The energy released if hydrogen atoms are combined to form ${ }_2^4 \mathrm{He}$ is $\_\_\_\_$ MeV .

(Take binding energies per nucleon of ${ }_1^2 \mathrm{H}$ and ${ }_2^4 \mathrm{He}$ as 1.1 MeV and 7.2 MeV , respectively)

The ratio of momentum of the photons of the $1^{\text {st }}$ and $2^{\text {nd }}$ line of Balmer series of Hydrogen atoms is $\alpha / \beta$. The possible values of $\alpha$ and $\beta$ are:-

Assuming the experimental mass of ${ }_6^{12} C$ as $12 u$, the mass defect of ${ }_6^{12} C$ atom is $\_\_\_\_$ $\mathrm{MeV} / \mathrm{c}^2$.

(Mass of proton $=1.00727 \mathrm{u}$. mass of neutron $=1.00866 \mathrm{u}, 1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^2$ and c is the speed of the light in vacuum).

In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit $(f)$. The ratio of the radius of the orbits in given by $r_i: r_f=16: 4$. The wavelength of photon emitted due to this transition is $\_\_\_\_$ nm.

(Given Rydberg constant $=1.0973 \times 10^7 / \mathrm{m}$ )