Question

The nuclide ${}_{\mathbf{92}}{}^{\mathbf{231}}\mathbf{U}$converts spontaneously to ${}_{\mathbf{91}}{}^{\mathbf{231}}\mathbf{Pa}$.

1. Write two balanced nuclear equations for this conversion, one if it proceeds by electron capture and the other if it proceeds by positron emission.

2. Using the nuclidic masses in Table 19.1, calculate the change in mass for each process. Explain why electron capture can occur spontaneously in this case but positron emission cannot.

Short answer

1. The balanced equation for electron capture is:${}_{92}{}^{231}\mathrm{U}+{}_{-1}{}^0\mathrm{e}^{-1}\to {}_{91}{}^{231}\mathrm{Pa}+\mathrm{\nu }$

The balanced equation for positron emission is:

${}_{92}{}^{231}\mathrm{U}\to {}_{91}{}^{231}\mathrm{Pa}+{}_1{}^0\mathrm{e}^{+}+\mathrm{\nu }$

b. The positron emission is non-spontaneous because $∆\mathrm{m}>0$while the electron capture process is spontaneous because$∆\mathrm{m}<0$.

Step-by-step solution

Nuclear equation representing electron capture

When a nucleus captures an electron from the environment, a proton is changed into a neutron. As a result, the mass number remains constant while the atomic number increases by one. A neutrino is emitted along with the electron capture. The balanced nuclear equation representing electron capture is given below.

${}_{92}{}^{231}\mathrm{U}+{}_{-1}{}^0\mathrm{e}^{-1}\to {}_{91}{}^{231}\mathrm{Pa}+\mathrm{\nu }$

(a) A nuclear equation representing positron emission

When the number of protons is more than the number of neutrons in a nucleus, positron emission takes place to stabilize the nucleus. In a positron emission, a proton is transformed into a neutron along with the emission of a high-energy positron accompanied by a neutrino. No change in mass number is observed for a positron emission. The atomic number of the daughter nucleus is one less than that of the parent nucleus. The balanced nuclear equation representing positron emission is:

${}_{92}{}^{231}\mathrm{U}\to {}_{91}{}^{231}\mathrm{Pa}+{}_1{}^0\mathrm{e}^{+}+\mathrm{\nu }$

Mass difference

The change in mass due to positron emission is:

$\begin{array}{rcl}∆\mathrm{m}& =& \mathrm{m}\left({}_{91}{}^{231}\mathrm{Pa}\right)+2\mathrm{m}\left({}_1{}^0\mathrm{e}^{+}\right)-\mathrm{m}\left({}_{92}{}^{231}\mathrm{U}\right)\\ & =& \left(231.0358+2\times 0.000548-231.0362\right)\mathrm{u}\\ & =& 0.000696\mathrm{u}\end{array}$

The change of mass due to electron capture is:

$\begin{array}{rcl}∆\mathrm{m}& =& \mathrm{m}\left({}_{91}{}^{231}\mathrm{Pa}\right)-\mathrm{m}\left({}_{92}{}^{231}\mathrm{U}\right)\\ & =& \left(231.0358-231.0362\right)\mathrm{u}\\ & =& -0.0004\mathrm{u}\end{array}$

(b) Spontaneity of a nuclear reaction

For a reaction to occur spontaneously, $∆\mathrm{E}<0$.

Since, $∆\mathrm{E}={\mathrm{c}}^2∆\mathrm{m}$, $∆\mathrm{m}<0$indicates that the reaction should be spontaneous.

In the case of positron emission $∆\mathrm{m}>0$and therefore the reaction will be non-spontaneous.

Whereas for electron capture $∆\mathrm{m}<0$and hence the reaction will be spontaneous.