A photon of light of wavelength `6000 overset@A` has energy E. What will be the wavelength of photon of a light which has energy of photon 4E?

Options

A. `1500 overset@A`
B. `6000 overset@A`
C. `2000 overset@A`
D. `750 overset@A`

Answer

To solve the problem, we need to find the wavelength of a photon that has energy 4E, given that a photon with a wavelength of 6000 Å has energy E.

### Step-by-Step Solution:

1. **Understand the relationship between energy and wavelength**:
The energy (E) of a photon is related to its wavelength (λ) by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) = energy of the photon
- \(h\) = Planck's constant (approximately \(6.626 \times 10^{-34} \, \text{Js}\))
- \(c\) = speed of light (approximately \(3.00 \times 10^8 \, \text{m/s}\))
- \(\lambda\) = wavelength of the photon

2. **Set up the equation for the given photon**:
For the photon with wavelength \(6000 \, \text{Å}\) (which is \(6000 \times 10^{-10} \, \text{m}\)):
\[ E = \frac{hc}{6000 \times 10^{-10}} \]

3. **Determine the energy for the second photon**:
We need to find the wavelength for a photon with energy \(4E\). Using the energy formula:
\[ 4E = \frac{hc}{\lambda_2} \]
where \(\lambda_2\) is the wavelength we want to find.

4. **Set up the ratio of energies**:
Since \(E\) and \(4E\) are related to their respective wavelengths, we can set up the following ratio:
\[ \frac{E}{4E} = \frac{\lambda_2}{6000 \, \text{Å}} \]
Simplifying this gives:
\[ \frac{1}{4} = \frac{\lambda_2}{6000} \]

5. **Cross-multiply to solve for \(\lambda_2\)**:
\[ \lambda_2 = \frac{6000}{4} \]

6. **Calculate \(\lambda_2\)**:
\[ \lambda_2 = 1500 \, \text{Å} \]

### Final Answer:
The wavelength of the photon with energy \(4E\) is \(1500 \, \text{Å}\).

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