Disintegration of the deuteron Considering the scattering of gamma rays on a deuteron, which leads to its break up acording to:
$$ \gamma+ d \longrightarrow  p +n $$
we can use the conservation of energy and momentum in order to determine the minimum photon energy in order to make this reaction possible, which happens to be very close to the binding energy of the deuteron nuclide. We assume that the speed of the proton and
neutron after scattering are highly non-relativistic.
So, the conservation of energy:
$$ p_{\gamma}c + m_dc^2=m_pc^2+m_nc^2+ \frac{1}{2}m_pv_p^2+\frac{1}{2}m_nv_n^2
\tag 1$$
and of momentum:
$$ \vec p_{\gamma} = m_p \vec v_{p}+m_n \vec v_{n} 
\tag 2$$
Squaring equation $(2)$ we obtain:
$$ p_{\gamma}^{ \space 2} = m_p^{ \space 2}v_{p}^{ \space 2}+m_n^{ \space 2}v_{n}^{ \space 2} + 2m_pm_n \vec v_{n}\cdot \vec v_p \tag 3$$
But if we are looking for the minimum photon energy, then we must find the case where
the proton and neutron speeds are also minimum. We assume $m_p \approx m_n$ and $v_n \approx v_p$.
With these approximations and the deuteron mass formula: $m_d = m_n + m_p -\frac{B_d}{c^2}$ we should be able to get to this equation:
$$ p_{\gamma}^2 = 2 m_d \left(  \frac{1}{2}m_pv_p^2 + \frac{1}{2}m_nv_n^2   \right)
\tag 4$$
but I'm having trouble getting there. Can someone show me how to do it?
 A: If you analyze the threshold reaction in the center-of-momentum frame, rather than the lab frame where the deuteron is initially at rest, the energy-conservation equation becomes
$$
pc + m_d c^2 + \frac{p^2}{2m_d} = m_p c^2 + m_n c^2
\tag{1*}
$$
Here $p$ is the magnitude of the initial momentum for both the photon and the deuteron (because that's how we get to the zero-momentum frame), and we have the minimum-energy configuration in the final state.  Therefore, when we boost back into the lab frame, a threshold disintegration will be one where $\vec v_p = \vec v_n = \vec v_\text{lab}$ are exactly the same, whether the fragments have similar masses or not.
If the fragments have the same velocity, your equation (3) becomes
\begin{align}
p_\gamma^2 
&= \left( m_p \vec v_\text{lab} + m_n \vec v_\text{lab} \right)^2
\tag{$3'$}
\\
&= (m_p + m_n)^2 v_\text{lab}^2
\\
&= 2(m_p + m_n) 
\left(\frac{m_p}2 + \frac{m_n}2\right) v_\text{lab}^2
\tag{$4'$}
\end{align}
which is the same as your (4) if you make the approximation that the binding energy is small and $m_p+m_n\approx m_d$.
Perhaps the author of the text you're following meant to write about temporarily neglecting the binding energy but thinko'd it into a hint that $m_p\approx m_n$ instead.
The point of the construction seems to be to set up $p_\gamma^2 / 2m_d$ to replace the ugly kinetic-energy term in (1).
It would be more correct to keep that term as $p_\gamma^2/2(m_p+m_n)$.  However the next thing the author does is a binomial expansion to find the first-order correction to the threshold energy, and that algebra is quite tedious enough without keeping track of two different mass terms.
Note that if you work in the center-of-momentum frame, starting with my (1*), you don't get quite the same quadratic equation for $p$ as your reference gets for $p_\gamma$ in the lab frame.  The first-order difference between the two results,
\begin{align}
p &= \frac{B}{c} \left(
1 - \frac12 \frac{B}{m_d c^2}
+ \mathcal O\left(\frac B{mc^2}\right)^2
\right)
\\
p_\gamma &= \frac{B}{c} \left(
1 + \frac12 \frac{B}{m_{(p+n)}c^2}
+ \mathcal O\left(\frac B{mc^2}\right)^2
\right)
\end{align}
is because, in the center-of-momentum frame, part of the energy required to destroy the deuteron is included in the deuteron's kinetic energy; meanwhile in the lab frame, the photon has to carry not only the binding energy but also enough kinetic energy to allow the fragments to carry the final momentum.  At first order, the two results are connected by the Doppler shift when you boost from the center-of-momentum frame to the lab frame, but the algebra is much more tractable in the $m_p+m_n=m_d$ approximation. At higher orders there starts to be plenty of higher-order funny business.
A: The mass of a system is constant in reactions and is a Lorentz invariant quantity, where the mass is defined to be $$E^2-p^2c^2.$$
$E$ is the total energy of the system, and $p$ is the magnitude of the net momentum of the system. For your initial system, in the lab, you have
$$E=p_{\gamma }c+m_d c^2\text{ and } p=p_{\gamma}.$$
In the center-of-momentum frame, after minimal  energy reaction, the neutron and proton are both at rest so $$E_{cm}=(m_n+m_p)c^2 \text{ and } p_{cm}=0.$$
Set these two mass calculations equal to each other and you can solve for $p_{\gamma}c$.
Now have a minimum photon energy in terms of the masses only without having to worry about the lab velocities with the binding energy concept built in because you will have terms in which it is important that $m_d \ne m_n+m_p$. This will also show you that if you approximate $m_d=m_n+m_p$, the minimum energy is zero. That's because the Q of this reaction with the approximation is zero, when in reality it is not.
If you want the lab velocities, at minimum photon energy, the neutron and proton will have identical velocities (they have zero velocity in the CoM), and their total momentum will be $(m_n+m_p)v=p_{\gamma}$. You can use that to solve for $v$. If you use the mass equality appoximation above, $v=0$.
