Light of wavelength lambda falls on a metal h | vedclass

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(C) The energy of an incident photon is given by $E = \frac{hc}{\lambda}$.For the photoelectric effect to occur,the energy of the incident photon must

Vedclass · Accepted Answer

$\lambda < \lambda_0$

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(C) The energy of an incident photon is given by $E = \frac{hc}{\lambda}$.For the photoelectric effect to occur,the energy of the incident photon must be greater than or equal to the work function $(\Phi)$ of the metal.The work function is given as $\Phi = \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.Therefore,the condition for the photoelectric effect is $E \geqslant \Phi$.Substituting the expressions,we get $\frac{hc}{\lambda} \geqslant \frac{hc}{\lambda_0}$.Dividing both sides by $hc$,we get $\frac{1}{\lambda} \geqslant \frac{1}{\lambda_0}$.Taking the reciprocal,the inequality sign reverses: $\lambda \leqslant \lambda_0$.Since the standard condition for emission is that the incident wavelength must be less than or equal to the threshold wavelength,the correct option is $\lambda < \lambda_0$ (or $\lambda \leqslant \lambda_0$).

Light of wavelength $\lambda$ falls on a metal having work function $\frac{hc}{\lambda_0}$. Photoelectric effect will take place only if ($\lambda_0$ is the threshold wavelength).

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