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Question: A \( 25 \) watt bulb emits monochromatic yellow light of a wavelength \( 0.57\mu m \) . What will be...

A 2525 watt bulb emits monochromatic yellow light of a wavelength 0.57μm0.57\mu m . What will be the rate of emission of quanta per second?
(A) 7.17×10107.17 \times {10^{10}}
(B) 7.17×10197.17 \times {10^{19}}
(C) 71.7×101971.7 \times {10^{19}}
(D) 71.7×10−1071.7 \times {10^{ - 10}}

Explanation

Solution

The given problem is very easy if we know the basics of quantum mechanics. We can easily calculate the rate of emission of quanta per second if we know the energy of the photon which is emitted. If we have the energy of photons and power, the rate of emission can be easily determined.

Formula Used
Energy of photon (E)\left( E \right)
E=hcλE = \dfrac{{hc}}{\lambda }
Where hh is Planck’s constant (h=6.626×10−34Js)\left( {h = 6.626 \times {{10}^{ - 34}}Js} \right) ,
cc is the speed of light (c=3×108m/s)\left( {c = 3 \times {{10}^8}m/s} \right)
λ\lambda is the wavelength of monochromatic light.
Rate of emission (R)\left( R \right)
R=PER = \dfrac{P}{E}
Where PP is the power and,
EE is the energy of the photon.

Complete step by step solution
First, we will understand some terms given in the question like wavelength and rate of emission of quanta per second.
Wavelength is defined as the ratio of the speed of light to the frequency. The rate of emission of quanta per second simply means the number of photons emitted per second.
First, we will write the given quantities from the question. We know that 1J=1W,1μm=10−6m1J = 1W,1\mu m = {10^{ - 6}}m so after converting the units we get,
(P=25W=25J,λ=0.57×10−6m)(P = 25W = 25J,\lambda = 0.57 \times {10^{ - 6}}m)
Now we will calculate the energy of one photon using the formula of the energy of the photon. Consider the values (h=6.626×10−34Js)\left( {h = 6.626 \times {{10}^{ - 34}}Js} \right) (c=3×108m/s)\left( {c = 3 \times {{10}^8}m/s} \right) and wavelength (λ=0.57×10−6m)(\lambda = 0.57 \times {10^{ - 6}}m) . Now substitute the values in the formula,
E=hcλE = \dfrac{{hc}}{\lambda }
⇒E=6.626×10−34Js×3×108m/s0.57×10−6m\Rightarrow E = \dfrac{{6.626 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m/s}}{{0.57 \times {{10}^{ - 6}}m}}
⇒E=3.487×10−19J\Rightarrow E = 3.487 \times {10^{ - 19}}J
Hence, the energy of one photon is E=3.487×10−19JE = 3.487 \times {10^{ - 19}}J
Now we will find the rate of emission of quanta per second using the formula of the rate of emission. We have E=3.487×10−19JE = 3.487 \times {10^{ - 19}}J and P=25JP = 25J . Substitute the value in the formula,
R=PER = \dfrac{P}{E}
⇒R=25J3.487×10−19\Rightarrow R = \dfrac{{25J}}{{3.487 \times {{10}^{ - 19}}}}
⇒R=7.17×1019\Rightarrow R = 7.17 \times {10^{19}}
Hence, the total number of photons emitted per second is R=7.17×1019R = 7.17 \times {10^{19}} .
Therefore, the correct option is (B).

Note:
The energy of a photon is energy carried by a single photon. It is directly proportional to the photon’s electromagnetic frequency and inversely proportional to the wavelength.
So the higher the photon’s frequency, the higher its energy.