JEE Physics 2023 Question with Solution

Given: Power of source = $15 \, \text{kW} = 15 \times 10^3 \, \text{W}$ and number of photons emitted per second = $10^{16}$.

Find: The part of the electromagnetic spectrum to which the radiation belongs.

Energy emitted per second is the power, so energy of one photon is

$$E = \frac{15 \times 10^3}{10^{16}} = 15 \times 10^{-13} \, \text{J}$$

Using

$$E = h\nu$$

with $h = 6 \times 10^{-34} \, \text{J s}$, we get

$$\nu = \frac{E}{h} = \frac{15 \times 10^{-13}}{6 \times 10^{-34}} = 2.5 \times 10^{21} \, \text{Hz}$$

This frequency lies in the gamma ray region of the electromagnetic spectrum.

Therefore, the correct option is C.

Given: The source emits $10^{16}$ photons each second and has total power $15 \, \text{kW}$.

Find: Which spectral region corresponds to the radiation.

First find the energy associated with each photon from total power divided by photons emitted per second:

$$\text{Energy per photon} = \frac{\text{Power}}{\text{photons per second}}$$

So,

$$E = \frac{15 \times 10^3}{10^{16}} = 15 \times 10^{-13} \, \text{J}$$

Now apply Planck's relation:

$$E = h\nu$$

Hence,

$$\nu = \frac{E}{h}$$

Substituting the values,

$$\nu = \frac{15 \times 10^{-13}}{6 \times 10^{-34}} = 2.5 \times 10^{21} \, \text{Hz}$$

Since gamma rays have extremely high frequencies, and this value is greater than $10^{19} \, \text{Hz}$, the radiation is gamma rays.

Therefore, the correct option is C.