The kinetic energy of an… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Modern Physics

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The kinetic energy of an electron, $\alpha -$ particle and a proton are given as $4K,2K$ and $K$ respectively. The de-Broglie wavelength associated with electron $({\lambda }_{e}),\alpha -$ particle $({\lambda }_{\alpha })$ and the proton $({\lambda }_{p})$ are as follows:

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Held on 6 Apr 2023 · Verified 6 Jul 2026.

Options

A
${\lambda }_{\alpha }={\lambda }_{p}>{\lambda }_{e}$
B
${\lambda }_{\alpha }<{\lambda }_{p}<{\lambda }_{e}$
C
${\lambda }_{\alpha }={\lambda }_{p}<{\lambda }_{e}$
D
${\lambda }_{\alpha }>{\lambda }_{p}>{\lambda }_{e}$

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Solution

The formula to calculate the de Broglie wavelength of a particle is given by

$\lambda =\frac{h}{\sqrt{2mK}}...(1)$

Now, the masses of the given particles are interrelated by the following relations:

$\begin{matrix}{m}_{e} & \approx & \frac{{m}_{p}}{1837} \\ {m}_{\alpha } & = & 4{m}_{p}\end{matrix}$

The de Broglie wavelength for electron is given by

$\begin{matrix}{\lambda }_{e} & = & \frac{h}{\sqrt{2{m}_{e}(4K)}} \\ & = & \frac{h}{\sqrt{\frac{8}{1837}{m}_{p}K}}...(2)\end{matrix}$

The de Broglie wavelength for proton is given by

$\begin{matrix}{\lambda }_{p} & = & \frac{h}{\sqrt{2{m}_{p}(2K)}} \\ & = & \frac{h}{\sqrt{4{m}_{p}K}}...(3)\end{matrix}$

And, the de Broglie wavelength of alpha particle is given by

$\begin{matrix}{\lambda }_{\alpha } & = & \frac{h}{\sqrt{2{m}_{\alpha }K}} \\ & = & \frac{h}{\sqrt{2(4{m}_{p})K}} \\ & = & \frac{h}{\sqrt{8{m}_{p}K}}...(4)\end{matrix}$

From equation (2), (3) and (4), it can be concluded that ${\lambda }_{e}>{\lambda }_{p}>{\lambda }_{\alpha }$ .

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Options

${\lambda }_{\alpha }={\lambda }_{p}>{\lambda }_{e}$

${\lambda }_{\alpha }<{\lambda }_{p}<{\lambda }_{e}$

${\lambda }_{\alpha }={\lambda }_{p}<{\lambda }_{e}$

${\lambda }_{\alpha }>{\lambda }_{p}>{\lambda }_{e}$

Solution

The formula to calculate the de Broglie wavelength of a particle is given by

$\lambda =\frac{h}{\sqrt{2mK}}...(1)$

Now, the masses of the given particles are interrelated by the following relations:

$\begin{matrix}{m}_{e} & \approx & \frac{{m}_{p}}{1837} \\ {m}_{\alpha } & = & 4{m}_{p}\end{matrix}$

The de Broglie wavelength for electron is given by

$\begin{matrix}{\lambda }_{e} & = & \frac{h}{\sqrt{2{m}_{e}(4K)}} \\ & = & \frac{h}{\sqrt{\frac{8}{1837}{m}_{p}K}}...(2)\end{matrix}$

The de Broglie wavelength for proton is given by

$\begin{matrix}{\lambda }_{p} & = & \frac{h}{\sqrt{2{m}_{p}(2K)}} \\ & = & \frac{h}{\sqrt{4{m}_{p}K}}...(3)\end{matrix}$

And, the de Broglie wavelength of alpha particle is given by

$\begin{matrix}{\lambda }_{\alpha } & = & \frac{h}{\sqrt{2{m}_{\alpha }K}} \\ & = & \frac{h}{\sqrt{2(4{m}_{p})K}} \\ & = & \frac{h}{\sqrt{8{m}_{p}K}}...(4)\end{matrix}$

From equation (2), (3) and (4), it can be concluded that ${\lambda }_{e}>{\lambda }_{p}>{\lambda }_{\alpha }$ .

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Solution

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JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

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Mechanics

Electromagnetism

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Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics