When a hydrogen atom emits a photon during a transition from n = 4 to n = 2, its recoil speed is about ,

Options

A. `4.28 ms^(-1)`
B. `0.814 ms^(-1)`
C. `0.07 ms^(-1)`
D. `0.407 ms^(-1)`

Answer

B

Explanation

We know tht momentum of a photon, ` p = (h)/(lambda)`
`rarr mv = (h)/(lambda) rArr v = (h)/(mlambda) " ". . . (i) [:. p = mv]`
Given, hydrogne atom emits a photon during a transition from n = 4 to n = 2 ,
According to Bohr.s formula for H-atom , the wavelength of the emitted photon in given by
`:. (1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
where, R = Rydberg.s constant ` = 1 . 097 xx 10^(7) m^(-1)`
`rArr (1)/(lambda) = 1 . 0 97 xx 10^(7) ((1)/(2^(2)) -(1)/(4^(2)))`
`rArr lambda = 4.86 xx 10^(-7) m`
`:. ` mass of a proton ` = 1.6 xx 10 ^(-27 kg`
Now , from Eq. (i), we get
Hence, recoil speed of photon,
` v = (6 . 62 xx 10^(-34))/( 1.6 xx 10 ^(-27) xx 4 . 86 xx 10^(-7))`
`= 0 . 8513m//s`
or `v~~ 0 . 814 ms^(-1)`