If the source of power` 4 kW` product`10^(20)` photons //second , the radiation belongs to a part spectrum called
If the source of power` 4 kW` product`10^(20)` photons //second , the radiation belongs to a part spectrum called
A
X-rays
B
ultraviolet rays
C
microwaves
D
`gamma`rays
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem, we need to determine the part of the electromagnetic spectrum to which the radiation from a power source of 4 kW producing \(10^{20}\) photons per second belongs. We will use the relationship between power, the number of photons, and the energy of each photon.
### Step-by-Step Solution:
1. **Understand the relationship between power, photons, and energy**:
The power \(P\) of a light source can be expressed in terms of the number of photons \(n\) produced per second and the energy \(E\) of each photon:
\[
P = n \cdot E
\]
where \(E\) (energy of one photon) is given by:
\[
E = h \cdot \nu
\]
Here, \(h\) is Planck's constant and \(\nu\) is the frequency of the light.
2. **Substituting the energy expression into the power equation**:
We can rewrite the power equation as:
\[
P = n \cdot h \cdot \nu
\]
3. **Rearranging to find the frequency**:
From the equation above, we can isolate the frequency \(\nu\):
\[
\nu = \frac{P}{n \cdot h}
\]
4. **Substituting the known values**:
- Given power \(P = 4 \text{ kW} = 4 \times 10^3 \text{ W}\)
- Number of photons per second \(n = 10^{20}\)
- Planck's constant \(h = 6.626 \times 10^{-34} \text{ J s}\)
Now substituting these values into the frequency formula:
\[
\nu = \frac{4 \times 10^3}{10^{20} \cdot 6.626 \times 10^{-34}}
\]
5. **Calculating the frequency**:
\[
\nu = \frac{4 \times 10^3}{6.626 \times 10^{-14}} \approx 6.036 \times 10^{16} \text{ Hz}
\]
6. **Identifying the part of the electromagnetic spectrum**:
The frequency \(6.036 \times 10^{16} \text{ Hz}\) falls within the ultraviolet (UV) range of the electromagnetic spectrum, which typically spans from about \(10^{15} \text{ Hz}\) to \(10^{20} \text{ Hz}\).
### Conclusion:
The radiation from the power source of 4 kW producing \(10^{20}\) photons per second belongs to the ultraviolet (UV) part of the electromagnetic spectrum.
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