Question

Find the expression for the energy of a photon in keV. If we want to express wavelength in angstrom.

Answer 1

Energy of a Photon:

$E_p=h\nu$

$=\frac{hc}{\lambda }J\{\because frequency=\nu =\frac{c}{\lambda }\}$

Here; $h=6.63\times {10}^{-34}J.s$

and $c=3\times {10}^{8}m/s$

Hence, $E_p=\frac{6.63\times {10}^{-24}J.s\times 3\times {10}^{8}m/s}{\lambda }$

Also, $1eV=1.6\times {10}^{-19}J$

$\therefore 1$ Joule $=\frac{1}{1.6\times {10}^{-19}}eV$

$\therefore E_p=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times \lambda }$

$=\frac{12400eV}{\lambda \left(A^o\right)}$

$=\frac{12.4\times {10}^3}{\lambda \left(A^o\right)}eV=\frac{12.4keV}{\lambda }$

Hence, $E_p=\frac{12.4keV}{\lambda \left(A^o\right)}$