neutron beam, in which each neutron has same kinetic energy, is passed through a sample of hydrogen like gas (but not hydrogen) in ground state and at rest. Due to collision of neutrons with the ions of the gas, ions are excited and then they emit photons. Six spectral lines are obtained in which one of the lines is of wavelength (6200/51) nm. The mass of neutron and proton can be assumed to be nearly same. Use hc = 12400 eV$\text{Å}.$.

The H-type atom is in the third excited state i.e. n = 4.
Energy corresponding to wave length $\frac{6200}{51}\mathrm{nm}$

$=\frac{12400\times 51}{62000}=10.2\mathrm{eV}$

$\text{ This is the }{E}_2-E_1\text{ for }H\text{ and }{E}_4-E_2\text{ for }H{e}^{+}\text{. }$

we get z = 2 for 4 $\to$ 2  radiation 

Hence the atom is Helium ion. 
Ans. He⁺ , (1) 

Let u be the speed of neutron before collision

At end of the deformation phase (when the kinetic energy of (neutron + He⁺) system is least)
 

Where $u_{cm}$ is velocity of centre of mass. From conservation of momentum

${\mathrm{u}}_{\mathrm{cm}}=\frac{\mathrm{mu}}{\mathrm{m}+4\mathrm{m}}=\frac{\mathrm{u}}{5}$

The loss of kinetic energy
$=\frac{1}{2}m{u}^2-\frac{1}{2}m{\left(\frac{u}{5}\right)}^2-\frac{1}{2}4m{\left(\frac{u}{5}\right)}^2=\frac{4}{5}\left(\frac{1}{2}m{u}^2\right)$

It K is the kinetic energy of neutron then the maximum loss in K.E. of system is 

$\frac{4}{5}\mathrm{K}=51\mathrm{eV} \mathrm{Or} \mathrm{K}=\frac{51\times 5}{4}=63.75\mathrm{eV}$

${\mathrm{K}}_{min}=\frac{255}{4}=63.75\mathrm{eV}$