Question

Ultraviolet light of wavelength 350 nm and intensity $ 1.00\text{ }W{{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 $ 1\text{ }c{{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=3.55\,eV $ $ =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}} $