Two sources of light emit X-rays of wavelengt | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The power $P$ of a light source is given by the formula $P = \frac{n E}{t} = \frac{n h c}{\lambda t}$,where $n$ is the number of photons emitted i

Vedclass · Accepted Answer

$1/500$

Vedclass · Answer

(A) The power $P$ of a light source is given by the formula $P = \frac{n E}{t} = \frac{n h c}{\lambda t}$,where $n$ is the number of photons emitted in time $t$,$h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.Let $N = n/t$ be the number of photons emitted per second.Then $P = N \frac{h c}{\lambda}$,which implies $N = \frac{P \lambda}{h c}$.Since both sources have the same power $P$,the ratio of the number of photons per second is $\frac{N_1}{N_2} = \frac{\lambda_1}{\lambda_2}$.Given $\lambda_1 = 1 \ nm$ and $\lambda_2 = 500 \ nm$,the ratio is $\frac{N_1}{N_2} = \frac{1}{500}$.Thus,the correct option is $A$.

Similar Questions

Which of the following will have the least value of $\frac{q}{m}$?

The wavelength of a $1 \;keV$ photon is $1.24 \times 10^{-9} \;m$. What is the frequency of a $1 \;MeV$ photon?

The energy of a photon with a wavelength of $6.20 \times 10^{-5} \ cm$ is ....... $eV$.

How is the intensity of radiation determined?

The energy of a photon with a wavelength of $\lambda = 4000 \ \mathring{A}$ is equal to ........ Joules.

Vedclass Test Series

Exam Paper Generator

Online Exam Module