Question

The nucleus decays byemission. Write down the decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

= 22.994466 u

= 22.989770 u.

Answer 1

In emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.

emission of the nucleus is given as:

It is given that:

Atomic mass of = 22.994466 u

Atomic mass of = 22.989770 u

Mass of an electron, m_e = 0.000548 u

Q-value of the given reaction is given as:

There are 10 electrons in and 11 electrons in. Hence, the mass of the electron is cancelled in the Q-value equation.

The daughter nucleus is too heavy as compared to and . Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV.