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Question: A beam of light has two wavelengths 4972 \(\mathop {\text{A}}\limits^0 \) and 6216 \(\mathop {\text{...

A beam of light has two wavelengths 4972 A0\mathop {\text{A}}\limits^0 and 6216 A0\mathop {\text{A}}\limits^0 with a total intensity of 3.6×10−3W/m23.6 \times {10^{ - 3}}W/{m^2} equally distributed among two wavelengths. The beam falls normally on an area of 1cm21c{m^2} of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo electrons liberated in 2 s is approximately:
A. 6×1011 B. 9×1011 C. 11×1011 D. 15×1011 \begin{aligned} &{\text{A}}{\text{. 6}} \times {\text{1}}{{\text{0}}^{11}} \\\ &{\text{B}}{\text{. 9}} \times {\text{1}}{{\text{0}}^{11}} \\\ &{\text{C}}{\text{. 11}} \times {\text{1}}{{\text{0}}^{11}} \\\ &{\text{D}}{\text{. 15}} \times {\text{1}}{{\text{0}}^{11}} \\\ \end{aligned}

Explanation

Solution

First we need to check which of the two wavelengths will produce photoelectrons which can be done by using the work function. Then using the appropriate wavelength, we can get the number of photoelectrons by using the intensity of light.
Formula Used: λ=hcϕ\lambda = \dfrac{{hc}}{\phi }
Complete step-by-step solution
The work function of the given metallic surface is given as
ϕ=2.3eV\phi = 2.3eV
Therefore, the threshold wavelength for the given work function can be calculated in the following way.
λ=hcϕ=6.6×10−34×3×1082.3×1.6×10−19=5.38×10−7m=5380A0\lambda = \dfrac{{hc}}{\phi } = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2.3 \times 1.6 \times {{10}^{ - 19}}}} = 5.38 \times {10^{ - 7}}m = 5380\mathop {\text{A}}\limits^0
Therefore, out of the two wavelengths, only 6216 A0\mathop {\text{A}}\limits^0 wavelength will be able to emit electrons from the given metallic surface.
We are given that the total intensity is equally distributed among the two wavelengths. Therefore, the intensity of each given wavelength is equal to
I=12×3.6×10−3W/m2=1.8×10−3W/m2I = \dfrac{1}{2} \times 3.6 \times {10^{ - 3}}W/{m^2} = 1.8 \times {10^{ - 3}}W/{m^2}
Now the number of photons incident on the metallic surface per second of the wavelength 6216 A0\mathop {\text{A}}\limits^0 can be calculated by dividing the intensity of light by the energy of the light. Therefore, we have
1.8×10−3=nhc6216×10−10 ⇒n=1.8×10−3×6216×10−106.6×10−34×3×108=565.09×1013 \begin{aligned} & 1.8 \times {10^{ - 3}} = \dfrac{{nhc}}{{6216 \times {{10}^{ - 10}}}} \\\ &\Rightarrow n = \dfrac{{1.8 \times {{10}^{ - 3}} \times 6216 \times {{10}^{ - 10}}}}{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} = 565.09 \times {10^{13}} \\\ \end{aligned}
Now finally the number of photo electrons liberated in 2 s from an area of 1cm21c{m^2} is given as
ne=nAt=565.09×1013×10−4×2=1130.18×109≃11×1011{n_e} = nAt = 565.09 \times {10^{13}} \times {10^{ - 4}} \times 2 = 1130.18 \times {10^9} \simeq 11 \times {10^{11}}
This is the required answer. Hence, the correct answer is option C.

Note: The number of photo-electrons emitted by a surface depends on the intensity of the light. The intensity of a given light is equal to the number of photons emitted per second and per unit area from a given source. The energy of the emitted photoelectrons does not depend on the intensity of the radiation, it depends on the energy of the radiation which in turn depends on the frequency or wavelength of light. The higher the frequency of radiation, the higher is its energy while the lower the wavelength of light, the higher is the energy of that light.