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  • A particle of mass  collides with a hydrogen atom at rest. Soon after the collision the particle comes to rest, and the atom recoils a

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} )______.
Option: 1 51
Option: 2 102
Option: 3 204
Option: 4 408

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best_answer

\begin{aligned} &M_{0}=200 \mathrm{MeV} / \mathrm{c}^{2} ; \mathrm{m}=1 \mathrm{GeV} / \mathrm{c}^{2}\\ &\text { Initial velocity }=v_{0}\\ \\ &\begin{array}{l} \mathrm{Mv}_{0}=\mathrm{mv} \\\\ \therefore \mathrm{v}=\left(\frac{M}{m}\right) \mathrm{v}_{0} \\\\ \begin{array}{l} \frac{1}{2} \mathrm{Mv}_{0}^{2}=\frac{1}{2} \mathrm{mv}^{2}+\frac{\Delta \mathrm{E}_{0} \times 3}{4} \quad\left(\Delta \mathrm{E}_{0}=13.6 \mathrm{eV}\right) \\\\ \Rightarrow \quad \frac{1}{2} \mathrm{Mv}_{0}^{2}\left(1-\frac{\mathrm{M}}{\mathrm{m}}\right)=\frac{3}{4} \Delta \mathrm{E}_{0} \\\\ \Rightarrow \frac{1}{2} \mathrm{Mv}_{0}^{2}=\frac{3}{4} \times \Delta \mathrm{E}_{0} \times \frac{10}{8}=\frac{1}{4}\left(3 \Delta \mathrm{E}_{0} \times \frac{10}{8}\right) \\\\ =\frac{51}{4}=\frac{51}{4} \mathrm{eV} \end{array} \\\\ \therefore \mathrm{N}=51 \end{array} \end{aligned}

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Deependra Verma

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