An electron in the third energy level of an excited `He^(o+)` ion returns back to the ground state.The photon emitted in the process is absorbed by a stationary hydrogen atom in the ground state. Determine the velocity of the photoelectron ejected from the hydrogen atom in metre per second.

Answer

A, C

Explanation

The difference in energy `(Delta E)` will be equal to the energy of the photon emitted
First line in Lyman series corresponds to the transition
`2 rarr 1`
` rArr Delta E = 2.18 xx 10^(-19)(2)^(2) ((1)/(1^(2)) - (1)/(2^(2))) 1 "atom"^(-1) `
`= 6.54 xx 10^(-19) J`
The photon of this much energy series a H atom in the ground state .Note the ionisation
Energy of H atom is `+ 2.18 xx 10^(-18) J` This will be the work function of H atom .Using the Einstein's photoelectric equation :
`KE = E - W = (1)/(mv^(2)`
`rArr v = sqrt((2(E_(1) - W))/(m))`
`rArr v = sqrt((2(6.54 xx 10^(-10) - 2.18 xx 10^(-19)))/(9.1 xx 10^(-23)))`
` rArr v= 3.09 xx 10^(6) m s^(-1)`