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Question: A \(100\,W\) point source emits monochromatic light of wavelength \(6000\mathop A\limits^o .\) Calcu...

A 100 W100\,W point source emits monochromatic light of wavelength 6000Ao.6000\mathop A\limits^o . Calculate the total number of photons emitted by the source per second.
A. 5×10205 \times {10^{20}}
B. 8×10208 \times {10^{20}}
C. 6×10206 \times {10^{20}}
D. 3×10203 \times {10^{20}}

Explanation

Solution

In order to solve this question, we should know that In particle nature of light, light is made of particles called photons and monochromatic light is made up of single wavelength and here, we will use the energy of a photon and power formula and then will calculate the number of photons by given point source.

Formula Used:
Energy of a photon is given by,
E=hcλE = \dfrac{{hc}}{\lambda }
where hh is known as Planck’s constant has a definite value of h=6.62×10−34Jsh = 6.62 \times {10^{ - 34}}Js c,λc,\lambda are the speed of light and wavelength of light.
Power of nn number of photons per second is given by P=nEP = nE.

Complete step by step answer:
According to the question, we have given these following parameters.
λ=6000Ao=6×10−7m\lambda = 6000\mathop A\limits^o = 6 \times {10^{ - 7}}m wavelength of monochromatic light. Then using formula of energy we have,
E=hcλE = \dfrac{{hc}}{\lambda } where c=3×108ms−1c = 3 \times {10^8}m{s^{ - 1}}.
⇒E=6.62×10−34×3×1086×10−7\Rightarrow E = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{6 \times {{10}^{ - 7}}}}
⇒E=3.31×10−19J\Rightarrow E = 3.31 \times {10^{ - 19}}J

Now, let us assume n is the total number of photons emitted by 100W100W point source then we have,’
P=100WP = 100W
⇒E=3.31×10−19J\Rightarrow E = 3.31 \times {10^{ - 19}}J
then, using the formula P=nEP = nE on putting the values, we get
100=n(3.31×10−19)100 = n(3.31 \times {10^{ - 19}})
⇒n=13.31×1021\Rightarrow n = \dfrac{1}{{3.31}} \times {10^{21}}
∴n=3×1020\therefore n = 3 \times {10^{20}}
So, the total number of photons emitted by point source of power are n=3×1020n = 3 \times {10^{20}}

Hence, the correct option is D.

Note: It should be noted that, basic unit of conversion used while solving given question is 1A0=10−10m1\mathop A\limits^0 = {10^{ - 10}}m and in terms of the frequency of the light, the energy of the photon is given by E=hνE = h\nu and power is simply the ratio of energy over time but in given problem we have calculated the number of photons per second so time is taken as 1sec⁡.1\sec .