Question 359: Dual Nature of Matter and Radiation

Question

When a beam of 10.6 eV photons of intensity $2.0 \mathrm{~W} / \mathrm{m}^2$ falls on a platinum surface of area $1.0 	imes 10^{-4} \mathrm{~m}^2$ and work function $5.6 \mathrm{eV} .0 .53 \%$ of the incident photons eject photoelectrons. The number of photoelectrons emitted per second is

Accepted Answer

Energy of incident photons,
$$ \begin{aligned} E_i & =10.6 \mathrm{eV} \ & =10.6 	imes 1.6 	imes 10^{-19} \mathrm{~J} \ & =16.96 	imes 10^{-19} \mathrm{~J} \end{aligned} $$
Energy incident per unit area per unit time (intensity) $=2 \mathrm{~J}$
Ã¢ÂÂ´ Number of photons incident on unit area in unit time
$$ =\frac{2}{16.96 	imes 10^{-19}}=1.18 	imes 10^{18} $$
Therefore, number of photons incident per unit time on given area
$$ \begin{aligned} & \left(1.0 	imes 10^{-4} \mathrm{~m}^2ight) \ & \quad=\left(1.18 	imes 10^{18}ight)\left(1.0 	imes 10^{-4}ight) \ & \quad=1.18 	imes 10^{14} \end{aligned} $$
But only $0.53 \%$ of incident photons emit photoelectrons
Ã¢ÂÂ´ Number of photoelectrons emitted per second ( $n$ )
$$ \begin{aligned} & n=\left(\frac{0.53}{100}ight)\left(1.18 	imes 10^{14}ight) \ & n=6.25 	imes 10^{11} \end{aligned} $$

Answer

$625 	imes 10^{11}$

Answer

$425 	imes 10^{10}$

Answer

$625 	imes 10^{10}$

Answer

$425 	imes 10^{11}$

Dual Nature of Matter and Radiation

When a beam of 10.6 eV photons of intensity 2.0 \mathrm{~W} / \mathrm{m}^2 falls on a platinum surface of area 1.0 \times 10^{-4} \mathrm{~m}^2 and work function 5.6 \mathrm{eV} .0 .53 \% of the incident photons eject photoelectrons. The number of photoelectrons emitted per second is

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