A particle of mass $200 MeV / c^{2}$ collides with a hydrogen atom at rest. Soon after the collision the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is $\frac{N}{4}$ . The value of N is_______ (Given the mass of the hydrogen atom to be 1 $GeV / c^{2}$ )

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answer is 51.

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Detailed Solution

From linear momentum conservation, $L_{i} = L_{f}$
$mV + 0 = 0 + 5 mV' \Rightarrow V' = \frac{v}{5}$
Loss of KE $= {KE}_{i} - {KE}_{f} = \frac{1}{2} {mv}^{2} - \frac{1}{2} (5 m) {(\frac{v}{5})}^{2}$
$= \frac{1}{2} {mv}^{2} (1 - \frac{1}{5}) = \frac{4}{5} (\frac{{mv}^{2}}{2})$
$= \frac{4}{5} {KE}_{i} = 10.2 eV$
[ $∵$ Energy in first excited state of atom = 10.2 eV]
${KE}_{i} = 12.75 eV = \frac{N}{4} \Rightarrow N = 51$
The value of N = 51.