A measurement of the energy Eof an intermediate nucleus must be made within the mean lifetime $\mathbf{∆}\mathit{t}$of the nucleus and necessarily carries an uncertainty $\mathbf{∆}\mathit{E}$according to the uncertainty principle

$\mathbf{∆}\mathit{E}\mathbf{·}\mathbf{∆}\mathit{t}\mathbf{=}\sout{\mathbf{h}}$.

(a) What is the uncertainty $\mathbf{∆}\mathit{E}$in the energy for an intermediate nucleus if the nucleus has a mean lifetime of${\mathbf{10}}^{\mathbf{-}\mathbf{22}}\mathbf{}\mathit{s}$? (b) Is the nucleus a compound nucleus?

Mean lifetime of the nucleus,$t_{avg}={10}^{-22}s$

The uncertainty principle given by Heinsberg states that an electron's position and velocity can be measured. At the same time, not even in theory.

A compound nucleus is an unstable nucleus formed by the coalescence of an atomic nucleus with a captured particle.

Formula:

The energy-time uncertainty relation,$∆E·∆t=\sout{h}or∆E·∆t=h/2\mathrm{\pi }........\left(1\right)$

where, h is the Planck’s constant.

Using the given data in equation (1), we can get the uncertainty in energy for an intermediate nucleus as follows:

$$\begin{gathered} ∆E\approx \frac{h/2\mathrm{\pi }}{t_{avg}} \\ =\frac{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)/2\mathrm{\pi }}{\left(1.6\times {10}^{-19}\mathrm{J}.\mathrm{eV}\right)\left(1.0\times {10}^{-22}\mathrm{s}\right)} \\ \approx 6.6\times {10}^6\mathrm{eV} \\ =6.6\mathrm{MeV} \end{gathered}$$

Hence, the uncertainty in energy is $6.6\mathrm{MeV}$.

In order to fully distribute the energy in a fairly large nucleus, and create a compound nucleus” equilibrium configuration, about ${10}^{-15}$sis typically required. A reaction state that exists no more than about ${10}^{-22}$sdoes not qualify as a compound nucleus.