A photoemissive substance is illuminated with a radiation of wavelength λi so that it releases electrons with de-Broglie wavelength λe. The longest wavelength of radiation that can emit photoelectron is λo. Expression for de-Broglie wavelength is given by: (m: mass of the electron, h: Planck's constant and c: speed of light) (a) λe = (√(hλi))/(√(2mc)) (b) λe = (h)/(√(2mc((1)/(λi)-(1)/(λo)))) (c) λe = √((h)/(2mc((1)/(λi)-(1)/(λo)))) (d) λe = √((hλ₀)/(2mc))

Answer: (c). Einstein: K = hc((1)/(λi) - (1)/(λo)) λe = (h)/(√(2mK)) = (h)/(√(2mhc((1)/(λi)-(1)/(λo)))) λe = √((h)/(2mc((1)/(λi)-(1)/(λo)))) So the answer is (c).

A photoemissive substance is illuminated with a radiation of… | JEE Main 2025 (7 Apr, Shift 2)

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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 $\lambda_o$ . Expression for de-Broglie wavelength is given by:

(m: mass of the electron, h: Planck's constant and c: speed of light)

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Answer: (c)

Einstein: $K = hc\left(\dfrac{1}{\lambda_i} - \dfrac{1}{\lambda_o}\right)$
$\lambda_e = \dfrac{h}{\sqrt{2mK}} = \dfrac{h}{\sqrt{2mhc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}$
$\lambda_e = \sqrt{\dfrac{h}{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}$
So the answer is (c).

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