An electron and a photon each has a wavelength of 1.0nm. Find (a) their momenta, (b) the energy of the photon , and © the kinetic energy electron.
Take `h=6.63xx10^(-34) Js`.

Explanation

Wavelength of an electron `(lambda_e)` and a photon `(lambda_p),lambda_e=lambda_p=lambda=1` nm =`1xx10^(-9)` nm
Planck's constant , `h=6.63xx 10^(-34)` Js
(a)The momentum of an elementary particle is given by de Broglie relation : `lambda=h/p`
`p=h/lambda`
It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths on an electron and a photon are equal , both have an equal momentum .
`therefore p=(6.63xx10^(-34))/(1xx10^(-9))=6.63xx10^(-25) kg m s^(-1)`
(b)The energy of a photon is given by the relation : `E=(hc)/lambda`
Where, Speed of light , `c=3xx10^8` m/s
`therefore E=(6.63xx10^(-34)xx3xx10^8)/(1xx10^(-9)xx1.6xx10^(-19))`
=1243.1 eV = 1.243 keV
Therefore, the energy of the photon is 1.243 keV.
(c) The kinetic energy (K) ofan electron having momentum p, is given by the relation : `K=1/2 p^2/m`
where, m=mass of the electron =`9.1xx10^(-31) kg`
`p=6.63xx10^(-25) kg m s^(-1)`
`therefore K=1/2 xx ((6.63xx10^(-25))^2)/(9.1xx10^(-31))=2.415xx10^(-19) J`
`=(2.415xx10^(-19))/(1.6xx10^(-19))=1.51 eV`
Hence, the kinetic energy of the electon is 1.51 eV.