An element \(A\) decays into element \(C\) by a two-step process
\(A\rightarrow B+_{2}^{4}\mathrm{He}\)
\(B\rightarrow C+2e^{-}\)
Then: 

1.	\(A\) and \(C\) are isotopes
2.	\(A\) and \(C\) are isobars
3.	\(A\) and \(B\) are isotopes
4.	\(A\) and \(B\) are isobars

The radius R of a nuclear matter varies with A as

1. $R \alpha A^{2/3}$

2. $R \alpha A^{1/3}$

3. $R \alpha A^0$ 

4. $R \alpha A$

Number of nuclei of a radioactive substance at time t = 0 are 2000 and 1800 at time t = 2s. Number of nuclei left after t = 6s is 

1. 1442 

2. 1554 

3. 1652 

4. 1458

In a radioactive material, the activity at time $t_1$ is $R_1$ and at a later time, $t_2$ is $R_2$. If the decay constant of the material is $\lambda$, then:

1. $R_1=R_2{e}^{-\lambda \left(t_1-t_2\right)}$   

2. $R_1=R_2{e}^{\lambda \left(t_1-t_2\right)}$

3. $R_1=R_2\left(\frac{t_2}{t_1}\right)$ 

4. $R_1=R_2$

Which of the following are suitable for the fusion process ? 

1. Light nuclei

2. Heavy nuclei

3. Elements lying in the middle of periodic table

4. Elements lying in the middle of binding energy curve

An archeologist analyses the wood in a prehistoric structure and finds that ${\mathrm{C}}^{14}$ (Half - life=5700 years) to ${\mathrm{C}}^{12}$ is only one - fourth of that found in the cells buried plants. The age of the wood is about

1. 5700 yr

2. 2850 yr

3. 11,400 yr

4. 22,800 yr

In the nuclear reaction: \(\mathrm{X}\left(n,\alpha\right){}_{3}^{7}\mathrm{Li}\) the term \({X}\) will be:

10 g of radioactive material of half-life 15 year is kept in store for 20 years. The disintegrated material mass is 

1. 12.5 g

2. 10.5 g

3. 6.03 g

4. 4.03 g

If a radioactive nucleus decays according to the following reaction

${}_{72}X_{180}\stackrel{\alpha }{\to }{X}_1\stackrel{\beta }{\to }{X}_2\stackrel{\alpha }{\to }{X}_3\stackrel{\gamma }{\to }{X}_4$

then the mass number and the atomic number of $X_4$ then the mass number and the atomic number of $X_4$ will be, respectively 

1. 172, 70 

2. 172, 69

3. 172, 68

4. 171, 69

Two radioactive materials $X_1$ and $X_2$ have decay constants 10$\lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_1$ to that of $X_2$ will be $\frac{1}{e}$ after a time

1. $\frac{1}{10 \lambda }$ 

2. $\frac{1}{11 \lambda }$

3. $\frac{11}{10 \lambda }$   

4. $\frac{1}{9 \lambda }$