If we express the energy of a photon in `KeV` and the wavelength in angstroms , then energy of a photon can be calculated from the relation

Answer

To solve the problem of expressing the energy of a photon in kilo electron volts (KeV) and the wavelength in angstroms, we can use the relationship between energy and wavelength.

### Step-by-Step Solution:

1. **Understand the relationship between energy and wavelength**:
The energy \( E \) of a photon is inversely related to its wavelength \( \lambda \). The general formula for the energy of a photon is given by:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

2. **Convert constants to appropriate units**:
- Planck's constant \( h \) is approximately \( 6.626 \times 10^{-34} \) Joule seconds.
- The speed of light \( c \) is approximately \( 3 \times 10^8 \) meters per second.
- To express energy in kilo electron volts (KeV), we need to convert Joules to electron volts. The conversion factor is \( 1 eV = 1.6 \times 10^{-19} \) Joules.

3. **Substituting values**:
To express energy in KeV, we can use the formula:
\[ E (\text{KeV}) = \frac{12.4}{\lambda (\text{Å})} \]
This formula is derived from the constants used in the original equation, where \( 12.4 \) comes from the conversion of \( hc \) into the units of KeV and angstroms.

4. **Final formula**:
Therefore, the energy of a photon can be calculated using the relation:
\[ E = \frac{12.4}{\lambda} \]
where \( E \) is in kilo electron volts (KeV) and \( \lambda \) is in angstroms (Å).