A particle of mass 200MeV c ^ - 2 collides with a hydrogen atom at...

Given, $M = 200MeV{c}^{- 2} and m = 1GeV{c}^{- 2}$, now use the concept of conservation of linear momentum, $M{v}_{0} = mv ⇒ v = \dfrac{M{v}_{0}}{m} . . . \left(1\right).</p><p>Now use$ the $conservation of energy, </p><p>\dfrac{1}{2}M{\left({v}_{0}\right)}$^{2} = $\dfrac{1}{2}$m{\left(v\right)}^{2} + ∆ E . . \left(2\right), here $∆ E = 13 . 6eV$, for the hydrogen atom in the ground state, now from equation, \left(1\right) \text{ and } \left(2\right), \dfrac{1}{2}M{\left({v}_{0}\right)}^{2} = \dfrac{1}{4}\left(3 ∆ E \times \dfrac{10}{8}\right)$eV = \dfrac{N}{4}$,

$⇒ \dfrac{N}{4} = \dfrac{51}{4}eV⇒ N = 51$.