$\mathrm{In} \mathrm{the} \mathrm{equation} {}_{13}{}^{27}\mathrm{AI} + {}_2{}^4\mathrm{He} \to {}_{15}{}^{30}\mathrm{P} + \mathrm{X} \mathrm{where} \mathrm{X} \mathrm{is}-$

1.  ${}_1{}^1\mathrm{P}$

2.  ${}_1{}^2\mathrm{H}$

3.  ${}_2{}^4\mathrm{He}$

4.  ${}_0{}^1\mathrm{n}$

If the nuclear radius of ${}_{27}Al$ is 3.6 Fermi, the approximate nuclear radius of ${}_{64}Cu$ in Fermi is

1. 2.4                                       

2. 1.2

3. 4.8                                       

4. 3.6

The half-life of a radioactive isotope X is 50 yr.

It decays to another element Y which is stable.

The two elements X and Y were found to be in 

the ratio of 1:15 in a sample of a given rock.

The age of rock was estimated to be

1. 200 yr

2. 250 yr

3. 100 yr

4. 150 yr

Fusion reaction takes place at a higher temperature because:

The decay constant of a radio isotope is $\mathrm{\lambda }.$ If ${\mathrm{A}}_1$ and ${\mathrm{A}}_2$ are its activities at times ${\mathrm{t}}_1$ and ${\mathrm{t}}_2$ respectively, the number of nuclei which have decayed during the time $({\mathrm{t}}_1-{\mathrm{t}}_2)$

1. ${\mathrm{A}}_1{\mathrm{t}}_1-{\mathrm{A}}_2{\mathrm{t}}_2$                                       

2. ${\mathrm{A}}_1-{\mathrm{A}}_2$

3. $\frac{({\mathrm{A}}_1-{\mathrm{A}}_2)}{\mathrm{\lambda }}$                                       

4. $\mathrm{\lambda }({\mathrm{A}}_1-{\mathrm{A}}_2)$ 

The binding energy per nucleon in deuterium and helium nuclei are $1.1 \mathrm{MeV}$ and $7.0 \mathrm{MeV},$ respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is 

1. $23.6 \mathrm{MeV}$                                         

2. $2.2 \mathrm{MeV}$

3. $28.0 \mathrm{MeV}$                                         

4. $30.2 \mathrm{MeV}$

Two radioactive material $X_1$ and $X_2$ have decay constants $5\lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei of $X_1$to that of $X_2 will be \frac{1}{e}$after a time 

1. $\lambda$                                     

2. $\frac{1}{2}\lambda$

3. $\frac{1}{4\lambda }$                                 

4. $\frac{e}{\lambda }$

One nanogram of matter converted into energy will give- 

1. 90 J                     

2. $9\times {10}^3 \mathrm{J}$ 

3. $9\times {10}^{10} \mathrm{J}$            

4. $9\times {10}^4 \mathrm{J}$

Fast neutrons can easily be slowed down by 

1. The use of lead shielding

2. Passing them through heavy water

3. Elastic collisions with heavy nuclei

4. Applying a strong electric field

The number of neutrons released when ${}_{92}\mathrm{U}_{235}$ undergoes fission by absorbing ${}_0{}^1\mathrm{n}$ and $\left({}_{56}{}^{144}\mathrm{Ba}+{}_{36}{}^{89}\mathrm{Kr}\right)$are formed, is 

1. 0           

2. 1

3. 2           

4. 3