JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2026
- Medium
- MCQ
An electron is travelling with a velocity \(v\) in free space and when it enters a medium, its velocity is reduced by \(20\%\). The de Broglie wavelength of electron in the medium is \(\alpha\lambda_0\), where \(\lambda_0\) is its de Broglie wavelength in free space. The value of \(\alpha\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(1.25\)
Step-by-step Solution
Detailed explanation
The de Broglie wavelength of an electron in free space is given by \(\lambda_0 = \dfrac{h}{mv}\).
When the electron enters the medium, its velocity is reduced by \(20\%\). The new velocity is \(v' = v - 0.20v = 0.80v\).
The de Broglie wavelength in the medium is \(\lambda' = \dfrac{h}{mv'}\).
Substituting \(v' = 0.80v\), we get \(\lambda' = \dfrac{h}{m(0.80v)} = \dfrac{1}{0.80} \left(\dfrac{h}{mv}\right)\).
Since 1 / 0.80 = 100 / 80 = 1.25,
we have λ' = 1.25λ_0.
Comparing this with λ' = αλ_0, we get α = 1.25. Answer: 1.25
When the electron enters the medium, its velocity is reduced by \(20\%\). The new velocity is \(v' = v - 0.20v = 0.80v\).
The de Broglie wavelength in the medium is \(\lambda' = \dfrac{h}{mv'}\).
Substituting \(v' = 0.80v\), we get \(\lambda' = \dfrac{h}{m(0.80v)} = \dfrac{1}{0.80} \left(\dfrac{h}{mv}\right)\).
Since 1 / 0.80 = 100 / 80 = 1.25,
we have λ' = 1.25λ_0.
Comparing this with λ' = αλ_0, we get α = 1.25. Answer: 1.25
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