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Found in: Page 1306

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

A measurement of the energy E of an intermediate nucleus must be made within the mean lifetime ${\mathbf{∆}}{\mathbit{t}}$ of the nucleus and necessarily carries an uncertainty ${\mathbf{∆}}{\mathbit{E}}$ according to the uncertainty principle ${\mathbf{∆}}{\mathbit{E}}{\mathbf{·}}{\mathbf{∆}}{\mathbit{t}}{\mathbf{=}}\overline{)\mathbf{h}}$.(a) What is the uncertainty ${\mathbf{∆}}{\mathbit{E}}$ in the energy for an intermediate nucleus if the nucleus has a mean lifetime of ${{\mathbf{10}}}^{\mathbf{-}\mathbf{22}}{\mathbf{}}{\mathbit{s}}$? (b) Is the nucleus a compound nucleus?

1. The uncertainty in the energy for an intermediate nucleus is 6.6 MeV .
2. The nucleus is not a compound nucleus.
See the step by step solution

Step 1: Given data

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

Step 2: Understanding the concept of uncertainty principle

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, $∆E·∆t=\overline{)h}or∆E·∆t=h/2\mathrm{\pi }........\left(1\right)$

where, h is the Planck’s constant.

Step 3: a) Calculation of the uncertainty in energy

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

$∆E\approx \frac{h/2\mathrm{\pi }}{{t}_{avg}}\phantom{\rule{0ex}{0ex}}=\frac{\left(6.626×{10}^{-34}\mathrm{J}.\mathrm{s}\right)/2\mathrm{\pi }}{\left(1.6×{10}^{-19}\mathrm{J}.\mathrm{eV}\right)\left(1.0×{10}^{-22}\mathrm{s}\right)}\phantom{\rule{0ex}{0ex}}\approx 6.6×{10}^{6}\mathrm{eV}\phantom{\rule{0ex}{0ex}}=6.6\mathrm{MeV}$

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

Step 4: b) Calculation for checking if the nucleus is a compound nucleus or not

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