Question

An unstable nucleus undergoes alpha decay with the release

of 5.52 MeV of energy. The combined mass of the parent and

daughter nuclei is 452 u. What was the parent nucleus?

Short answer

Therefore, the parent$\text{nucleus with mass of}228\mathrm{u}\text{is}{}^{228}\mathrm{Th}\text{.}$

Step-by-step solution

Given information

$$\begin{gathered} K_{\alpha }=5.52\mathrm{MeV} \\ m\left({}^{4}H_e\right)=4.002602\mathrm{u} \\ {\mathrm{m}}_{\mathrm{X}}+{\mathrm{m}}_{\mathrm{Y}}=452\mathrm{u} \end{gathered}$$

Explanation

We use the formula:

$$\begin{gathered} K_{\alpha }=\Delta m{c}^2 \\ {K}_{\alpha }=\left(m_X-m_Y-m{H}_e\right)c^2 \\ 5.52\mathrm{MeV}=\left(m_X-m_Y-m{H}_e\right) \\ 5.52\frac{\mathrm{MeV}}{{\mathrm{c}}^2}\left(\frac{1\mathrm{u}}{931.5\frac{\mathrm{MeV}}{{\mathrm{c}}^2}}\right)=\left(m_X-m_Y-m{H}_e\right) \\ 0.0059259\mathrm{u}=\left(m_X-m_Y-m{H}_e\right) \\ {m}_X-m_Y=0.0059259\mathrm{u}+m\left({}^{4}H_e\right) \\ {m}_X-m_Y=0.0059259\mathrm{u}+4.002602\mathrm{u} \\ {m}_X-m_Y=4.0085279\mathrm{u} \end{gathered}$$

We got a system of two equations:

$$\begin{gathered} m_X+m_Y=452\mathrm{u} \\ {\mathrm{m}}_{\mathrm{X}}-{\mathrm{m}}_{\mathrm{Y}}=4.0085279\mathrm{u} \end{gathered}$$

Calculations

We will find the value of m_X:

$$\begin{gathered} 2m_X=452\mathrm{u}+4.0085279\mathrm{u} \\ 2{\mathrm{m}}_{\mathrm{X}}=456.008529\mathrm{u} \\ {\mathrm{m}}_{\mathrm{X}}=\frac{456.008529\mathrm{u}}{2} \\ \Rightarrow {\mathrm{m}}_{\mathrm{X}}=228.0042645\mathrm{u} \end{gathered}$$

$\text{So, nucleus with mass of}228\mathrm{u}\text{is}{}^{228}\mathrm{Th}\text{.}$