Question

A nucleus whose mass is $\mathbf{3}\mathbf{.}\mathbf{499612}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{25}}\mathit{k}\mathit{g}$undergoes spontaneous alpha decay. The original nucleus disappears and there appear two new particles: a He-4 nucleus of mass $\mathbf{6}\mathbf{.}\mathbf{640678}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathit{k}\mathit{g}$(an “alpha particle” consisting of two protons and two neutrons) and a new nucleus of mass$\mathbf{3}\mathbf{.}\mathbf{433132}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{25}}\mathit{k}\mathit{g}$ (note that the new nucleus has less mass than the original nucleus, and it has two fewer protons and two fewer neutrons). (a) When the alpha particle has moved far away from the new nucleus (so the electric interactions are negligible), what is the combined kinetic energy of the alpha particle and new nucleus? (b) How many electron volts is this? In contrast to this nuclear reaction, chemical reactions typically involve only a few eV.

Short answer

1. The combined kinetic energy of the alpha particle and new nucleus is$6.5898\times {10}^{-13}J$.
2. The combined kinetic energy of alpha particle and the new nucleus in terms of electron-volt unit is$4.1186\times {10}^{5}eV$ .

Step-by-step solution

Given data

The value of nucleus mass = $3.499612\times {10}^{-25}kg$

The value of He-4 nucleus mass = $6.640678\times {10}^{-27}kg$

The value of a new nucleus mass = $3.433132\times {10}^{-25}kg$

Step 2:(a) Determine the combined kinetic energy of the alpha particle and the new nucleus

The expression of the mass difference between alpha particle and new nucleus is expressed as follows:

$m=Nucleusmass-\left[He-4nucleusmass+newnucleusmass\right]$

Here, represents the mass difference between alpha particle and new nucleus.

Substitute all the known values in the above expression.

$$\begin{gathered} m=\left(3.499612\times {10}^{-25}kg\right)-\left[\left(6.640678\times {10}^{-27}kg\right)+\left(3.433132\times {10}^{-25}\right)\right] \\ =7.322\times {10}^{-30}kg \end{gathered}$$

The expression of the combined kinetic energy of the alpha particle and the new nucleus is expressed as follows:

$E=m{c}^2$

Here, $E$ represents the combined kinetic energy of the alpha particle and the new nucleus and $C$ represents the speed of light in vacuum whose value is$c=3\times {10}^{8}m/s$ .

Substitute all the known values in the above expression.

$$\begin{gathered} E=\left(7.322\times {10}^{-30}\right){\left(3\times {10}^{8}m/s\right)}^2 \\ =6.5898\times {10}^{13}kg{m}^2|s^2 \\ =\left(6.5898\times {10}^{-13}kg{m}^2|s^2\right)\times \left(\frac{1J}{1kg{m}^2|s^2}\right) \\ =6.5898\times {10}^{-130}J \end{gathered}$$

Thus,the combined kinetic energy of the alpha particle and new nucleus is$6.5898\times {10}^{-13}J$ .

(b) Determine the combined kinetic energy of alpha particle and the new nucleus in terms of electron-volt unit.

The relation between electron-volt and joule is expressed as follows:

$1eV=1.6\times {10}^{-13}J$

The combined kinetic energy of alpha particle and the new nucleus in terms of electron-volt unit can be calculated as,

$E=6.5898\times {10}^{-13}J$

role="math" localid="1654102911061" $$\begin{gathered} =\left(6.5898\times {10}^{-13}J\right)\left(\frac{{\displaystyle \frac{1}{1.6\times {10}^{-19}}}eV}{1J}\right) \\ =4.1186\times {10}^{5}eV \end{gathered}$$

Thus, the combined kinetic energy of alpha particle and the new nucleus in terms of electron-volt unit is$4.1186\times {10}^{5}eV$ .