If in a nuclear fusion  process, the masses of the fusion nuclei be $m_1 and m_2$ and the mass of the resultant nucleus be $m_3$, then 

1. $m_3=m_1+m_2$

2. $m_3=\left|m_1-m_2\right|$

3. $m_3<\left(m_1+m_2\right)$ 

4. $m_3>\left(m_1+m_2\right)$

The graph between the square root of the frequency of a specific line of the characteristic spectrum of X-rays and the atomic number of the target will be:

1.
2.
3.
4.

The nuclei of which one of the following pairs of nuclei are isotones? 

1. ${}_{34}S{e}^{74}, {}_{31}G{a}^{71}$ 

2. ${}_{42}M{o}^{92}, {}_{40}Z{r}^{92}$

3. ${}_{38}S{r}^{84}, {}_{38}S{r}^{86}$ 

4. ${}_{20}C{a}^{40}, {}_{16}S_{32}$

The counting rate observed from a radioactive source at t = 0 second was 1600 counts per second and at t = 8 seconds it was 100 counts per second. The counting rate observed, as counts per second, at t = 6 seconds will be:

1. 400

2. 300

3. 200

4. 150

If $N_0$ is the original mass of the substance of half life period $T_{1/2}$= 5 years, then the amount of substance left after 15 years is

1. $N_0$/8 

2. $N_0$/16

3. $N_0$/2

4. $N_0$/4

The mass number of He is 4 and that for sulphur is 32. The radius of sulphur nucleus is larger than that of helium, by times

1. $\sqrt{8}$ 

2. 4

3. 2 

4.  8

The numbers of nuclei of a radioactive substance at time t = 0 are 1000 and 900 at time t = 2 sec. Then the number of nuclei at time t = 4 sec will be:

1. 800

2. 810 

3. 790 

4. 700

A nucleus ${}_Z{}^{A}X$ has mass represented by m(A, Z). If $m_p$ and $m_n$ denote the mass of proton and neutron respectively and BE is the binding energy (in MeV), then:

1. BE = [m(A, Z) - Z$m_p$- (A - Z)$m_n$] $c^2$

2. BE = [Z$m_p$+ (A - Z)$m_n$- m(A,Z)] $c^2$

3. BE = [Z$m_p$+ A$m_n$- m(A, Z)] $c^2$

4. BE = m(A, Z) - Z$m_p$- (A, Z)$m_n$

Half-lives of two radioactive substances A and B respectively are 20 min and 40 min.  Initially, the samples of A and B have an equal number of nuclei. After 80 min the ratio of the remaining number of A and B nuclei is:

1. 1 : 16

2. 4 : 1

3. 1 : 4 

4. 1 : 1

Radioactive ${}_{27}{}^{60}Co$ is transformed into stable ${}_{28}{}^{60}Ni$ by emitting two $\gamma$-rays of energies

1. 1.33 MeV and 1.17 MeV in succession

2. 1.17 MeV and 1.33 MeV in succession

3. 1.37 MeV and 1.13 MeV in succession

4. 1.17 MeV and 1.13 MeV in succession