Question

A radioactive nucleus at rest decays into a second nucleus, an electron, and a neutrino. The electron and neutrino are emitted at right angles and have momenta of \(9.6 \times {10^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}\) and \(6.2 \times {10^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}\), respectively. Determine the magnitude and the direction of the momentum of the second (recoiling) nucleus.

Short answer

The momentum of the second (recoil) nucleus is \(11.4 \times {10^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}\) at an angle \({147.14^ \circ }\) with the direction of the electron.

Step-by-step solution

Given data

Initially, the mother nucleus was at rest; therefore, the momentum after decay is zero according to momentum conservation.

The momentum of the electron is \({P_{\rm{e}}} = 9.6 \times {10^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}\) and the momentum of the neutrino is \({P_{\rm{n}}} = 6.2 \times {10^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}\).

Let \({P_{\rm{r}}}\) be the momentum of the second recoiling nucleus at an angle \(\theta \) with the negative y-axis (clockwise), .i.e. the angle of the recoil nucleus is \(\left( {{{90}^ \circ } + \theta } \right)\)with the emitted electron.

Momentum conservation

The momentum before the decay at any direction is zero as the parent nucleus is at rest.

Using the momentum conservation along the x-axis, you get:

\(\begin{array}{c}{P_{\rm{e}}} - {P_{\rm{r}}}\sin \theta = 0\\{P_{\rm{r}}}\sin \theta = {P_{\rm{e}}}\end{array}\) … (i)

Using the momentum conservation along the y-axis, you get:

\(\begin{array}{c}{P_{\rm{n}}} - {P_{\rm{r}}}\cos \theta = 0\\{P_{\rm{r}}}\cos \theta = {P_{\rm{n}}}\end{array}\) … (ii)

Now, dividing equation (i) by (ii), you get:

\(\begin{array}{c}\frac{{{P_{\rm{r}}}\sin \theta }}{{{P_{\rm{r}}}\cos \theta }} = \frac{{{P_{\rm{e}}}}}{{{P_{\rm{n}}}}}\\\tan \theta = \frac{{9.6 \times {{10}^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}}}{{6.2 \times {{10}^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}}}\\\theta = {\tan ^{ - 1}}\left( {\frac{{9.6}}{{6.2}}} \right)\\\theta = {57.14^ \circ }\end{array}\)

Now, substituting the value of \(\theta \) in equation (i), you get:

\(\begin{array}{c}{P_{\rm{r}}}\sin {57.14^ \circ } = 9.6 \times {10^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}\\{P_{\rm{r}}} = 11.4 \times {10^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}\end{array}\)

Hence, the momentum of the second (recoil) nucleus is \(11.4 \times {10^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}\) at an angle \({147.14^ \circ }\) with the direction of the electron.