Chapter 30, Problem 25P

To determine

The maximum kinetic energy of an electron emitted, in the ${\text{β}}^{\text{-}}$ decay of a free neutron.

Answer to Problem 25P

Solution:

The energy released, when tritium ${}_{\text{1}}{}^3\text{H}$, decays by ${\text{β}}^{\text{-}}$ emission is $0.782\text{MeV}$.

Explanation of Solution

Formula used:

Binding energy is $\Delta m{c}^2$.

When a neutron ${}_0{}^1\text{n}$ decays by ${\text{β}}^{\text{-}}$ emission, the decay equation is ${}_0{}^1\text{n}\to {}_1{}^1\text{H}+{\beta }^{-}$.

Calculation:

Mass of neutron ${}_0{}^1\text{n}$ is $1.008665\text{u}$. Mass of proton ${}_1{}^1\text{H}$ is $1.007825\text{u}$. Mass defect is $\Delta m=1.008665\text{u}-1.007825\text{u}=\text{0}\text{.00084}\text{u}$

Energy released is $\text{0}\text{.00084}\text{u}\times \frac{931.5\text{MeV}}{\text{u}}=0.782\text{MeV}$