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
A) $E = 12.4\ hv$
B) $E = 12.4\ h/\lambda$
C) $E = 12.4/\lambda$
D) $E = hv$
Explanation
(c) Energy of photon $E = \frac{hc}{\lambda}(Joules) = \frac{hc}{e\lambda}(eV)$
$\Rightarrow \ \begin{matrix}
E \\
(eV)
\end{matrix} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times \lambda_{(m)}\mathring{\mathrm{A}}} = \frac{12375}{\lambda(\mathring{\mathrm{A}})}$
$\Rightarrow E(keV) = \frac{12.37}{\lambda(\mathring{\mathrm{A}})} = \frac{12.4}{\lambda}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.