Question

Light rays of wavelengths $6000\stackrel{\text{o}}{A}$ and photon intensity $39.6watt/m^2$ 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 $=6.64\times {10}^{-34}J-s,$ 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 $=12\times {10}^{19}\times \frac{1}{100}$ $=12\times {10}^{17}$