Question

Ultraviolet light of wavelength 350 nm and intensity $1.00W{m}^{-2}$ is incident on a potassium surface. If 0.5% of the photons participate in ejecting the photoelectrons, how many photo electrons, are emitted per second, if the potassium surface has an area of $1c{m}^2$ ?

• A. $1.76\times {10}^{18}$ photoelectrons/s
• B. $1.76\times {10}^{14}$ photoelectrons/s
• C. $8.8\times {10}^{11}$ photoelectrons/s
• D. The value of work function is required to complete the value of emitted photoelectrons/s

Answer 1

Answer: (C)

Energy of photon, $E=\frac{hc}{\lambda }$ $=\frac{1242}{350}eV=3.55eV$ $=5.68\times {10}^{-19}J$ Let n photons, per unit area per unit time are reaching the potassium surface, then $n=\frac{1.00}{5.68\times {10}^{-19}}$ $=1.76\times {10}^{18}$ So, number of photons received by potassium surface per unit time is, $n\times$ Area of potassium surface $=1.76\times {10}^{18}\times 1\times {10}^{-4}$ $=1.76\times {10}^{14}$ Required number of photoelectrons emitted per unit time $=1.76\times {10}^{14}\times \frac{0.5}{100}$ $=8.8\times {10}^{11}$