Question

a) Calculate the energy released in the $\alpha$ decay of ${}^{238}\mathrm{U}$ . (b) What fraction of the mass of a single ${}^{238}\mathrm{U}$ is destroyed in the decay? The mass of ${}^{234}\mathrm{T}\mathrm{h}$ is $234.043593\mathrm{u}$. (c) Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?

Short answer

a) The energy released is$4.2774\mathrm{M}\mathrm{e}\mathrm{V}$.

b) The fraction of mass destroyed is $1.92\times {10}^{-5}$.

c) The mass loss is almost unnoticeable.

Step-by-step solution

Define energy

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

Evaluating the energy

a) The daughter nucleus of ${}^{238}\mathrm{U}$, $\alpha$decay is ${}^{234}\mathrm{T}\mathrm{h}$.The nuclei's masses are,

$\begin{array}{rl}m({}^{238}U)& =238.0507882\mathrm{u}\\ m({}^{234}Th)& =234.043593\mathrm{u}\\ m(\alpha )& =4.00260325415\mathrm{u}\end{array}$

As a result, the mass difference is,

$\begin{array}{rl}\mathrm{\Delta }m& =m({}^{238}U)-\{(m({}^{234}Th))+(m(\alpha ))\}\\ & =\left(238.0507882\mathrm{u}\right)-\{(234.043593\mathrm{u})+(4.00260325415\mathrm{u})\}\\ & =0.0045919\mathrm{u}\end{array}$

We can calculate the energy released during the alpha decay.

$\begin{array}{rl}E& =\mathrm{\Delta }m{c}^2\\ E& =(0.0045919\mathrm{u})(931.5\mathrm{M}\mathrm{e}\mathrm{V}/\mathrm{u}{\mathrm{c}}^2)\times c^2\\ & =4.2774\mathrm{M}\mathrm{e}\mathrm{V}\end{array}$

Therefore, the energy released is $4.2774\mathrm{M}\mathrm{e}\mathrm{V}$

Evaluating the fraction of mass

b) The mass destroyed as a percentage of the total mass is,

$\begin{array}{c}\frac{\mathrm{\Delta }m}{m({}^{238}U)}=\frac{(0.0045919\mathrm{u})}{\left(238.0507882\mathrm{u}\right)}\\ =1.92\times {10}^{-5}\end{array}$

Therefore, the fraction of mass destroyed is $1.92\times {10}^{-5}$.

Explanation

c) Since the half-life of${}^{238}\mathrm{U}$ is so long $\left(4.468\times {10}^9\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}\right)$, just a few atoms will decay in a macroscopic sample of Uranium. As a result, the decrease in mass is barely noticeable.