$\mathrm{If}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{binding}<mspace\; width="1em"></mspace>\mathrm{energy}<mspace\; width="1em"></mspace>\mathrm{per}<mspace\; width="1em"></mspace>\mathrm{nucleon}<mspace\; width="1em"></mspace>\mathrm{in}<mspace\; width="1em"></mspace>{}_3\mathrm{Li}_7<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{}_2\mathrm{He}_4<mspace\; width="1em"></mspace>\mathrm{nuclei}<mspace\; width="1em"></mspace>\mathrm{are}<mspace\; width="1em"></mspace>\mathrm{respectively}<mspace\; width="1em"></mspace>5.60<mspace\; width="1em"></mspace>\mathrm{MeV}<mspace\; width="1em"></mspace>\mathrm{and}$
$7.06<mspace\; width="1em"></mspace>\mathrm{MeV},<mspace\; width="1em"></mspace>\mathrm{then}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{energy}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{proton}<mspace\; width="1em"></mspace>\mathrm{in}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{reaction}<mspace\; width="1em"></mspace>{}_2\mathrm{Li}_7<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{p}<mspace\; width="1em"></mspace>\to 2_2{\mathrm{He}}^4<mspace\; width="1em"></mspace>\mathrm{is}<mspace\; width="1em"></mspace>:$

1.  19.6 MeV

2.  2.4 MeV

3.  8.4 MeV

4.  17.3 MeV

$\mathrm{In}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{equation}<mspace\; width="1em"></mspace>{}_{13}{}^{27}\mathrm{AI}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{}_2{}^4\mathrm{He}<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>{}_{15}{}^{30}\mathrm{P}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{X}<mspace\; width="1em"></mspace>\mathrm{where}<mspace\; width="1em"></mspace>\mathrm{X}<mspace\; width="1em"></mspace>\mathrm{is}-$

1.  ${}_1{}^1\mathrm{P}<mspace\; width="1em"></mspace>$

2.  ${}_1{}^2\mathrm{H}<mspace\; width="1em"></mspace>$

3.  ${}_2{}^4\mathrm{He}<mspace\; width="1em"></mspace>$

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 $1.1<mspace\; width="1em"></mspace>\mathrm{MeV}$ and $7.0<mspace\; width="1em"></mspace>\mathrm{MeV},$ respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is 

1. $23.6<mspace\; width="1em"></mspace>\mathrm{MeV}$                                         

2. $2.2<mspace\; width="1em"></mspace>\mathrm{MeV}$

3. $28.0<mspace\; width="1em"></mspace>\mathrm{MeV}$                                         

4. $30.2<mspace\; width="1em"></mspace>\mathrm{MeV}$

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 $X_2<mspace\; width="1em"></mspace>will<mspace\; width="1em"></mspace>be<mspace\; width="1em"></mspace>\frac{1}{e}$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. $9\times {10}^3<mspace\; width="1em"></mspace>\mathrm{J}$ 

3. $9\times {10}^{10}<mspace\; width="1em"></mspace>\mathrm{J}$            

4. $9\times {10}^4<mspace\; width="1em"></mspace>\mathrm{J}$

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