Dual Nature of Matter

JEE Main level, single correct, medium

6 of 200

Question 6 of 200

An electron is travelling with a velocity vv in free space and when it enters a medium, its velocity is reduced by 20%20\%. The de Broglie wavelength of electron in the medium is αλ0\alpha\lambda_0, where λ0\lambda_0 is its de Broglie wavelength in free space. The value of α\alpha is _______.

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Answer: (C) 1.251.25

The de Broglie wavelength of an electron in free space is given by λ0=hmv\lambda_0 = \dfrac{h}{mv}.

When the electron enters the medium, its velocity is reduced by 20%20\%. The new velocity is v′=v−0.20v=0.80vv' = v - 0.20v = 0.80v.

The de Broglie wavelength in the medium is λ′=hmv′\lambda' = \dfrac{h}{mv'}.

Substituting v′=0.80vv' = 0.80v, we get λ′=hm(0.80v)=10.80(hmv)\lambda' = \dfrac{h}{m(0.80v)} = \dfrac{1}{0.80} \left(\dfrac{h}{mv}\right).

Since 10.80=10080=1.25\dfrac{1}{0.80} = \dfrac{100}{80} = 1.25, we have λ′=1.25λ0\lambda' = 1.25\lambda_0.

Comparing this with λ′=αλ0\lambda' = \alpha\lambda_0, we get α=1.25\alpha = 1.25.

Answer: 1.251.25

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