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. $E=\frac{12.4h}{\lambda }$
• C. $E=\frac{12.4}{\lambda }$
• D. $E=hv$

Answer 1

Answer: (C)

. $E=\frac{12.4}{\lambda }$.

Energy of photon $E=\frac{hc}{\lambda }\left(Joules\right)=\frac{hc}{e\lambda }\left(eV\right)$

$\Rightarrow E_{eV}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times \lambda \left(\stackrel{\circ }{A}\right)}=\frac{12375}{\lambda \left(\stackrel{\circ }{A}\right)}$
$\Rightarrow E\left(keV\right)=\frac{12.37}{\lambda \left(\stackrel{\circ }{A}\right)}\approx \frac{12.4}{\lambda }$