Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A laser diode emits light of wavelength \lambda=780 \mathrm{~nm} with a power of P = 20 mW. How many photons are emitted per second from the laser diode?

Option: 1

1.94 \times 10^{15} \mathrm{~s}^{-1}


Option: 2

2.45 \times 10^{15} s^{-1}


Option: 3

3.67 \times 10^{15} s^{-1}


Option: 4

3.67 \times 10^{15} s^{-1}


Answers (1)

best_answer

The energy of each photon can be calculated using the formula

E=\frac{h c}{\lambda} , where h is Planck's constant and c is the speed of light.

Substituting the values, we get:

\begin{aligned} E & =\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{780 \times 10^{-9} \mathrm{~m}}= 8.06 \times 10^{-20} \mathrm{~J} \end{aligned}

The number of photons emitted per second can be calculated using the formula n=\frac{P}{E} .

Substituting the values, we get:

n=\frac{20 \times 10^{-3} \mathrm{~W}}{8.06 \times 10^{-20} \mathrm{~J}}=2.48 \times 10^{15} \mathrm{~s}^{-1}

Rounding off to the correct number of significant figures, we get the answer as (2)

2.45 \times 10^{15} s^{-1}

Therefore, option (2) is the correct answer.

 

Posted by

sudhir.kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024: NTA debars 39 candidates for 3 years on account of using unfair means
anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

Latest Question