Q31.4-38PE - College Physics (Urone)

Question

(a) Write the complete \({{\rm{\beta }}^{\rm{ - }}}\) decay equation for \({}^{{\rm{90}}}{\rm{Sr}}\) , a major waste product of nuclear reactors. (b) Find the energy released in the decay.

Step-by-Step Solution

Answer

(a) The equation is \({}_{38}^{90}S{r_{62}} \to {}_{39}^{90}{Y_{61}} + \beta \).

(b) The energy released is \(0.0477\,{\rm{MeV}}\).

Define energy

Energy is a measurable attribute that may be transferred from one thing to another in order for it to do work.

Explanation

(a) The atomic number of the nucleus increases by one during \({\rm{\beta }}\) decay, whereas the mass number stays same. 

As a result, the \({\rm{\beta }}\) decay equation is,

\({}_{38}^{90}S{r_{62}} \to {}_{39}^{90}{Y_{61}} + \beta \)

Evaluating the energy

(b) The \({}^{{\rm{90}}}{\rm{Sr}}\)has a mass of \(89.9077\,{\rm{u}}\).

\({}^{{\rm{90}}}{\rm{Y}}\)has a mass of \(89.9071\,{\rm{u}}\).

While an electron has a mass of \(0.0005485\,{\rm{u}}\).

As a result, the mass difference is

\(\begin{array}{c}\Delta m = (89.9077\,{\rm{u}}) - [(89.9071\,{\rm{u}}) + (0.00054858\,{\rm{u}})]\\ = 5.12 \times {10^{ - 5}}\,{\rm{u}}\end{array}\)

 

The energy released is

\(\begin{array}{c}E = \Delta m{c^2}\\E = (5.12 \times {10^{ - 5}}\,{\rm{u}})(931.5\,{\rm{MeV/u}}{{\rm{c}}^{\rm{2}}}) \times {c^2}\\ = 0.0477\,{\rm{MeV}}\end{array}\)

Therefore, the energy released is \(0.0477\,{\rm{MeV}}\).