One milliwatt of light of wavelength $λ = 4560 o A$ is incident on a cesium metal surface. Calculate the electron current liberated. 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

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

1 mW 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 photo electrons.

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 electron current $= 1.15 \times 10^{13} \times 1.6 \times 10^{- 19} A = 1.84 μ A$

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One milliwatt of light of wavelength λ=4560 oA is incident on a cesium metal surface. Calculate the electron current liberated. 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 $λ = 4560 o A$ is incident on a cesium metal surface. Calculate the electron current liberated. Assume a quantum efficiency of $η = 0.5 %$ . [Work function for cesium $= 1.89 e V$ ] Take $h c = 12400 e V - A$ .