The value of minimum kinetic energy must a hydrogen atom possess so that for its inelastic head on c

Question

The value of minimum kinetic energy must a hydrogen atom possess so that for its inelastic head on collision with another stationary hydrogen atom one of them emits a photon, is $\_\_\_\_$ eV . (Both atoms are supposed to be in the ground state prior to the collision.)

Professor PC Thomas Classes · Answer

(20.4) : The loss in kinetic energy should be 10.2 eV \so that the hydrogen atom may be excited to first excited state. \For minimum kinetic energy case, the two atoms will move together. \ $ m v_0=(m+m) v \Rightarrow v=\frac{v_0}{2} $ \ Loss in kinetic energy $ \  =\frac{1}{2} m v_0^2-\frac{1}{2}(2 m)\left(\frac{v_0}{2}ight)^2 ; 10.2 \mathrm{eV} \ =\frac{1}{4} m v_0^2=\frac{1}{2}(K) \ 	herefore \quad  K=2 	imes 10.2 \mathrm{eV}=20.4 \mathrm{eV}  $