Chapter 30, Problem 31P

To determine

Energy release in the decay of ${}_{84}{}^{218}\text{P}\text{o}$ by a ${\text{α}}_{}^{}$ emission and ${\text{β}}^{\text{-}}$ emission.

Answer to Problem 31P

Solution:

Energy release in the decay of ${}_{84}{}^{218}\text{P}\text{o}$ by a ${\text{α}}_{}^{}$ emission is $6.114\text{MeV}$. Energy release in the decay of ${}_{84}{}^{218}\text{P}\text{o}$ by a ${\text{β}}^{\text{-}}$ emission is $0.259\text{MeV}$.

Explanation of Solution

Given Info:

The isotope ${}_{84}{}^{218}\text{P}\text{o}$ can decay either by a $\text{α}$ emission or ${\text{β}}^{\text{-}}$ emission. Mass of ${}_{84}{}^{218}\text{P}\text{o}$ is $218.008973\text{u}$.

Formula used:

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

Calculation:

${}_{84}{}^{218}\text{P}\text{o}$ decays by ${}_2{}^4\text{H}\text{e}$ emission, the decay equation is ${}_{84}{}^{218}\text{P}\text{o}\to {}_{82}{}^{214}\text{P}\text{b}+{}_2{}^4\text{H}\text{e}$.

${}_{84}{}^{218}\text{P}\text{o}$ decays by ${\text{β}}^{\text{-}}$ emission, the decay equation is ${}_{84}{}^{218}\text{P}\text{o}\to {}_{85}{}^{218}\text{A}\text{t}+{\text{β}}^{\text{-}}+\overline{\nu }$.

Mass of ${}_{84}{}^{218}\text{P}\text{o}$ is $218.008973\text{u}$. Mass of ${}_{82}{}^{214}\text{P}\text{b}$ is $213.999806\text{u}$. Mass of ${}_2{}^4\text{H}\text{e}$ is $4.002603\text{u}$. Mass defect is $\Delta m=218.008973\text{u}-\left(213.999806\text{u}+4.002603\text{u}\right)=\text{0}\text{.006564}\text{u}$

Energy released is $\text{0}\text{.006564}\text{u}\times \frac{931.5\text{MeV}}{\text{u}}=6.114\text{MeV}$

Mass of ${}_{84}{}^{218}\text{P}\text{o}$ is $218.008973\text{u}$. Mass of ${}_{85}{}^{218}\text{A}\text{t}$ is $218.008695\text{u}$. Mass of electron is $0.0005486\text{u}$ Mass defect is $\Delta m=218.008973\text{u}-218.008695\text{u}=\text{0}\text{.000278}\text{u}$

Energy released is $\text{0}\text{.000278}\text{u}\times \frac{931.5\text{MeV}}{\text{u}}=0.259\text{MeV}$