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

Dual Nature of Radiation PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q76 with Solution

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Correct answer: C

For a particle,

\lambda = \frac{h}{p}

If the proton and electron have the same de-Broglie wavelength, then they have the same momentum:

p_p = p_e

For non-relativistic motion,

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

Since both particles have the same momentum $p$ ,

K \propto \frac{1}{m}

Let $K_p$ be the kinetic energy of proton and $K_e$ that of electron. Then

\frac{K_p}{K_e} = \frac{\frac{p^2}{2m_p}}{\frac{p^2}{2m_e}} = \frac{m_e}{m_p}

Given,

m_p = 1836\, m_e

So,

\frac{K_p}{K_e} = \frac{m_e}{1836 m_e} = \frac{1}{1836}

Thus,

K_p : K_e = 1 : 1836

This corresponds to Option C.

Stored correct answer = C

Our derived answer = C

So they agree.