Atom Emitting Photon: Energy Conservation

[71GA]

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 alowed to write energy conservation for an atom which emitts a photon (when his electron changes energy for a value ##\Delta E##) like this (The atom is kicked back when it emmits an 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 normaly}\\\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
\end{align}

 

[mfb]

##p_2 c = E_\gamma## requires that the initial atom is at rest in your coordinate system, so you can keep E_1 at the left side (and I think I would not use E_0 at all, as it can be confusing), you don't need that p₁ at all.
That is possible, indeed.