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 $h =6 \times 10^{-34} Js$ )
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 $h =6 \times 10^{-34} Js$ )
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- Micro waves
Radio waves
Gamma rays
- Ultraviolet rays
The Correct Option is C
Approach Solution - 1
So gamma Rays. Option 3
Approach Solution -2
\[ E = \frac{\text{Power}}{\text{Photon frequency}} \]
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.Top JEE Main Physics Questions
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