If we assume that penetrating power of any radiation particle is inversely proportional to its de-Broglie wavelength of the particle then

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Proton's penetrating power will be less than penetrating power of an electron which has been accelerated by the same potential difference.

Penetrating powers cannot be compared as all these are particles having no wavelength or wave nature.

A proton and an alpha-particle after getting accelerated through same potential difference will have equal penetrating Power.

Penetrating power of alpha-particle will be greater than that of proton which have been accelerated by same potential difference.

answer is B.

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Detailed Solution

$\begin{array}{l} λ = \frac{h}{p} = \frac{h}{\sqrt{2 mE}} \Rightarrow \frac{h}{\sqrt{2 m (Vq)}} \\ ∴ For higher m and q, \\ λ will be smaller . \\ For an' α' particle, both' m' and' q' are higher . \\ Hence, lesser is the wavelength . \\ As, (Penetrating power) \propto Energy \propto \frac{1}{λ} \\ From above, penetrating power of an "α" particle is more than that of a proton . \end{array}$