• # question_answer If we assume that penetrating power of any radiation/particle Is inversely proportional to its De-broglie wavelength of the particle then: A) a proton and an $\alpha$-particle after getting accelerated through same potential difference will have equal penetrating power, B) penetrating power of $\alpha$-particle will be greater than that of proton which have been accelerated by same potential difference. C) proton's penetrating power will be less than penetrating power of an electron which has been accelerated by the same potential difference. D) penetrating powers can not be compared as all these are particles having no wavelength or wave nature.

 $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}$              $\Rightarrow$   $\frac{h}{\sqrt{2m\,(Vq)}}$ $\therefore$ For higher m and q; $\lambda$ will be smaller. For an $'\alpha '$ particle; both 'm' and 'q' are higher. hence lesser is the wavelength. As, (penetrating power) $\propto$ Energy $\propto \frac{1}{\lambda }$ From above; penetrating power of an $\alpha$-particle is more than that of a proton.