Question

Light rays of wavelengths $ 6000\,\overset{\text{o}}{\mathop{\text{A}}}\, $ and photon intensity $ 39.6\text{ }watt/{{m}^{2}} $ is incident on a metal surface. If only one percent of photons incident on the surface emit photoelectron, then the number of electrons emitted per second per unit area from the surface will be: (Planck constant $ =\text{ }6.64\text{ }\times \text{ }{{10}^{-34}}J-s, $ Velocity of light $ =3\times {{10}^{8}}m/s $ )

• A. $ 12\times {{10}^{18}} $
• B. $ 10\times {{10}^{18}} $
• C. $ 12\times {{10}^{17}} $
• D. $ 12\times {{10}^{16}} $

Answer 1

Answer: (C)

Energy of each photon is $ E=\frac{hc}{\lambda } $ $ =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6000\times {{10}^{-10}}} $ $ =3.3\times {{10}^{-19}}J $ Now, power of source $ P=IA=39.6\times 1=39.6\,\text{watt} $ Number of photons emitted per second $ =\frac{p}{E}=\frac{39.6}{3.3\times {{10}^{-9}}} $ $ =12\times {{10}^{19}} $ Only 1% of photon eject elements, so no. of photoelectrons $ =12\times {{10}^{19}}\times \frac{1}{100} $ $ =12\times {{10}^{17}} $