JEE Main 2022 · 25 Jun · Shift 2 · Physics Q46

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PhysicsSingle correct+4 / -1Question 46 of 90

A proton, a neutron, an electron and an $\alpha$ -particle have same energy. If $\lambda_p$ , $\lambda_n$ , $\lambda_e$ and $\lambda_\alpha$ are the de Broglie's wavelengths of proton, neutron, electron and $\alpha$ particle respectively, then choose the correct relation from the following :

\lambda_p=\lambda_n>\lambda_e>\lambda_\alpha

\lambda_\alpha<\lambda_n<\lambda_p<\lambda_e

\lambda_e<\lambda_p=\lambda_n>\lambda_\alpha

\lambda_e=\lambda_p=\lambda_n=\lambda_\alpha

Answer and worked solution

Official answer: B

Same energy. $\lambda=\dfrac h{\sqrt{2mE}}\propto\dfrac1{\sqrt m}$ .

Masses. $m_\alpha>m_n>m_p>m_e$ (the neutron is slightly heavier than the proton), so $\lambda_\alpha<\lambda_n<\lambda_p<\lambda_e$ .

(B)

Topics:de Broglie waves

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JEE Main 2022 · 25 Jun · Shift 2, Physics question 46