Question

Ultraviolet light of wavelength $300nm$ and intensity $1.0W/m^2$ falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons, then the number of photo electrons emitted from an area of $1.0{cm}^2$ of the surface is nearly

• A. $19.61\times {10}^{12}{s}^{-1}$
• B. $4.12\times {10}^{12}{s}^{-1}$
• C. $1.51\times {10}^{12}{s}^{-1}$
• D. $2.13\times {10}^{12}{s}^{-1}$

Answer 1

The correct option is C $1.51\times {10}^{12}{s}^{-1}$

Energy incident over $1m^2:$ $=1\times {10}^{-4}J$

Energy required to produce photoelectron

$=1\times {10}^{-4}\times {10}^{-2}={10}^{-6}J$

Number of photoelectrons ejected $=$ Number of photons which can produce photoelectrons $=$ (energy required for producing electron)/(energy of photon) $=\frac{{10}^{-6}}{h\left(c/\lambda \right)}=\frac{{10}^{-6}\times 300\times {10}^{-9}}{6.6\times {10}^{-34}\times 3\times {10}^8}=1.51\times {10}^{12}{s}^{-1}$