A stationary hydrogen atom emits a photon corresponding to first line of the Lyman series. What velocity does the atom acquire?

Options

A. 3.25
B. 3.5
C. 1.5
D. `4.0`

Answer

A

Explanation

The problem is similar to gun firing. The momentum of photon is `h/lamda`
According to conservation principle of momentum,

`vecP_(i) = vecP_(f)`
`0 = mv - h/lamda`
`:. " " mv = h/lamda`
Here v is recoiling speed of H-atom. According to energy conservation principle,
`E_i - E_f`
The initial energy of atom is `E_(i) = -(13.6)/(n^2)eV = -(13.6)/((1)^2) eV`
`:. E_(i) = - (13.6)/n^2 eV`
The final energy of the system (atom + photon) is `E_f = K.E.` of atom + energy of electron + photon energy
`=1/2 mv^2 - (13.6)/((1)^2)eV+(hc)/lamda`
`:. E_i=E_f`
`:. - (13.6)/4eV =1/2mv^2-13.6eV+(hc)/lamda`
or `(13.6-(13.6)/(4))eV =1/2mv^2+(hc)/lamda`
or `(13.6xx3/4)eV=1/2 mv^2 +mvc`
[From eqn(i)]
or `10.2 xx1.6 xx10^(-19)` joule `=1/2 mv^(2) +mvc`
Here `m = 1.67 xx10^(-27) kg, c = 3 xx 10^(8)m//s`
After solving `v = 3.25m//s`