is a proton.

3. A neutron does not exert electric repulsion.
4. Coulomb forces have longer range compared to the nuclear forces.

Explanation:

This is because in heavy nuclei, the $\frac{\text{N}}{\text{Z}}$ ratio becomes larger in order to maintain their stability and reduce instability caused due to the repulsion among the protons. The neutrons exert only attractive short-range nuclear forces on each other as well as on the neighbouring protons, whereas the protons exert attractive short-range nuclear forces on each other as well as the electrostatic repulsive force. Thus, the nuclei with high mass number, in order to be stable, have large neutron to proton ratio $\frac{\text{N}}{\text{Z}}.$

1. $\alpha-\text{decay}$

2. $\beta^+-\text{decay}$

Explanation:

In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number (atomic number) Z as well as neutron number N by 2.

$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}-4}_{\text{Z}-2}\text{Y}+\text{ }^4_2\text{He}$

During $\beta^--\text{decay},$ a neutron is converted to a proton​, an electron and an antineutrino. Thus, there is an increase in the atomic number.

$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}+1}\text{Y}+\text{e}^-+\bar{\text{v}}$

During $\beta^+-\text{decay},$ a proton in the nucleus is converted to a neutron​, a positron and a neutrino in order to maintain the stability of the nucleus. Thus, there is a decrease in the atomic number. ​

$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}-1}\text{Y}+\beta^++\text{v}$

When a nucleus is in higher excited state or has excess of energy, it comes to the lower state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. The element in the gamma decay doesn't change.

Therefore, alpha and beta plus decay suffer decrease in atomic number.

4. $\gamma-\text{decay}$

Explanation:

Alpha rays, beta-plus and beta-minus rays carry charged particles that show particle behaviour. On the other hand, gamma rays carry photons that show particle as well as wave behaviour. Hence, only gamma rays are electromagnetic waves.

1.  A straight line.

Explanation:

The average nuclear radius (R) and the mass number of the element (A) has the following relation:

$\text{R}=\text{R}_{0}\text{A}^{\frac{1}{2}}$

$\frac{\text{R}}{\text{R}_{0}}=\text{A}^{\frac{1}{3}}$

In $\Big(\frac{\text{R}}{\text{R}_0}\Big)=\frac{1}{3}$ In A

Therefore, the graph of ln$\Big(\frac{\text{R}}{\text{R}_0}\Big)$ versus ln A is a straight line passing through the origin with slope $\frac{1}{3}.$

1. Exact 12u

Explanation:

In nuclear physics, a unit used for measurement of mass is unified atomic mass unit, which is denoted by u.

It is defined such that

$1\text{u}=\frac{1}{12}\times$ (Mass of neutral carbon atom in its ground state)

Mass of neutral carbon atom in its ground state = 12 × 1u = 12u

Thus, the mass of neutral carbon atom in its ground state is exactly 12u.

3. Density.

Explanation:

Radius of a nucleus with mass number A is given as

$\text{R}=\text{R}_{\text{0}}\text{A}^{\frac{1}{3}}$

Here, $\text{R}_0=1.2\text{fm}$

$\therefore$ Volume of the nucleus $=\frac{4\pi\text{R}^3}{3}=\frac{4\pi\text{R}^3\text{A}}{3}$

This depends on A. With an increase in A, V increases proportionally.

Mass of the nucleus $\simeq\text{Am}_{\text{N}}$

Here, m_N is the mass of a nucleon.

Therefore, mass of the nucleus also increases with the increasing mass number. Binding energy also depends on mass number (number of nucleons) as it is the difference between the total mass of the constituent nucleons and the nucleus. Therefore, it also varies with the changing mass number.

On the other hand,

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

$=\frac{\text{Am}_{\text{N}}}{\frac{4\pi\text{R}3}{3}}=\frac{\text{Am}_{\text{N}}}{\frac{4\pi\text{R}_0^3\text{A}}{3}}=\frac{\text{m}_{\text{N}}}{\frac{4\pi\text{R}_0^3}{3}}=\frac{3\text{m}_{\text{N}}}{4\pi\text{R}_{0}^3}$

This is independent of A and hence does not change as mass number increases.

2.  F_pp = F_pn = F_nn

Explanation:

Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear force on each other. These forces are equal in magnitude, irrespective of the charge present on the nucleons.

$\therefore$ F_pp = F_pn = F_nn

Here, F_pp = F_pn = F_nn denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.

3. More than half the active nuclei decay.

Explanation:

The average life is the mean life time for a nuclei to decay.

It is given as

$\tau=\frac{1}{\lambda}=\frac{\text{T}_{\frac{1}{2}}}{0.693}$

Here, $\tau,\lambda$ and $\text{T}_{\frac{1}{2}}$ are the average life, decay constant and half life-time of the active nuclei, respectively. The value of the average lifetime comes to be more than the average lifetime. Therefore, in one average life, more than half the active nuclei decay.

3. A neutron in the nucleus decays emitting an electron.

Explanation:

Negative beta decay is given as

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

Neutron decays to produce proton, electron and anti-neutrino.