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  4. 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

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Q. 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

Dual Nature of Radiation and Matter

Solution:

Energy of photon $E=\frac{h c}{\lambda}$ (Joules)
$=\frac{h c}{e \lambda}( eV )$
$\Rightarrow E( eV )=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times \lambda(\mathring{A})}$
$=\frac{12375}{\lambda(\mathring{A})}$
$\Rightarrow E( keV )=\frac{12.37}{\lambda(\mathring{A})} \approx \frac{12.4}{\lambda}$