Question

Light of intensity $I$ & frequency $\nu$ is incident perpendicularly on a metal surface of unit square area. The quantum efficiency, defined as the ratio of no. of ejected electrons to the no. of incident photons, is $x$ and the work function of the metal is $\varphi$.Then:

• A. no. of photons striking the surface per unit time is $\frac{I}{hv}$
• B. no. of electrons coming out of the surface per unit time is $\frac{xI}{hv}$
• C. the maximum kinetic energy of the ejected electron is $xhv-\varphi$
• D. the power of light used in liberating electrons is $\frac{xI\varphi }{hv}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A no. of photons striking the surface per unit time is $\frac{I}{hv}$
B no. of electrons coming out of the surface per unit time is $\frac{xI}{hv}$
D the power of light used in liberating electrons is $\frac{xI\varphi }{hv}$

Intensity of light is $I=nhv$ $(A=1m^2)$
So, no. of photons striking the surface per unit time is $n=\frac{I}{hv}$.
Since quantum efficiency is $x$ hence no. of electrons coming out of the surface per unit time is $\frac{xI}{hv}$.
By photoelectric equation $K_{max}=hv-\varphi$
Since minimum energy required to bring out an electron from metal surface is equal to the work function of the metal hence the power of light used in liberating electrons is $=$ no. of photoelectrons emitted $\times$ work function $=\frac{xI\varphi }{hv}$.