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One milliwatt of light of wavelength of 4560 $A^{\circ}$ is incident on a cesium surface. Calculate the electron current liberated. Assume a quantum efficiency of 0.5%.

$\begin {array} {1 1} (a)\;0.184 \: \mu A & \quad (b)\;184 \: \mu Aabc \\ (c)\;Can’t\: be\: determined & \quad (d)\;1.84 \mu A \end {array}$


1 Answer

Ans : (d)
$E=hv = \large\frac{hc}{\lambda} = \large\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{ 4560 \times 10^{10}J}$
$= 4.34 \times 10^{-19} J$
$1\: mW$ of light energy is equivalent to $\large\frac{ 10^{-3}}{(4.34 \times 10^{-19} )} = 2.3 \times 10^{15}$ photons/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 0.5/100$
$= 1.15 \times 10^{13 }$ electrons/s
The electron current = $1.15 \times 10^{13} \times 1.6 \times 10^{-19} A = 1.84 \: \mu A$


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