Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to `1 Å`, which is of the order of inter-atomic spacing in the lattice) `(m_(e) =9.11 × 10^(-31) kg).`

Explanation

Wavelength of photon of X-ray
`lambda=1Å=10^(-10)m`
`h=6.63xx10^(-34)Js,c=3xx10^(8)ms^(-1)`
Energy of one photon,
`E=hv=(hc)/(lambda)`
`therefore E=(6.63xx10^(-34)xx3xx10^(8))/(10^(-10))`
`therefore E=19.89xx10^(-16)J`
`therefore E=(19.89xx10^(-16))/(1.6xx10^(-19))eV`
`=12.43xx10^(3)eV~~12.4KeV` ........(1)
`implies` For electron
`lambda=1Å=10^(-10)m`
`m=9.1xx10^(-31) kg`
`therefore "momentum p"=(h)/(lambda)=(6.63x10^(-34))/(10^(-10))`
`therefore p=6.63xx10^(-24)kg ms^(-1)`
`implies` Kinetic energy of eleectron,
`K=(p^(2))/(2m)`
`therefore K=((6.63xx10^(-24))^(2))/(2xx9.1xx10^(-31))`
`therefore K=2.4152xx10^(-17)J`
`therefore K=(2.4152xx10^(-17))/(1.6xx10^(-19))`
`therefore K=1.5-95xx10^(2)`
`therefore K=150.9 eV`
From equation (1) and (2) For equal (same) wavelength energy of photon is very ,arge compared to energy of electron.