Question

When a beam of 10.6 eV photons of intensity 2.0 W/$m^2$ falls on a platinum surface of area $1.0\times {10}^{-4}{m}^2$ and work function 5.6 eV, 0.53% of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second

• A. $6.25\times {10}^8$
• B. $1.25\times {10}^9$
• C. $1.25\times {10}^6$
• D. $6.25\times {10}^{11}$

Answer 1

Answer: (D)

$6.25\times {10}^{11}$

Incident energy
$E=10.6eV=10.6\times (1.6\times {10}^{-19})J=16.96\times {10}^{-19}J$

Given : $\frac{Energyincident}{area\times time}=2W/m^2$

$\therefore \frac{Numberofincidentphotons}{area\times time}=\frac{2}{16.96\times {10}^{-19}}=1.18\times {10}^{18}$

$\therefore \frac{Incidentphotons}{time}=(1.18\times {10}^8)\times area$

$=1.18\times {10}^{18}\times (1.0\times {10}^{-4})=1.18\times {10}^{14}$

$\therefore \frac{Numberofphotoelectrons}{time}=\left(\frac{0.53}{100}\right)\times (1.18\times {10}^{14})$

or $n=6.25\times {10}^{11}$