When light of wavelength ' $\lambda$ ' is incident on photosensitive surface, photons of power 'P' are…

When light of wavelength ' $\lambda$ ' is incident on photosensitive surface, photons of power 'P' are emitted. The number of photons (n) emitted in 't' second is (h= Planck's constant, $\mathrm{c}=$ velocity of light in vacuum)
  1. $\frac{h \mathrm{C}}{\mathrm{P} \lambda t}$
  2. $\frac{P \lambda t}{h_{C}}$
  3. $\frac{\mathrm{P} \lambda}{\operatorname{htc}}$
  4. $\frac{h \mathrm{P}}{\lambda t \mathbf{C}}$

Solution

Energy of each photon $=\frac{h c}{\lambda}$ If $n$ photons are emitted in time $t$ then the energy emitted in time $t$ is $\frac{\text { nhc }}{\lambda}$ $\begin{aligned} \text { Power P } &=\text { Energy emitted per second } \\ &=\frac{\mathrm{nhc}}{\lambda \mathrm{t}} \\ \therefore \mathrm{n} &=\frac{\mathrm{P} \lambda \mathrm{t}}{\mathrm{hc}} \end{aligned}$ .

Asked in: MHT CET 2020 (14 Oct Shift 2)

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