Find recoil speed When a hydrogen atom emits a photon during the transition from n = 5 to n=1

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We know that momentum is given by equation

p= h/λ

Hydrogen atom emits photon during transition from n= 5 to n=1 

Substituting in equation (1) we get

This is the recoil speed of hydrogen atom. 

Answered by Expert 10th September 2018, 11:46 AM

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Text Solution

Answer : c

Solution : For photon emitted from hydrogen atom, the wavelength is <br> `(1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` (i) <br> But according to de Broglie concept <br> `lambda = (h)/(p) implies (1)/(lambda) = (p)/(h)` (ii) <br> From (i) and (ii) <br> `(p)/(n) = R((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) implies p = Rh ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` (iii) <br> Since the momentum of the hydrogen atom initially was zero . therefore finally the momentum of photon is equal to momentum of the hydrogen atom in magnitude (By law of conservation of momentum ). Let the momentum of hydrogen atom be `m_(H) V_(H)`. Then, from (iii) <br> `m_(H) V_(H) = R_(h) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` <br> implies `V_(H) = (Rh)/(m_(H)) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` <br> `= (1.097 xx 10^(7) xx 6.63 xx 10^(-34))/(1.67 xx 10^(-27)) ((1)/(1^(2)) - (1)/(5^(2)))` <br> `implies V_(H) = 4.178 m s^(-1)`