Fusion reaction takes place at high temperature because:
 

1.	atoms get ionized at high temperature
2.	kinetic energy is high enough to overcome the Coulomb repulsion between nuclei
3.	molecules break up at high temperature
4.	nuclei break up at high temperature

A nucleus \({ }_{{n}}^{{m}} \mathrm{X}\) emits one $\mathrm{}$\(\alpha\text -\text{particle}\) and two \(\beta\text- \text{particle}\) The resulting nucleus is:

The mass of a ${}_3{}^7\mathrm{Li}$ nucleus is \(0.042~\text{u}\) less than the sum of the masses of all its nucleons. The binding energy per nucleon of the ${}_3{}^7\mathrm{Li}$ nucleus is near:
1. \(4.6~\text{MeV}\)
2. \(5.6~\text{MeV}\)
3. \(3.9~\text{MeV}\)
4. \(23~\text{MeV}\)

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:

The decay constants of two radioactive materials X₁ and X₂ are \(5\lambda\) and \(\lambda\) respectively. Initially, they have the same number of nuclei.  The ratio of the number of nuclei of X₁ to that of X₂  will be \(1/e\) after a time:
1. \(\lambda\)

2. \(\frac{1}{2\lambda }\)

3. \(\frac{1}{4\lambda }\)

4. \(\frac{e}{\lambda }\)

If \(M(A,~Z)\), \(M_p\)${}_{}$, and \(M_n\) denote the masses of the nucleus \(^{A}_{Z}X,\) proton, and neutron respectively in units of \(u\) \((1~u=931.5~\text{MeV/c}^2)\) and represent its binding energy \((BE)\) in \(\text{MeV}\). Then:

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. \(1:1\)

In the radioactive decay process, the negatively charged emitted β-particles are:

A nucleus ${}_Z{X}^A$ has a mass represented by \(M(A, Z).\) If \(M_P\) and \(M_n\) denote the mass of proton and neutron respectively and BE the binding energy, then:

1. $BE=\left[M\right(A,Z)-Z{M}_p-(A-Z\left)M_n\right]c^2$

2. $BE=[Z{M}_p+(A-Z)M_n-M(A,$ $Z\left)\right]c^2$

3. $BE=[Z{M}_p+A{M}_n-M-(A,$ $Z\left)\right]c^2$

4. $BE=M(A,$ $Z)-Z{M}_p-(A-Z)M_n$