daughter nucleus

logistic_guy

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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.
 
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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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:

\(\displaystyle {}^{228}_{\ \ 90}\text{Th}\)
 
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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:

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