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 x ${10}^{-4}{m}^2$ and work function 5.6 eV, 53% of the incident photons eject photo electrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies (in eV).
Take 1 eV = 1.6 x ${10}^{-19}$ J

• A. 6.25 x ${10}^{11}$, 0, 5
• B. 6.25 x ${10}^{11}$, 0, 2
• C. 6.25 x ${10}^{11}$, 4, 5
• D. 6.25 x ${10}^{12}$, 2, 5

Answer 1

The correct option is A 6.25 x ${10}^{11}$, 0, 5

Number of photo-electrons emitted per second $=\frac{\left(Intensity\right)\left(Area\right)}{\left(Energyofeachphoton\right)}\times \frac{0.53}{100}$

$=\frac{\left(2.0\right)(1\times {10}^{-4})}{10.6\times 1.6\times {10}^{-19})}\times \frac{0.53}{100}=6.25\times {10}^{11}$

Minimum kinetic energy of photo-electrons, $K_{min}=0$ and

maximum kinetic energy is, $K_{max}=E-W=(10.6-5.6)eV=5eV$