When `0.50` Å X-ray strike a material , the photoelectron from the `K` shell are observed to move in a circle of radius `2.3 nm` in a magnetic field of `2 xx 10^(-2)` tesle acting perpendicular to direct of emission of photoelectron . What is the binding elergy of `k-shell` electron?

Explanation

The velocity of the photoelectrons is found by the relation :
`e vB = m (v^(2))/( R)` or `v = (e)/(m) BR`
The kinetic energy of the hotoelectrons is
`K = (1)/(2) m v^(2) = (1)/(2) (e^(2) B^(2) R^(2))/(m)`
`= (1)/(2) ((1.6 xx 10^(-19)) ^(2) (2 xx 101^(-2))^(2) (23 xx 10^(-3))^(2))/((9.1 xx 10^(-31)))`
`= 2.97 xx 10^(-15) J`
`= (2.97 xx 10^(-15)) (1)/(1.6 xx 10^(-19)) = 18.36 KeV`
The energy of the incident photon is `E_(v) = (hc)/(lambda) = (12.4)/(0.50) = 24.8 Ke V`
The binding energy is the difference between these two values:
`BE = E_(v) - K = 24.8 - 18.6 - 6.2 k e V`