# Calculating frequency shift due to atomic recoil momentum [closed]

The question: An atom of hydrogen emits a photon of energy $$2eV = hf_0$$. As a result, the H atom recoils causing the frequency of the photon to be changed to $$f$$. Write an expression for the change in frequency of the photon.

I have solved this question two ways, but get different answers, I want to know why.

$$f_0$$ = initial frequency of emitted photon; $$f$$ = measured frequency of photon; $$M_p$$ = proton mass

(i) via energy and momentum

$$p_{photon} = \frac{h v_0}{c}$$

Recoil energy of H atom = $$p_{photon}^2/2M_p = \frac{h^2 f_0^2}{2M_pc^2}$$

New photon energy, accounting for recoil energy loss = $$hf_0 - \frac{h^2 f_0^2}{2M_pc^2} = hf$$

Rearrange $$\to$$ $$f_0 - f= \frac{h f_0^2}{2M_pc^2}$$

This is the correct formula, quoted from Bransden and Joachain. Now inserting the numbers (SI units) the answer is $$\approx$$ $$5$$ x $$10^5$$

(ii) via recoil velocity and doppler shift
$$H_{vel} = -\frac{hf_0}{cM_p}$$

Doppler Shift: $$f = f_0 \left( \frac{\sqrt{1 - \frac{H_{vel}^2}{c^2}}}{1 - \frac{H_{vel}}{c}}\right)$$

$$f_0 - f$$ = $$f_0 - f_0\left( \frac{\sqrt{1 - \frac{H_{vel}^2}{c^2}}}{1 - \frac{H_{vel}}{c}}\right)$$

Evaluating this gives me $$\approx 1$$ x $$10^6$$

Can someone explain my mistake?