An electron emits a photon and the core is pushed (recoiled) back!

Question

I have come across a problem which is a homework indeed, but i tried to pack this question up so that it is more theoretical.

What I want to know is: If I am allowed to write energy conservation for an atom which emits a photon (when its electron changes energy for a value $\Delta E$) like this (the atom is kicked back when it emits a photon):

\begin{align} E_1 &= E_2\\ E_{ \text{H atom 1}} &= E_{ \text{H atom 2} } + E_\gamma\\ \sqrt{ \!\!\!\!\!\!\!\!\!\!\smash{\underbrace{(E_0 + \Delta E)^2}_{\substack{\text{I am not sure about}\\\text{this part where normally}\\\text{we write only ${E_0}^2$. Should I}\\\text{put $\Delta E$ somewhere else?}}}}\!\!\!\!\!\!\!\!\!\!\!\! + {p_1}^2c^2} &= \sqrt{ {E_0}^2 + {p_2}^2c^2 } + E_\gamma \longleftarrow \substack{\text{momentum $p_1=0$ and because of}\\\text{the momentum conservation}\\\text{$p_2 = p_\gamma = E_\gamma/c$}}\\ \phantom{1}\\ \phantom{1}\\ \phantom{1}\\ \sqrt{{(E_0 + \Delta E)}^2} &= \sqrt{{E_0}^2 + {E_\gamma}^2} + E_\gamma\\\ E_0 + \Delta E &= \sqrt{{E_0}^2 + {E_\gamma}^2} + E_\gamma\\\ \end{align}

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EDIT: Is this a better way? I know we get the same result this time, but what if momentum $p_1$ wasn't $0$? Then it would come out differently right?

\begin{align} E_1 &= E_2\\ E_{ \text{H atom 1}} + \Delta E &= E_{ \text{H atom 2} } + E_\gamma\\ \sqrt{{E_0}^2 + {p_1}^2c^2} + \Delta E &= \sqrt{ {E_0}^2 + {p_2}^2c^2 } + E_\gamma \longleftarrow \substack{\text{momentum $p_1=0$ and because of}\\\text{the momentum conservation}\\\text{$p_2 = p_\gamma = E_\gamma/c$}}\\ E_0 + \Delta E &= \sqrt{{E_0}^2 + {E_\gamma}^2} + E_\gamma \end{align}

Answer 1

Your edit is the correct way to handle the $\Delta E$.

You can think of it exactly like a conservation of energy problem from freshman mechanics but with a different definition of kinetic energy:

Initially

$E = K_H + U$

The kinetic energy is

$K_H = \sqrt{(m_H c^2)^2 + p^2 c^2} = m_H c^2$

The potential energy is

$U = \Delta E$

Finally

$E = K_H + K_{\gamma} + U$

where $K_H$ of the atom is

$K_H = \sqrt{(m_{H}c^2)^2 + p^2 c^2}$

$K_{\gamma} = \sqrt{(m_{\gamma}c^2)^2 + p^2 c^2} = pc$

and now there is no potential energy.

To find $p$ you use conservation of momentum as you suggested.

Equating initial and final energies you get what you wrote in your edit.