Question

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. $E=12.4hv$
• B. $E=12.4h/\lambda$
• C. $E=12.4/\lambda$
• D. $E=hv$

Answer 1

Answer: (C)

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