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 photoelectrons emitted from an area of $ 1.0c{m^2}\; $ 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

Hint: Ultraviolet rays are a form of electromagnetic radiation that makes black-light posters glow and are also responsible for summer tans. The intensity of the light is given so from the data the Power can be calculated as it is the ratio of the intensity to the area which is also given. The wavelength of the Ultraviolet light is also known here. The number of photons is the ratio of the power to the energy. try to use the energy equation in terms of wavelength and the required terms can be calculated.

Complete Step By Step Answer:
Given that wavelength of ultraviolet $ \lambda = 300nm $
Intensity $ I = 1.0W/{m^2}\; $
Area $ A = 1.0c{m^2}\; = \dfrac{1}{{{{10}^4}}}{m^2} $
The formula for power, $ P = \dfrac{I}{A} $
Here, $ I $ is the intensity
 $ A $ is the area.
Substituting the values we get,
 $ P = \dfrac{1}{{\dfrac{1}{{{{10}^4}}}}} $
 $ P = {10^4}W $
The number of photoelectrons can be calculated using the formula,
 $ N = \dfrac{P}{E} $
Where,
 $ E = \dfrac{{hc}}{\lambda } $
Here, $ N $ is the number of photons per unit $ {m^2} $ area per unit time.
 $ \lambda $ is the wavelength of the ultraviolet rays.
 $ h $ is the Planck’s constant in $ J.s $ . $ h = 6.6 \times {10^{ - 34}}J.s $
 $ c $ is the velocity of light in $ m/s $
So,
 $ N = \dfrac{P}{{\dfrac{{hc}}{\lambda }}} = \dfrac{{P\lambda }}{{hc}} $
Therefore substituting all the values,
 $ N = \dfrac{{{{10}^4} \times 300 \times {{10}^{ - 9}}}}{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} $
Solving this we will get,
 $ N = 1.51 \times {10^{12}}{s^{ - 1}} $
Therefore the correct option is C.

Note:
Intensity is the amount of energy the wave exerts per unit of time across a unit surface area. This is equal to the energy density that is multiplied by the speed of the wave. It is usually measured in units of watts per square meter. Intensity will always depend on the wave’s strength and amplitude of a wave. Intensity is represented as $ I $ .