A moving neutron collides with stationary H-atom in ground state. As a result it excites and then de excites. The corresponding radiation fall on a surface, having work function $σ$ . The minimum value of required kinetic energy for neutron is $E_{0}$ and possible minimum value of de Broglie wavelength of emitted photoelectrons is $λ_{0}$ . If neutron hits stationary $H e^{+}$ ion instead of stationary H atom, then minimum value of kinetic energy of neutron is $E_{1}$ . Then

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The value of energy transferred from neutron to H-atom is $\frac{3 E_{0}}{4}$ .

The value of $^{λ_{0}}$ is $\frac{h}{\sqrt{m_{e} (E_{0} - 2 σ)}}$ .

The value of $E_{1}$ is $\frac{5 E_{0}}{2}$ .

The value of $E_{1}$ is $4 E_{0}$ .

answer is A, B, D.

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Detailed Solution

$E_{e x c i t a t i o n} = \frac{1}{2} \frac{m}{2} u^{2} = E_{2} - E_{1} = \frac{E_{0}}{2}$
$λ_{0} = \frac{h c}{\sqrt{2 m_{e} [(E_{2} - E_{1}) - o]}}$
$= \frac{h c}{\sqrt{2 m_{e} [\frac{E_{0}}{2} - o]}} = \frac{h c}{\sqrt{m_{0} E_{0} - 2 σ m_{e}}}$
Also $\Rightarrow$ Energy transferred
$\Rightarrow \frac{1}{2} m u^{2} - \frac{1}{2} m {(\frac{u}{2})}^{2} = m u^{2} (\frac{1}{2} - \frac{1}{8}) = m u^{2} (\frac{3}{8}) = \frac{m u^{2}}{4}, \frac{3}{2} = \frac{E_{0}}{2} . \frac{3}{2} = \frac{3 E_{0}}{4}$
Also For second case,
$E_{e x c i t a t i o n} = \frac{1}{2} \frac{4 m}{5} u'^{2} = E_{2} - E_{1} = 4 (E_{2} - E_{1}) = \frac{4}{5} E_{0}$
$E_{0} = 5 (E_{2} - E_{1}) = \frac{5 E_{0}}{2} z$