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Dual Nature of Radiation question

2024 · 8 Apr · Shift 2 · Q76

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Dual Nature of Radiation
2024 · 8 Apr · Shift 2 · Q76

Dual Nature of Radiation question

2024 · 8 Apr · Shift 2 · Q76

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1

A proton and an electron have the same de Broglie wavelength. If

\mathrm{K}_{\mathrm{p}}

and

\mathrm{K}_{\mathrm{e}}

be the kinetic energies of proton and electron respectively, then choose the correct relation :

A
$\mathrm{K_p>K_e}$
B
$\mathrm{K_p=K_e}$
C
$\mathrm{K}_{\mathrm{p}}\lt \mathrm{K}_{\mathrm{e}}$
D
$\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{e}}{ }^2$

View written solutionFree

Correct answer: C

Use de Broglie relation

For any particle,

\lambda = \frac{h}{p}

If a proton and an electron have the same de Broglie wavelength, then their momenta must be equal:

p_p = p_e

Relate kinetic energy to momentum

For non-relativistic motion,

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

Since both particles have the same momentum $p$ ,

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

Compare the two kinetic energies

Now,

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

Since

m_p \gg m_e

we have

\frac{m_e}{m_p} < 1

Therefore,

K_p < K_e

Check options

A: $K_p > K_e$ — Incorrect
B: $K_p = K_e$ — Incorrect
C: $K_p < K_e$ — Correct
D: $K_p = K_e^2$ — Dimensionally incorrect / Incorrect

Hence, the correct relation is

\boxed{K_p < K_e}

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