Chapter 30, Problem 29P

To determine

The maximum kinetic energy of the $\text{β}$ particle emitted, during the decay of ${}_{27}{}^{60}\text{C}\text{o}$.

Answer to Problem 29P

Solution:

The maximum kinetic energy of the $\text{β}$ particle emitted, during the decay of ${}_{27}{}^{60}\text{C}\text{o}$

is $2.822\text{MeV}$.

Explanation of Solution

Formula used:

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

Solution:

When a neutron ${}_{27}{}^{60}\text{C}\text{o}$ decays by ${\text{β}}^{\text{-}}$ emission, the decay equation is ${}_{27}{}^{60}\text{C}\text{o}\to {}_{28}{}^{60}\text{N}\text{i}+{\beta }^{-}$.

Calculation:

Mass of ${}_{27}{}^{60}\text{C}\text{o}$ is $59.933816\text{u}$. Mass of ${}_{28}{}^{60}\text{N}\text{i}$ is $59.930786\text{u}$. Mass defect is $\Delta m=59.933816\text{u}-59.930786\text{u}=\text{0}\text{.00303}\text{u}$

Energy released is $\text{0}\text{.00303}\text{u}\times \frac{931.5\text{MeV}}{\text{u}}=2.822\text{MeV}$

Conclusion:

Mass is equivalent to energy. Mass in a decay reaction cannot increase. The change in mass appears as energy.The maximum kinetic energy of the $\text{β}$ particle emitted, during the decay of ${}_{27}{}^{60}\text{C}\text{o}$ is $2.822\text{MeV}$.