One milliwatt of light of wavelength $λ = 4560 ˚ A$ is incident on a cesium metal surface. Calculate the electrons liberaated. Assume a quantum efficiency of $η = 0.5 %$ . [work function for cesium $= 1.89 e V$ ] Take $h c = 12400 e V - ˚ A$

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Solution

We know that

$E = h v = \frac{h c}{λ} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{4560 \times 10^{10} J} = 4.34 \times 10^{- 19} J$

$1 m w$ of light energy is equivalent to $\frac{10^{- 3}}{4.34 \times 10^{- 19}} = 2.3 \times 10^{15} P h o t o n s / s$

The quantum efficiency is $0.5$

This means that only $0.5$ of these photons release photoelectrons.
So, the number of electrons released from the surface per second

$= 2.3 \times 10^{15} \times \frac{0.5}{100} = 1.15 \times 10^{13} e l e c t r o n s / s$
The electrons curcent $= 1.15 \times 10^{13} \times 1.6 \times 10^{- 18} A$

$= 1.84 μ A$

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One milliwatt of light of wavelength λ=4560 ˚A is incident on a cesium metal surface. Calculate the electrons liberaated. Assume a quantum efficiency of η=0.5%. [work function for cesium=1.89 eV] Take hc=12400 eV−˚A

One milliwatt of light of wavelength $λ = 4560 ˚ A$ is incident on a cesium metal surface. Calculate the electrons liberaated. Assume a quantum efficiency of $η = 0.5 %$ . [work function for cesium $= 1.89 e V$ ] Take $h c = 12400 e V - ˚ A$