One of the spectral lines of atomis hydrogen has the wavelength `lambda = 656.3 nm`. Find the Doppler shifl `Delta lambda` of that line when observed at right angles to the beam of hydrogen atoms with kinetic enegry `T = 1.0 Mev`(the transverse Doppler effect).

Explanation

In transverse Doppler effect.
`omega = omega_(0) sqrt(1-beta^(2)) = omega_(0) (1-(1)/(2)beta^(2))`
so `lambda = (c )/(omega) = (c )/(omega_(0)) (1+(1)/(2)beta^(2)) = lambda_(0)(1+(1)/(2)beta^(2))`
Hence `Delta lambda = (1)/(2)beta^(2)lambda`
Using `beta^(2) = (v^(2))/(c^(2)) = (2T)/(mc^(2))`, where `T = K.E` of `H` atoms
`Delta lambda = (T)/(mc^(2)) lambda = (1)/(938) xx 656.3mm = 0.70 nm`