A photo-emissive substance is illuminated with a radiation of wavelength λᵢ so that it releases electrons with…

Question

A **photo-emissive 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 given by :

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

Accepted Answer

A. $$\lambda_e = \sqrt{\frac{h}{2mc \left( \frac{1}{\lambda_i} - \frac{1}{\lambda_0} \right)}}$$

Answer

B. $$\lambda_e = \sqrt{\frac{h \lambda_0}{2mc}}$$

Answer

C. $$\lambda_e = \frac{h}{\sqrt{2mc \left( \frac{1}{\lambda_i} - \frac{1}{\lambda_0} \right)}}$$

Answer

D. $$\lambda_e = \sqrt{\frac{h \lambda_i}{2mc}}$$

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