Chapter 30, Problem 32P

To determine

Part (a):

The daughter nucleus.

Answer to Problem 32P

Solution:

The daughter nucleus is ${}_{16}{}^{32}\text{S}$.

Explanation of Solution

Given Info:

The nuclide ${}_{15}{}^{32}\text{P}$ decays by emitting an electron whose kinetic energy can be $1.71\text{MeV}$. Ignore the recoil of the daughter nucleus.

Formula used:

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

${}_{15}{}^{32}\text{P}$ decays by ${\text{β}}^{\text{-}}$ emission, the decay equation is ${}_{15}{}^{32}\text{P}\to {}_{16}{}^{32}\text{S}+{\text{β}}^{\text{-}}+\overline{\nu }$. The daughter nucleus is ${}_{16}{}^{32}\text{S}$

To determine

Part (b):

The approximate mass in u of the daughter atom.

Answer to Problem 32P

Solution:

The approximate mass in u of the daughter atom is $\text{31}\text{.972072}\text{u}$

Explanation of Solution

Given Info:

The nuclide ${}_{15}{}^{32}\text{P}$ decays by emitting an electron whose kinetic energy can be $1.71\text{MeV}$. Ignore the recoil of the daughter nucleus.

Formula used:

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

Calculation:

${}_{15}{}^{32}\text{P}$ decays by ${\text{β}}^{\text{-}}$ emission, the decay equation is ${}_{15}{}^{32}\text{P}\to {}_{16}{}^{32}\text{S}+{\text{β}}^{\text{-}}+\overline{\nu }$. The daughter nucleus is ${}_{16}{}^{32}\text{S}$

Mass of ${}_{15}{}^{32}\text{P}$ is $31.973908\text{u}$. $1.71\text{MeV}$ of kinetic energy is equivalent to mass $\frac{1.71\text{MeV}}{\frac{931.5\text{MeV}}{\text{u}}}=0.001836\text{u}$. Mass of daughter nucleus ${}_{16}{}^{32}\text{S}$ is $31.973908\text{u}-0.001836\text{u}=\text{31}\text{.972072}\text{u}$.