Question

Show with calculation, which one has greater energy? An X-ray photon with $\lambda$ = 1 $\stackrel{o}{A}$ or an electron with $\lambda$ = 1 $\stackrel{o}{A}$. ($m_e$ = 9.1 x 10${}^{-31}$ kg, c = 3 x 10${}^8$ m/s)

Answer 1

Energy of photon/light $=\frac{hc}{{\lambda }_p}$

Energy of electron i.e kinetic energy $=\frac{1}{2}{M}_e{V}_e^2$

Let say $V_e=C$ (speed of light)

$E_e=\frac{1}{2}{M}_e{c}^2=\frac{1}{2}\times 9.1\times {10}^{-31}\times {(3\times {10}^8)}^{2}J$

$E_e=40.95\times {10}^{-15}J$

& $E_P=\frac{hc}{{\lambda }_P}=\frac{6.63\times {10}^{-24}\times 3\times {10}^8}{{10}^{-10}}$

$=19.89\times {10}^{-16}J$

But as both photon & electron have same de-broghe wavelength $=1A^0={10}^{-10}m$

$E_c=\frac{1}{2}\left(\frac{h}{{\lambda }_e.c}\right)c^2[{\lambda }_e=\frac{h}{mc}]$

$=\frac{hc}{2{\lambda }_e}$

So, $E_{proton}>E_{electron}$ for same $\lambda$.