What should be the order of arrangement of de-Broglie wavelength of electron ($\lambda_{\mathrm{e}}$), an $\alpha$-particle $(\lambda_{\alpha})$ and proton $(\lambda_{\mathrm{p}})$ given that all have the same kinetic energy?

(D) $\lambda_{e}>\lambda_{p}>\lambda_{\alpha}$

(A) $\lambda_{e}=\lambda_{p}=\lambda_{\alpha}$

(B) $\lambda_{e}<\lambda_{p}<\lambda_{\alpha}$

(C) $\lambda_{e}=\lambda_{p}>\lambda_{\alpha}$

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Modern Physics

What should be the order of arrangement of de-Broglie wavelength of electron (

λ_{e}

), an

α

-particle

(λ_{α})

and proton

(λ_{p})

given that all have the same kinetic energy?

(A) $λ_{e} = λ_{p} = λ_{α}$
(B) $λ_{e} < λ_{p} < λ_{α}$
(C) $λ_{e} = λ_{p} > λ_{α}$
(D) $λ_{e} > λ_{p} > λ_{α}$ ✓ Correct

Answer: (D)

Solution

A particle of kinetic energy

K

has momentum

p = 2 m K

, so

λ = \frac{h}{p} = \frac{h}{2 m K}

At a common kinetic energy the wavelength varies as

\frac{1}{m}

. The masses increase in the order electron, proton, alpha particle, so the wavelengths decrease in that same order and

λ_{e} > λ_{p} > λ_{α}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option

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