Animated Solution for Physics - Atoms and Nuclei: Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron $=1.6725\times 10^{-27}$ kg, mass of proton $=1.6725\times 10^{-27}$ kg, mass of electron $=9\times 10^{-31}$ kg)

A free neutron is a fascinating particle. While neutrons are perfectly stable when bound inside a stable atomic nucleus, a solitary, free neutron is radioactive. It undergoes beta decay, spontaneously transforming into a proton, an electron, and an electron antineutrino.

Because the combined mass of the resulting proton and electron is slightly less than the original mass of the neutron, this missing mass—known as the mass defect—is converted into pure kinetic energy according to Einstein's legendary mass-energy equivalence principle, $E=\Delta m{c}^2$.

A Test of Concepts

This specific question from AIEEE 2012 contains a notorious trap—a typographical error in the provided data. The question states that the mass of the neutron is $1.6725\times 10^{-27}$ kg, and the mass of the proton is also $1.6725\times 10^{-27}$ kg.

If we were to blindly use these values, the mass defect $\Delta m=m_n-(m_p+m_e)$ would be negative! A negative mass defect implies that energy must be supplied for the reaction to occur, meaning spontaneous decay would be physically impossible. To solve this problem correctly, we must rely on our conceptual understanding and use the standard, accepted mass of a neutron: $m_n=1.6747\times 10^{-27}$ kg.

Let's calculate the true mass defect using the standard neutron mass. To make the subtraction seamless, we express the electron's mass with the same power of ten as the nucleons:

$m_n=1.6747\times 10^{-27}$ kg $m_p=1.6725\times 10^{-27}$ kg $m_e=9\times 10^{-31}\text{ kg}=0.0009\times 10^{-27}$ kg

The mass defect $\Delta m$ is:

Now, we unleash $E=\Delta m{c}^2$ to find the energy released ($Q$-value) in Joules:

Finally, we convert this energy from Joules to Mega electron-volts (MeV). Recall the conversion factor: $1\text{ MeV}=1.6\times 10^{-13}\text{ J}$.

This energy is shared as kinetic energy among the proton, the electron, and the nearly massless antineutrino, explaining why the emitted electrons in beta decay exhibit a continuous energy spectrum!