Question

If we assume that penetrating power of any radiation/particle is inversely proportional to the De-broglie wavelength of the particle, then

• A. Penetrating power of $\alpha -$particle will be greater than that of proton which have been accelerated by same potential difference.
• B. Proton's penetrating power will be less than penetrating power of an electron which has been accelerated by the same potential difference.
• C. Penetrating powers can not be compared as all these are particles having no wavelength or wave nature.
• D. A proton and an $\alpha -$particle after getting accelerated through same potential difference will have equal penetrating power.

Answer 1

Answer: (A)

$\lambda =\frac{h}{P}=\frac{h}{\sqrt{2mE}}\Rightarrow \frac{h}{\sqrt{2m\left(Vq\right)}}$

$\therefore$ 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.

So, the answer is option (B).