The de-Broglie wavelength of an electron is the same as that of a 50 ke X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is ( the energy equivalent of electron mass of 0.5 MeV)

Options

A. `1 : 50`
B. `1 : 20`
C. `20 : 1`
D. `50 : 1`

Answer

C

Explanation

de Broglie wavelength of an electron
`lambda_(e) = (h)/(sqrt(2 m_e K_e))`
where subscript e refers to electron.
`therefore` Kinetic energy of the electron.
`K_(e) = (h^2)/(2m_e lambda_e^2)`
Energy of a photon, `E_(ph) = (hc)/(lambda_(ph))`
where subscript ph refers to photon.
`therefore (E_(ph))/(K_e) = (hc)/(lambda_(ph)) xx (2m_e lambda_e^2)/(h^2) = (2m_ec^2)/(hc//lambda_(ph)) = ((2xx 0.5 xx 10^6)eV)/(50 xx 10^(3) eV) = 20 /1`
or `E_(ph) : K_e = 20 :1`