Question

A small metal plate (work function$=2eV$) is placed at a distance of $2m$ from a monochromatic light source of wavelength $4.8\times {10}^{-7}$ m and power $1.0$ watt. The light falls normally on the plate. The number of photons striking the metal plate per second per unit area will be

• A. $4.82\times {10}^{12}$
• B. $4.82\times {10}^{14}$
• C. $4.82\times {10}^{16}$
• D. $4.82\times {10}^{18}$

Answer 1

Answer: (C)

$4.82\times {10}^{16}$

Power of source $P=1.0$ watt

Distance of metal plate from source $r=2m$

Thus intensity of incident radiation at metal plate $I=\frac{P}{4\pi r^2}$

$\therefore$ $I=\frac{1.0}{4\pi (2{)}^2}=0.0199$ $W{m}^{-2}$

Wavelength of source of light $\lambda =4.8\times {10}^{-7}m$

Thus energy of incident photon $E=\frac{hc}{\lambda }$

$\therefore$ $E=\frac{(6.626\times {10}^{-34})(3\times {10}^8)}{4.8\times {10}^{-7}}=4.14\times {10}^{-19}J$

Number of photons striking per second per unit area $N=\frac{I}{E}$

$\therefore$ $N=\frac{0.0199}{4.14\times {10}^{-19}}=4.82\times {10}^{16}$ photons per second per unit area.