Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? An X-ray photon or the electron? (For quantitative comparison, take the wavelength of the probe equal to `1Å`, which is of the order of interatomic spacing in the lattice), `m_(e)=9.11xx10^(-31)kg`.

Explanation

An X-ray probe has a greater energy than a electron probe for the same wavelength.
Wavelength of light emitted from the probe, `lambda = 1, Å = 10^(-10) m`
Mass of an electron, `m_(e) = 9.11xx10^(-34) kg`
Planck's constant, `h = 6.6xx10^(-19)C`
The kinetic energy of the electron is given as:
`E = (1)/(2)m_(e)v^(2)`
`m_(e)v = sqrt(2Em_(e))`
Where, v = velocity of the electron
`m_(e)v` = Momentum (p) of the electron
According to the de Broglie principle, the de Brogile wavelength is given as:
`lambda = (h)/(P) = (h)/(m_(e)y) = (h)/(sqrt(2Em_(e)))`
`:. E = (h^(2))/(2lamda^(2)m_(e))`
`= (6.6xx10^(-34))^(2)/(2xx(10^(-10))^(2)xx9.11xx10^(-31)) = 2.39xx10^(-17)J`
`= (2.39xx10^(-17))/(1.6xx10^(-19)) = 149.375 eV`
`E' = (hc)/(lambdae)eV`
Energy of a photon,
`= (6.6xx10^(-34)xx3xx10^(8))/(10^(-10)xx1.6xx10^(-19))`
`= 12.375xx10^(3) eV = 12.375 keV`
Hence, a photon has a greater than an electron for the same wavelength.