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MCQEasyJEE 2026Einstein's Equation

JEE Physics 2026 Question with Solution

Number of photons of equal energy emitted per second by a 6 mW6 \, \text{mW} laser source operating at wavelength 663 nm663 \, \text{nm} is _____. (Given: h=6.63×10−34 J⋅sh=6.63\times10^{-34} \, \text{J}\cdot\text{s} and c=3×108 m/sc=3\times10^8 \, \text{m/s})

  • A

    10×101510\times10^{15}

  • B

    5×10165\times10^{16}

  • C

    5×10155\times10^{15}

  • D

    2×10162\times10^{16}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Power of laser source is P=6×10−3 J/sP=6\times10^{-3} \, \text{J/s} and wavelength is λ=663×10−9 m\lambda=663\times10^{-9} \, \text{m}.

Find: Number of photons emitted per second.

Concept: The energy of a single photon is

E=hcλE=\frac{hc}{\lambda}

The power of the source gives the energy emitted per second.

Step 1: Convert given quantities

P=6 mW=6×10−3 J/sP=6~\text{mW}=6\times10^{-3}~\text{J/s} λ=663 nm=663×10−9 m\lambda=663~\text{nm}=663\times10^{-9}~\text{m}

Step 2: Energy of one photon

E=(6.63×10−34)(3×108)663×10−9≈3.0×10−19 JE=\frac{(6.63\times10^{-34})(3\times10^8)}{663\times10^{-9}} \approx3.0\times10^{-19}~\text{J}

Step 3: Number of photons emitted per second

N=PE=6×10−33.0×10−19=2×1016N=\frac{P}{E} =\frac{6\times10^{-3}}{3.0\times10^{-19}} =2\times10^{16}

Conclusion: Therefore, the number of photons emitted per second is 2×10162\times10^{16}. The correct option is D.

Energy per photon then divide power

Given: A laser of power 6 mW6 \, \text{mW} operates at wavelength 663 nm663 \, \text{nm}.

Find: Photon emission rate.

First compute the energy carried by one photon using

E=hcλE=\frac{hc}{\lambda}

Then divide the energy emitted each second by this value, because power is energy per unit time.

Substituting the given values,

E=(6.63×10−34)(3×108)663×10−9E=\frac{(6.63\times10^{-34})(3\times10^8)}{663\times10^{-9}}

This gives

E≈3.0×10−19 JE\approx3.0\times10^{-19}~\text{J}

Now use

N=PEN=\frac{P}{E}

with

P=6×10−3 J/sP=6\times10^{-3}~\text{J/s}

So,

N=6×10−33.0×10−19=2×1016N=\frac{6\times10^{-3}}{3.0\times10^{-19}}=2\times10^{16}

Conclusion: Hence the laser emits 2×10162\times10^{16} photons each second, so the correct option is D.

Common mistakes

  • Using the wavelength directly in nm\text{nm} without converting it to m\text{m}. This makes the photon energy incorrect by a factor of 10910^9. Always convert 663 nm663 \, \text{nm} to 663×10−9 m663\times10^{-9} \, \text{m} first.

  • Forgetting that power is energy emitted per second. The question asks for photons emitted per second, so after finding energy of one photon, divide PP by EE. Do not multiply PP and EE.

  • Not converting 6 mW6 \, \text{mW} into J/s\text{J/s} correctly. Since 1 mW=10−3 W1 \, \text{mW}=10^{-3} \, \text{W}, the power is 6×10−3 J/s6\times10^{-3} \, \text{J/s}, not 6 J/s6 \, \text{J/s}.

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