If a source of electromagnetic radiation having power `15kW` produces `10^(16)` photons per second,the radiation belongs to a part of spectrum is.(Take Planck constant `h=6times10^(-34)Js` )

Options

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

Answer

To solve the problem, we need to determine the wavelength of the electromagnetic radiation based on the given power and the number of photons emitted per second. We can then identify which part of the electromagnetic spectrum this wavelength corresponds to.

### Step-by-Step Solution:

1. **Understand the relationship between power, energy, and photons:**
The power \( P \) of the radiation can be expressed in terms of the energy \( E \) of the photons and the number of photons \( N \) emitted per second:
\[ P = \frac{E}{T} = N \cdot \frac{E}{T} = N \cdot \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.

2. **Substitute the known values:**
Given:
- Power \( P = 15 \, \text{kW} = 15 \times 10^3 \, \text{W} \)
- Number of photons per second \( N = 10^{16} \, \text{photons/s} \)
- Planck's constant \( h = 6 \times 10^{-34} \, \text{Js} \)
- Speed of light \( c = 3 \times 10^8 \, \text{m/s} \)

Substitute these values into the equation:
\[ 15 \times 10^3 = 10^{16} \cdot \frac{(6 \times 10^{-34})(3 \times 10^8)}{\lambda} \]

3. **Rearrange to solve for \( \lambda \):**
Rearranging gives:
\[ \lambda = \frac{(10^{16})(6 \times 10^{-34})(3 \times 10^8)}{15 \times 10^3} \]

4. **Calculate the numerator:**
First, calculate the numerator:
\[ 10^{16} \cdot 6 \cdot 3 = 18 \times 10^{16} \times 10^{-34} \times 10^8 = 18 \times 10^{-10} \]

5. **Calculate the denominator:**
The denominator is:
\[ 15 \times 10^3 \]

6. **Combine to find \( \lambda \):**
Now substituting back gives:
\[ \lambda = \frac{18 \times 10^{-10}}{15 \times 10^3} = \frac{18}{15} \times 10^{-10 - 3} = 1.2 \times 10^{-13} \, \text{m} \]

7. **Convert to nanometers:**
To convert meters to nanometers (1 m = \( 10^9 \) nm):
\[ \lambda = 1.2 \times 10^{-13} \, \text{m} = 1.2 \times 10^{-4} \, \text{nm} \]

8. **Identify the part of the spectrum:**
The wavelength \( 1.2 \times 10^{-4} \, \text{nm} \) is much less than \( 10^{-12} \, \text{m} \), which indicates that it falls in the gamma rays region of the electromagnetic spectrum.

### Final Answer:
The radiation belongs to the gamma rays part of the spectrum.

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