A beam of light has three $λ, 4144 Å, 4972 Å$ and $6216 Å$ with a total intensity of $3.6 \times 10^{- 3} {Wm}^{- 2}$ equally distributed amongst the three $λ$ . The beam falls normally on an area $1.0 {cm}^{2}$ of a clean metallic surface of work function $2.3 eV$ . Assume that there is no loss of light by reflection etc. Calculate the no. of photoelectrons emitted in 2 sec, in scientific notation, $x \times 10^{y}$ find the value of y.

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Detailed Solution

The threshold wavelength will be given by

$\begin{aligned} λ = \frac{hc}{ω_{0}} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{2.3 \times 1.6 \times 10^{- 19}} \\ = 5.396 \times 10^{- 7} m = 5396 Å \end{aligned}$

Hence light of wavelength 6216 A will not emit photoelectrons from the metal surface. Since the total intensity of light is equally distributed among the wavelengths, the intensity which each wavelength will have is

$\frac{3.6 \times 10^{- 3}}{3} = 1.2 \times 10^{- 3} {Wm}^{- 2}$