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MHT CET 2026 (17 Apr, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from MHT CET 2026 (17 Apr, Shift 1)

Q3MHT CET 2026 · 17 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength

The kinetic energy of a free electron increases to 3 times the previous K.E. The ratio of new de- Broglie wavelength to previous de-Broglie wavelength is

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Answer: (a)

$\lambda = \dfrac{h}{\sqrt{2mK}}$ , so $\lambda \propto \dfrac{1}{\sqrt{K}}$ .
$K' = 3K \Rightarrow \dfrac{\lambda'}{\lambda} = \sqrt{\dfrac{K}{3K}} = \dfrac{1}{\sqrt{3}}$ .
So the answer is (a) $\dfrac{1}{\sqrt{3}}$ .

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