A proton and an electron have the same de Broglie wavelength. If K_ p and K_ e be the kinetic energies of proton and electron respectively, then choose the correct relation :

Correct answer: B. K_ p < K_ e De Broglie wavelength of proton \& electron =λ aligned & λ= h p & p_ proton =p_ electron & KE= p^2 2 m & KE_ proton < KE_ electron & [K_ p < K_ c ] aligned

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2024 · 8 April · Evening Shift · Physics · Dual Nature of Matter

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

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 :

Options

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

Correct answer

B. $\mathrm{K}_{\mathrm{p}} < \mathrm{K}_{\mathrm{e}}$

Step-by-step solution

De Broglie wavelength of proton \& electron

=\lambda

\begin{aligned} & \because \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\ & \therefore \mathrm{p}_{\text{proton }}=\mathrm{p}_{\text{electron }} \\ & \because \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{ m}} \\ & \therefore \mathrm{KE}_{\text{proton }} < \mathrm{KE}_{\text{electron }} \\ & {\left[\mathrm{K}_{\mathrm{p}} < \mathrm{K}_{\mathrm{c}}\right]}\end{aligned}

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