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Q.

An electron, a doubly ionized helium ion $(H e^{+ +})$ and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths $λ_{e}, λ_{H e^{+ +}} a n d λ_{p}$ is:

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a

$λ_{e} > λ_{H e^{+ +}} > λ_{P}$

b

$λ_{e} > λ_{P} > λ_{H e^{+ +}}$

c

$λ_{e} < λ_{P} < λ_{H e^{+ +}}$

d

$λ_{e} < λ_{H e^{+ +}} = λ_{P}$

answer is C.

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Detailed Solution

$\begin{array}{l} λ = \frac{h}{\sqrt{2 K m}} \\ \Rightarrow λ \propto \frac{1}{\sqrt{m}} \\ λ_{e} : λ_{H e^{+ +}} : λ_{p} = \frac{1}{\sqrt{m_{e}}} : \frac{1}{\sqrt{m_{H e}}} : \frac{1}{\sqrt{m_{p}}} \\ \Rightarrow λ_{e} > λ_{p} > λ_{H e^{+ +}} \end{array}$

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