What happens to the mass number and the atomic number of an element when it emits \(\gamma\text{-}\)radiation?

1.	mass number decreases by four and atomic number decreases by two.
2.	mass number and atomic number remain unchanged.
3.	mass number remains unchanged while the atomic number decreases by one.
4.	mass number increases by four and the atomic number increases by two.

The binding energy per nucleon in deuterium and helium nuclei are \(1.1\) MeV and \(7.0\) MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is:
1. \(2.2\) MeV
2. \(28.0\) MeV
3. \(30.2\) MeV
4. \(23.6\) MeV

Boron has two isotopes ${}_{5}B^{10}$ and ${}_{5}B^{11}$. If atomic weight of Boron is 10.81 then ratio of ${}_{5}B^{10}$ to ${}_{5}B^{11}$in nature will be: 

1. 15 : 16 

2. 19: 81

3. 81 : 19 

4. 20: 53

For the given reaction, the particle \(X\) is:
\({ }_6^{11} \mathrm{C}\rightarrow { }_5^{11}\mathrm{B}+\beta^{+}+X\)
1. neutron
2. anti neutrino
3. neutrino
4. proton

In any fission process the ratio

$\frac{\text{ mass of fission products }}{\text{ mass of parent nucleus }}\text{ is:}$

1. Greater than 1

2. Depends on the mass of the parent nucleus

3. Equal to 1

4. Less than 1

Fission of nuclei is possible because the binding energy per nucleon in them:
 

Nuclear – fission is best explained by:

1. Liquid droplet theory

2. Yukawa $\pi$ - meson theory

3. Independent particle model of the nucleus

4. Proton-proton cycle

For the nuclear reaction:

${}_{92}U_{235}$ $+$ ${}_{0}n_1$ $\to$ ${}_{56}B{a}^{144}$ $$ $+$ $...$ $$ $+$ $3_0{n}^1$

The blank space can be filled by:

1. ${}_{26}K{r}^{89}$

2.  ${}_{36}K{r}^{89}$

3. ${}_{26}S{r}^{90}$

4.  ${}_{38}S{r}^{89}$

${}_{n}X_m$ emitted one $\alpha and\mathrm{2 }\beta$ particles, then it will become: 

1. ${}_{n}X_{m-4}$

2. ${}_{n-1}X_{m-1}$

3. ${}_{n}Z_{m-4}$

4. None of these