An electrons of a stationary hydrogen aton passes form the fifth enegry level to the ground level. The velocity that the atom acquired as a result of photon emission will be
`(m` is the mass of the electron, `R`, Rydberg constanrt and `h`, Planck's constant)

Options

A. `(24hR)/(25m)`
B. `(25hR)/(24m)`
C. `(25m)/(24hr)`
D. `(24m)/(25hR)`

Explanation

According to Rydberg formula
`(1)/(lambda)=R[(1)/(n_(f)^(2))-(1)/(n_(i)^(2))]`
Here, `n_(f)=1, n_(i)=5`
`therefore" "(1)/(lambda)=R[(1)/(1^(2))-(1)/(5^(2))]=R[(1)/(1)-(1)/(25)]=(24)/(25)R`
According to conservation of linear momentum, we get
Momentum of photon = Momentum of atom
`(h)/(lambda)=mv or v =(h)/(mlambda)=(h)/(m)((24R)/(25))=(24hR)/(25m)`