A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume $\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg}$ and $\mathrm{m}_{\mathrm{p}}=1836$ times $\mathrm{m}_{\mathrm{e}}$)

(C) $1: \frac{1}{\sqrt{1836}}$

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Modern Physics

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:
(Assume

h = 6.63 \times 1 0^{- 34} J s, m_{e} = 9.0 \times 1 0^{- 31} kg

and

m_{p} = 1836

times

m_{e}

)

(A) $1 : 1836$ ✓ Correct
(B) $1 : \frac{1}{1836}$
(C) $1 : \frac{1}{1836}$
(D) $1 : 1836$

Answer: (A)

Solution

Equal de-Broglie wavelength

λ = \frac{h}{p}

means equal momentum

p

. Since

K E = \frac{p ^{2}}{2 m}

, at equal

p

we have

K E \propto \frac{1}{m}

\frac{K E _{p}}{K E _{e}} = \frac{m _{e}}{m _{p}} = \frac{1}{1836} \Rightarrow K E_{p} : K E_{e} = 1 : 1836

Option (A).

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option

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