Question

If a source of electromagnetic radiation having power $15 kW$ produces $10^{16}$ photons per second, the radiation belongs to a part of spectrum is(Take Planck constant $h =6 \times 10^{-34} Js$ )

• A. Micro waves
• B. Radio waves
• C. Gamma rays
• D. Ultraviolet rays

Hint:

Gamma rays have frequencies greater than \( 10^{19} \, \text{Hz} \), which are much higher than visible light or radio waves.

Answer 1

The Correct Option is C

$\text{Energy of one photon}=\frac{\text{Power}}{\text{Photon frequency}}$
$E=hv=\frac{15\times 10^3}{10^{16}}$
$v=\frac{15\times 10^{-13}}{6\times 10^{-34}}=2.5\times 10^{21}$
So gamma Rays. Option 3

Answer 2

The energy of one photon is given by the formula:

\[ E = \frac{\text{Power}}{\text{Photon frequency}} \]

We know that:

\[ E = h \nu \quad (\text{where } h = 6 \times 10^{-34} \, \text{Js}) \]

Given:

\[ \text{Power} = 15 \, \text{kW} = 15 \times 10^3 \, \text{W}, \quad \text{Photon frequency} = 10^{16} \, \text{photons/second} \]

So, the energy of one photon is:

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

Now, using \( E = h \nu \), we calculate the frequency:

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

Since this frequency lies in the range of gamma rays, the correct answer is gamma rays.