QJEEM20Ph8328: JEE Main 2020 Physics - Class 12 JEEM20Ph8328 Solution: A particle of mass 200 MeV/c 2 collides with a hydrogen atom at …

Given, m_particle = 200 MeV/c² = m(Assume)

and m_H = 1 GeV/c² = 1000 MeV/c² = 5 $\times$ 200 = 5m

Applying momentum conservation,

p_i = p_f

$\Rightarrow$ mv₀ + 0 = 0 + 5 mv'

$\Rightarrow$ v' = ${{{v_0}} \over 5}$

Initial kinetic energy, k_i = ${1 \over 2}mv_0^2$

Final kinetic energy, k_f = ${1 \over 2}\left( {5m} \right){\left( {{{{v_0}} \over 5}} \right)^2}$

$\therefore$ Loss in KE

= ${1 \over 2}mv_0^2$ - ${1 \over 2}\left( {5m} \right){\left( {{{{v_0}} \over 5}} \right)^2}$

= ${4 \over 5}\left( {{1 \over 2}mv_0^2} \right)$ = ${4 \over 5}\left( {{k_i}} \right)$

This lost energy is used by the hydrogen atom to move from ground state to the first excited state. We know the the energy required by the hydrogen atom to move from ground state to first excited state is 10.2 eV.

$\therefore$ ${4 \over 5}\left( {{k_i}} \right)$ = 10.2

$\Rightarrow$ k_i = ${{5 \times 10.2} \over 4} = {{51} \over 4}$

$\therefore$ N = 51