Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A point source of light emits photons of energy 3.5  eV each. If the power of the source is 2.0 W, then the number of photons emitted per second is:  

A point source of light emits photons of energy 3.5  eV each. If the power of the source is 2.0 W, then the number of photons emitted per second is:

 

Option: 1

1.23 \times 10^{17}$ photons $/ \mathrm{s}


Option: 2

5.71 \times 10^{18}$ photons $/ \mathrm{s}


Option: 3

3.57 \times 10^{18}$ photons $/ \mathrm{s}


Option: 4

8.33 \times 10^{22}$ photons $/ \mathrm{s}


Answers (1)

best_answer

The energy of each photon is given as E=3.5 \mathrm{eV}. We need to convert this to Joules to use the formula for the number of photons emitted per second.
1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}
Therefore, the energy of each photon in Joules is E=3.5 \times 1.6 \times 10^{-19}=5.6 \times 10^{-19} \mathrm{~J}.


The power of the source is given as P=2.0 \mathrm{~W}.


Using the formula for the number of photons emitted per second, we get:


\begin{aligned} & n=\frac{P}{E}=\frac{2.0}{5.6 \times 10^{-19}}=3.57 \times & 10^{18} \text { photons } / s \end{aligned}

Posted by

Sayak

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