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Dual Nature of Radiation question

2023 · 15 Apr · Shift 1 · Q49

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Dual Nature of Radiation
2023 · 15 Apr · Shift 1 · Q49

Dual Nature of Radiation question

2023 · 15 Apr · Shift 1 · Q49

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1

The de Broglie wavelength of an electron having kinetic energy

\mathrm{E}

is

\lambda

. If the kinetic energy of electron becomes

\frac{E}{4}

, then its de-Broglie wavelength will be :

A
$\sqrt{2} \lambda$
B
$2 \lambda$
C
$\frac{\lambda}{2}$
D
$\frac{\lambda}{\sqrt{2}}$

View written solutionFree

Correct answer: B

Use de Broglie relation

For a particle,

\lambda = \frac{h}{p}

where $h$ is Planck's constant and $p$ is momentum.

Relate momentum to kinetic energy

For a non-relativistic electron,

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

So,

p = \sqrt{2mE}

Substitute into de Broglie equation:

\lambda = \frac{h}{\sqrt{2mE}}

Thus,

\lambda \propto \frac{1}{\sqrt{E}}

Change in kinetic energy

If the kinetic energy becomes

E' = \frac{E}{4}

then new wavelength $\lambda'$ is

\lambda' \propto \frac{1}{\sqrt{E/4}} = \frac{1}{\frac{\sqrt{E}}{2}} = \frac{2}{\sqrt{E}}

So compared to the original wavelength,

\lambda' = 2\lambda

Check options

A: $\sqrt{2}\lambda$ ❌
B: $2\lambda$ ✅
C: $\frac{\lambda}{2}$ ❌
D: $\frac{\lambda}{\sqrt{2}}$ ❌

Therefore, the correct answer is:

\boxed{2\lambda}

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