Solution ⟹
Let u be velocity of hydrogen atom before collision and V the ve locity of two atoms moving together after collision By principal of conservation of momentum , we have
[tex]Mu + M \times 0 = 2VM[/tex]
[tex]V = \frac{u}{2} \\ [/tex]
The loss in kinetic energy Δϵ due to collision is
[tex]Δϵ= \frac{1}{2}Mu {}^{2} − \frac{1}{2} (2M)V {}^{2} \\ [/tex]
[tex]As, V = \frac{u}{2} \: we \: have \: Δϵ= \frac{1}{2}Mu {}^{2} − \frac{1}{2} (2M)( \frac{u}{2}) {}^{2} \\ [/tex]
[tex] = \frac{1}{2} Mu {}^{2} - \frac{1}{4}Mu { }^{2} = \frac{1}{4} Mu {}^{2} \\ [/tex]
Rest answer is in the pic
