A neutron with kinetic energy 70 eV collides in-elastically with a stationary $H{e}^{+}$  atom (in ground state) and neutron is scattered at right angle to the original direction of its motion. After collision the electron in $H{e}^{+}$  atom is transferred to the third excited state, then after collision  $\left(take{M}_{He}\text{= }4m_N\right)$. Choose the correct option. 

$$\begin{gathered} CLM(x-axis) \sqrt{2m_n\left(70ev\right)}=p_x \\ CLM(y-axis) \sqrt{2m_{n}K{n}_f}=p_y \\ {K}_{\alpha }=\frac{{\left(\sqrt{p_x^2+p_y^2}\right)}^2}{2m_a}=\frac{2m_n\left(70ev\right)+2m_n\left(K{n}_f\right)}{2\times \left(4m_n\right)} \\ \Rightarrow 4K_{\alpha }-K{n}_f=70eV\mathrm{.................}\left(1\right) \\ Also, K{n}_i-K_{\alpha }-K{n}_f=51eV \\ \Rightarrow K_{\alpha }-K{n}_f=19eV\mathrm{.................}\left(2\right) \end{gathered}$$

$\therefore$  Using (1) and (2)     

$\begin{array}{l}{K}_{\alpha }=\frac{89}{5}eV=17.8eV\\ K_{n_f}=1.2eV\end{array}$

On de-excitation of  $e^{-}$ of  He-ion
No of wavelength emitted  $=\frac{n(n-1)}{2}=6$