# Question #e9c21

Mar 16, 2016

The extra mass is converted into energy. In this case, most of the missing mass is emitted as a photon.

#### Explanation:

$n \to p + e + \gamma + \nu$

Masses of the particles are:
${m}_{n} = 939.565 \frac{M e V}{c} ^ 2$
${m}_{p} = 938.272 \frac{M e V}{c} ^ 2$
${m}_{e} = 0.511 \frac{M e V}{c} ^ 2$
${m}_{\nu} = 0.32 \frac{e V}{c} ^ 2$

The mass of the neutrino ($\nu$) is very small in comparison to the other masses; it is a million times smaller than the mass of the tiny electron.

Notice that I've listed the masses in terms of their energies. This is done using Einstein's famous equation which shows how mass and energy are equivalent: $E = m {c}^{2}$ .

The missing energy goes into forming a photon (or gamma ray: $\gamma$) of up to $0.722 M e V$. This is the upper limit. Since the final state of the decay results in 4 particles, kinetic energy can be shared among them in different ways.