What is the recoil velocity of a carbon atom in pair production?

[Paradox101]

Q. If a photon travels in an electric field(usually by a nucleus,such as ^12C),it can spontaneously disintegrate into an electron and a positron--known as pair production.
A)Calculate the smallest possible photon frequency that produces pair production by assuming that both electron and positron are at rest.
B)Calculate the momentum of the photon from A)
C)Find the recoil velocity of the Carbon atom as a result of pair production.
SOLUTION: i got the energy of the positron=energy of electron,so:
E(e)=E(p)=mc^2=9.11×10^-31×(3×10^8)
=8.2×10^-14 J
= 0.511 MeV
⇒E of incident photon= E(e)+E(p)=1.04 MeV
Finally,frequency will be:-
f=E/h=2×8.2×10^-14/6.63×10^-34
=2.5×10^20 Hz
and the momentum:-
p=E/c=2×8.2×10^-14/3×10^8
=5.5×10^-22 kg.m/s
Now, what i can't get is the recoil velocity of the carbon atom.I have no idea how, all i CAN get is the carbon atom's mass,which is:
M=12×1.66×10^-27
≈2.0×10^-26 kg.
in the book the answer is 2.75*10^4 m/s.
Pls Help!

 

[BruceW]

It looks good so far. And you are told to assume the electron and positron are at rest. But before the pair production happened, the photon had some momentum, so what happened to the momentum?

 

[Paradox101]

Wait...should I divide the momentum of the photon by the mass of the carbon atom?, because it seems dimensionally correct.

 

[Paradox101]

never mind, i got it-thanks anyway!

 

[Paradox101]

\begin{align} V_C &= \frac{p}{M_C} \end{align}

 

[BruceW]

yep. nice work. that's why the carbon atom must be nearby - so that the process can conserve momentum of the system.