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

In summary, the conversation discusses pair production, where a photon can spontaneously disintegrate into an electron and a positron in the presence of an electric field. The smallest possible photon frequency for pair production is calculated assuming both electron and positron are at rest. The momentum of the photon and the recoil velocity of the carbon atom as a result of pair production are also calculated. The carbon atom must be nearby in order to conserve the momentum of the system.
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!

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?

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

never mind, i got it-thanks anyway!

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

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

What is a photon in an electric field?

A photon in an electric field is a quantum of electromagnetic radiation that interacts with electric charges. It is considered to be both a particle and a wave, and its behavior is described by quantum mechanics.

How does an electric field affect a photon?

An electric field affects a photon by causing it to accelerate and change direction. This is because the electric field exerts a force on the electric charge of the photon. The strength of the electric field determines how much the photon will be affected.

Can a photon exist without an electric field?

Yes, a photon can exist without an electric field. Photons can travel through vacuum, which has no electric field. However, the presence of an electric field can affect the behavior and properties of the photon.

What is the relationship between photons and electric fields?

Photons and electric fields are closely related as photons are the particles that make up the electromagnetic field. Electric fields also affect the behavior and properties of photons, as they interact with the electric charge of the photon.

What is the significance of studying photons in electric fields?

Studying photons in electric fields allows scientists to better understand the behavior of light and electromagnetic radiation. This can lead to advancements in various fields such as optics, electronics, and telecommunications. Additionally, it can also provide insights into the fundamental nature of quantum mechanics.

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