A photoemissive substance is illuminated with a radiation of… | MHT CET 2026 (16 Apr, Shift 2)

Q24MHT CET 2026 · 16 APR · SHIFT 2Difficult

Matter Waves: de Broglie Wavelength

A photoemissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_e$ . The longest wavelength of radiation that can emit photoelectron is $\lambda_0$ . Expression for de-Broglie wavelength is (m = mass of electron, h = Planck's constant, C = Speed of light)

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Answer: (c)

Einstein: $K = hc\left(\dfrac{1}{\lambda_i} - \dfrac{1}{\lambda_0}\right)$
$\lambda_e = \dfrac{h}{\sqrt{2mK}} = \dfrac{h}{\sqrt{2mhc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right)}}$
$\lambda_e = \left[\dfrac{h}{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right)}\right]^{1/2}$
So the answer is (c), with the whole bracket in the denominator.

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Matter Waves: de Broglie Wavelength question from MHT CET 2026 (16 Apr, Shift 2)

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