An excited hydrogen atom in state n travels with a velocity v ( v c ). It emits a photon in the direction of its motion and transitions to a lower state m . Applying the principles of momentum and energy conservation, determine the frequency of the emitted radiation. Express the result in terms of the frequency ₀ that would be emitted if the atom were at rest.

Step-by-step solution

Let the mass of the hydrogen atom be M . From the conservation of momentum: Mv = Mv' + h c where v' is the velocity of the atom after emitting the photon. v' = v - h Mc From the conservation of energy: E_n + 1 2 Mv^2 = E_m + 1 2 Mv'^2 + h E_n - E_m = 1 2 Mv'^2 - 1 2 Mv^2 + h The energy difference between the states is given by E_n - E_m = h ₀ . Substituting this and the expression for v' : h ₀ = 1 2 M (v - h Mc )^2 - 1 2 Mv^2 + h h ₀ = 1 2 M (v^2 - 2vh Mc + h^2 ^2 M^2c^2 ) - 1 2 Mv^2 + h h ₀ = - vh c + h^2 ^2 2Mc^2