Two nuclei have their mass numbers in the ratio of \(1:3.\) The ratio of their nuclear densities would be:
1. \(1:3\)
2. \(3:1\)
3. \((3)^{1/3}:1\)
4. \(1:1\)

The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an:
1. Isobar of a parent.
2. Isomer of a parent.
3. Isotone of a parent.
4. Isotope of a parent.

A nucleus \({ }_{{n}}^{{m}} {X}\) emits one $\mathrm{}$\(\alpha\text -\text{particle}\) and two \(\beta\text- \text{particle}\) The resulting nucleus is:
1. \(^{m-}{}_n^6 Z \)
2. \(^{m-}{}_{n}^{4} X \)
3. \(^{m-4}_{n-2}Y\)
4. \(^{m-6}_{n-4} Z \)

An element \(X\) decays, first by positron emission, and then two \(\alpha\text-\)particles are emitted in successive radioactive decay. If the product nuclei have a mass number \(229\) and atomic number \(89\), the mass number and the atomic number of element \(X\) are:
1. \(237,~93\) 
2. \(237,~94\)
3. \(221,~84\)
4. \(237,~92\)

A nucleus ${}^{\mathrm{z}}{\mathrm{X}}_{\mathrm{A}}$ emits 9 $\mathrm{\alpha }$-particles and 5 $\beta -$ particles. The ratio of total protons and neutrons in the final nucleus is:

1. $\frac{\mathrm{Z}-13}{\left(\mathrm{A}-\mathrm{Z}-23\right)}$

2. $\frac{\left(\mathrm{Z}-18\right)}{\left(\mathrm{A}-36\right)}$

3. $\frac{\left(\mathrm{Z}-13\right)}{\left(\mathrm{A}-36\right)}$

4. $\frac{\left(\mathrm{Z}-13\right)}{\left(\mathrm{A}-\mathrm{Z}-13\right)}$

Calculate the Q-value of the nuclear reaction:
\(2~{ }_{6}^{12} \mathrm{C}\rightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_2^4 \mathrm{He}\)
The following data are given:
\(m({ }_{6}^{12} \mathrm{C})=12.000000~\text{u}\)
\(m({ }_{10}^{20} \mathrm{Ne})=19.992439~\text{u}\)
\(m({ }_{2}^{4} \mathrm{He})=4.002603~\text{u}\)
1. \(3.16~\text{MeV}\)
2. \(5.25~\text{MeV}\)
3. \(3.91~\text{MeV}\)
4. \(4.65~\text{MeV}\)

If in nuclear reactor using U²³⁵ as fuel, the power output is 4.8 MW, the number of fissions per second is:
(Energy released per fission of U²³⁵ = 200 MeV watts, 1 eV = 1.6 X 10^–19 J)

1. 1.5×10¹⁷           

2. 3×10¹⁹           

3. 1.5×10²⁵           

4. 3×10²⁵

The Binding energy per nucleon of \(^{7}_{3}\mathrm{Li}\) and \(^{4}_{2}\mathrm{He}\) nucleon are \(5.60~\text{MeV}\) and \(7.06~\text{MeV}\), respectively. In the nuclear reaction \(^{7}_{3}\mathrm{Li} + ^{1}_{1}\mathrm{H} \rightarrow ^{4}_{2}\mathrm{He} + ^{4}_{2}\mathrm{He} +Q\), the value of energy \(Q\) released is:
1. \(19.6~\text{MeV}\)
 2. \(-2.4~\text{MeV}\)
 3. \(8.4~\text{MeV}\)
4. \(17.3~\text{MeV}\)

If $\frac{N}{Z}$ ratio in a nucleus is smaller than the required value for stability, then:

Determine the energy released in the process:

${}_{1}H_2+{}_{1}H_2\to {}_{2}H{e}^4+Q$

Given: M $\left({}_{1}H_2\right)$= 2.01471 amu

            M$\left({}_{2}H{e}^4\right)$= 4.00388 amu

1. 3.79 MeV
2.13.79 MeV
3. 0.79 MeV 
4. 23.79 MeV