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Light of Intensity $10^{-5}W/m^2$ falls on a sodium photo-cell of surface area $2\: cm^2$ .Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave – picture of radiation.( Effective atomic area of a sodium atom = $10^{-20}m^2$ )

$\begin {array} {1 1} (a)\;0.607\: years & \quad (b)\;0.507\: years \\ (c)\;9\: months & \quad (d)\;1208 \: days \end {array}$


1 Answer

Ans : (b)
$I = 10^{-5}W/m^2\: \: \: \: \: A= 2cm^2 = 2 \times 10^{-4}m^2$
$P = I \times A$
$= 10^{-5} \times 2 \times 10^{-4}$
$= 2 \times10^{-9} J$
No. of layers of sodium that absorbs the incident energy, $n = 5$
Effective atomic area of a sodium atom, $A_e = 10^{-20}m^2$
Hence,no. of conduction electrons in $n$ layers,
$n’ = n \times \large\frac{A}{A_e}$
$= \large\frac{5 \times 2 \times 10^{-4}}{ 10^{-20}} = 10^{17}$
Hence, the amount of energy absorbed per electron, $E = \large\frac{P}{ n’}$
$= \large\frac{2 \times 10^{-9}}{10^{17}} = 2 \times 10^{-26} J/s$
Time required for photoelectric emission, $t = \large\frac{\phi_{\circ}}{E}$
$= \large\frac{3.2 \times 10^{-19}}{ 2 \times 10^{-26} } = 1.6 \times 10^7s ≈ 0.507$ years


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