Question

Light of wavelength $5000\mathring{A}$ and intensity $3.96\times {10}^{-3}W/c{m}^2$ is incident on the surface of a photosensitive material. If 1 percent of incident photons only emit photoelectrons, then the number of electrons emitted per unit area from the surface will be

• A. ${10}^{16}$
• B. ${10}^{18}$
• C. ${10}^{20}$
• D. ${10}^{22}$

Answer 1

Answer: (B)

${10}^{18}$

Intensity $=3.96\ast {10}^{-3}W/c{m}^2=3.96\ast {10}^{+1}W/m^2$

Energy of one photon $=\frac{hc}{\lambda }=\frac{6.64\ast {10}^{-34}\ast 3\ast {10}^8}{5000\ast {10}^{-10}}=3.96\ast {10}^{-19}J.$

i.e. number of photons incident is equal to

$\frac{3.96\ast {10}^{+1}}{3.96\ast {10}^{-19}}={10}^{20}$ photons.

But only 1% eject photo electrons.
Hence number of electrons ejected per unit area per second $={10}^{18}$ electrons.

So, the answer is option (B).