Dual Nature of Matter

JEE Main level, single correct, medium

25 of 200

Question 25 of 200

The de Broglie wavelength of an electron having kinetic energy EE is λ\lambda. If the kinetic energy of electron becomes E4\dfrac{E}{4}, then its de-Broglie wavelength will be:

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Answer: (D) 2λ2\lambda

The de-Broglie wavelength in the first case is
λ=h2mE\lambda =\dfrac{h}{\sqrt{2mE}}
In the second case the kinetic energy is K=E4K=\dfrac{E}{4}.
Let the de-Broglie wavelength be λ′\lambda '.
The value of wavelength is
λ′=h2mE4=2h2mE=2λ\lambda '=\dfrac{h}{\sqrt{2m\dfrac{E}{4}}}=\dfrac{2h}{\sqrt{2mE}}=2\lambda

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