A stationary hydrogen atom emits photon corresponding to the first line of Lyman series. If R is the Rydberg constant and M is the mass of the atom, then the velocity acquired by the atom is

Explanation

The difference in energy `(Delta E)` will be equal to the energy of the photon emitted
The first line in Lyman series corresponds to the transition
`2 rarr 1`
`Delta E = 2.18 xx 10^(-18)(2)^(2) ((1)/(1^(2)) - (1)/(2^(2))) 1 ` per atom
`= 6.54 xx 10^(-19) J`
A photon xcarying energy strickes a H atom in the ground state .Note that the ionisation energy of H atom is ` + 2.18 xx 10^(-19) J` .This will be the work function of H atom Using the Einstein's photoelectric equation
`KE = E_(1) - W_(0) = (1)/(2) m_(a)v_(e)^(2)``[E_(1) = ` Incident energy]
`rArr v_(0) sqrt((2(E_(1) - W_(0)))/(m_(e)))`
`= sqrt((2(6.54 xx 10^(-19) - 2.18 xx 10^(-18)))/(9.1 xx 10^(-31)))`
`= 3.09 xx 10^(6) ms^(-1)`
We can calculate the wavelength of electrion ejected out to be `2.36 xx 10^(-19) m = 2.36 Å`
`lambda_(e)= (h)/(m_(e)v_(e)) = (6.626 xx 10^(-34))/(9.7 xx 10^(-38) xx 3.09 xx 10^(6)) m`
`= 2.36Å`