daughter nucleus

[logistic_guy]

A $\displaystyle {}^{232}_{\ \ 92}\text{U}$ nucleus emits an $\displaystyle \alpha$ particle with kinetic energy $\displaystyle = 5.32 \ \text{MeV}$. What is the daughter nucleus and what is the approximate atomic mass (in u) of the daughter atom? Ignore recoil of the daughter nucleus.

 

[khansaheb]

A $\displaystyle {}^{232}_{\ \ 92}\text{U}$ nucleus emits an $\displaystyle \alpha$ particle with kinetic energy $\displaystyle = 5.32 \ \text{MeV}$. What is the daughter nucleus and what is the approximate atomic mass (in u) of the daughter atom? Ignore recoil of the daughter nucleus.

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[logistic_guy]

A $\displaystyle {}^{232}_{\ \ 92}\text{U}$ nucleus emits an $\displaystyle \alpha$ particle with kinetic energy $\displaystyle = 5.32 \ \text{MeV}$. What is the daughter nucleus and what is the approximate atomic mass (in u) of the daughter atom? Ignore recoil of the daughter nucleus.

An $\displaystyle \alpha$ particle is just a helium nucleus. In other words $\displaystyle \alpha = {}^{4}_{2}\text{He}^{2+}$. For shortcut, they just write it like a helium atom, ie, $\displaystyle {}^{4}_{2}\text{He}$.

Let $\displaystyle {}^{A}_{Z}\text{X}$ be the daughter nucleus, then, we have this reaction:

$\displaystyle {}^{232}_{\ \ 92}\text{U} \rightarrow {}^{A}_{Z}\text{X} + \alpha$

Or

$\displaystyle {}^{232}_{\ \ 92}\text{U} \rightarrow {}^{A}_{Z}\text{X} + {}^{4}_{2}\text{He}$

This gives:

$\displaystyle {}^{232}_{\ \ 92}\text{U} \rightarrow {}^{228}_{\ \ 90}\text{X} + {}^{4}_{2}\text{He}$

So the daughter nucleus $\displaystyle {}^{228}_{\ \ 90}\text{X}$ has an atomic number $\displaystyle Z = 90$. Let us look at the periodic table and find out what is this element.



It is thorium $\displaystyle (\text{Th})$. Then the daughter nucleus is:

$\displaystyle {}^{228}_{\ \ 90}\text{Th}$

 

[logistic_guy]

what is the approximate atomic mass (in u) of the daughter atom? Ignore recoil of the daughter nucleus.

Since recoil is ignored, we use only the Q-value equation.

$\displaystyle K = \bigg[M\left({}^{232}\text{U}\right) - M\left({}^{228}\text{Th}\right) - M\left({}^{4}\text{He}\right)\bigg]c^2$

$\displaystyle 5.32 \ \text{MeV} = \bigg[(232.037146 \ \text{u}) - M\left({}^{228}\text{Th}\right) - (4.002603 \ \text{u})\bigg]c^2 \times \frac{931.5 \ \text{MeV}}{\text{u}c^2}$

This gives:

$\displaystyle M\left({}^{228}\text{Th}\right) = \textcolor{blue}{228.028832 \ \text{u}}$