A 2 mW laser operates at a wavelength of 500 nm. The number of photons that will be emitted per second is:

[Given Planck’s constant h = 6.6 × 10This question was previously asked in

JEE Mains Previous Paper 1 (Held On: 10 Apr 2019 Shift 2)

Option 1 : 5 × 10^{15}

Junior Executive (ATC) Official Paper 1: Held on Nov 2018 - Shift 1

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__Concept__:

**Energy of photon (E):**

A photon is a quantum of EM radiation. Its energy is given by E = hf and is related to the frequency f and wavelength λ of the radiation by

\({\rm{E}} = \frac{{{\rm{hc}}}}{{\rm{\lambda }}}\)

Where, E is the energy of a single photon and c is the speed of light. When working with small systems, energy in eV is often useful. Note that Planck’s constant in these units is h = 4.14 × 10^{−15} eV · s.

Number of photons of wavelength λ emitted in t second from laser of power P is given by the following formula

\({\rm{P}} = \frac{{{\rm{nhc}}}}{{{\rm{t\lambda }}}}\)

\({\rm{n}} = \frac{{{\rm{Pt\lambda }}}}{{{\rm{hc}}}}\)

__Calculation__:

Given:

Power of the laser, P = 2 mW

Wavelength, λ = 500 nm

Planck’s constant, h = 6.6 × 10^{-34} Js

Speed of light, c = 3.0 × 10^{8} m/s

Number of photons of wavelength λ emitted in t second from laser of power P is given by:

\({\rm{n}} = \frac{{{\rm{Pt\lambda }}}}{{{\rm{hc}}}}\)

\({\rm{n}} = \frac{{2 \times {{10}^{ - 3}} \times 1 \times 500 \times {{10}^{ - 9}}}}{{6.6 \times {{10}^{ - 34}} \times 3.0 \times {{10}^8}}}\)

\({\rm{n}} = \frac{{2 \times 5 \times {{10}^{ - 10}}}}{{6.6 \times 3 \times {{10}^{ - 26}}}}\)

\({\rm{n}} = \frac{{{{10}^{ - 9}} \times {{10}^{26}}}}{{19.8}} = \frac{{{{10}^{17}}}}{{19.8}}\)

\({\rm{n}} = \frac{{{{10}^{17}}}}{{1.98 \times {{10}^1}}} = \frac{{{{10}^{16}}}}{2} \)

n = 5 × 10^{15}