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**Another Nuclear Physics problem: Minimum photon energy for deuteron dissociation**

## Homework Statement

What is the minimum photon energy necessary to dissociate a deuteron. Take the binding energy to be 2.224589 MeV

## Homework Equations

[tex]\gamma[/tex] +

^{2}H [tex]\rightarrow[/tex]

^{1}H + n

[tex]\vec{P}[/tex] = [tex]\vec{P_n}[/tex] + [tex]\vec{P_p}[/tex]

where P = momentum of incident gamma ray

P_p = momentum of recoiling proton

P_n = momentum of recoiling neutron

E + m(

^{2}H)*c

^{2}= E_p + E_n

where E = energy of incident gamma ray (what I need to find) = P/c, c = speed of light

E_p = energy of recoiling proton = [m(

^{1}H)

^{2}*c

^{4}+ P_p

^{2}*c

^{2}]

^{1/2}

E_n = energy of recoiling neutron = [m(n)

^{2}*c

^{4}+ P_n

^{2}*c

^{2}]

^{1/2}

B = [m(

^{1}H) + m(n) - m(

^{2}H)]*c

^{2}= 2.224589 MeV

## The Attempt at a Solution

I tried solving all the above equations simultaneously, but then of course ran into problems writing the magnitudes of the momentum for the neutron and proton in terms of the momentum of the incoming photon. How do I know the neutron and proton recoil at the same angle and how could I go about calculating that angle?

Thanks for any help.

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