Question

Characteristic nuclear time is a useful but loosely defined quantity, taken to be the time required for a nucleon with a few million electron-volts of kinetic energy to travel a distance equal to the diameter of a middle-mass nuclide. What is the order of magnitude of this quantity? Consider 5 MeVneutrons traversing a nuclear diameter ${\mathbf{}}^{\mathbf{197}}\mathbf{Au}$of; use Eq. 42-3.

Short answer

The order of magnitude of this quantity is ${10}^{-22}$.

Step-by-step solution

The given data

Kinetic energy of the neutrons, K = 5 MeV

Understanding the concept of classical physics

In the classical concept of motion, the time taken to cover a distance is given by distance per speed of the body. Now, as the body is in motion, it exhibits kinetic motion. Thus kinetic energy can be used o calculated the speed of the motion.

Formulae:

The time taken by the body to travel,$t=\frac{d}{v}.....\left(1\right)$

The kinetic energy of body in motion,$K=\frac{1}{2}m{v}^2$

The radius of an atom of mass number, A in a nucleus,

$r=r_0{A}^{1/3}.......\left(3\right)\mathrm{where}{r}_0=1.2\mathrm{fm}$

a) Calculation of the order of magnitude of time

Here, the distance travelled by the neutrons is the nuclear diameter of the nuclide gold, thus, substituting this value with the speed equation from equation (2) in equation (1), the time taken for the travel can be given as follows:

$$\begin{gathered} \mathrm{t}=\frac{2\mathrm{r}}{\sqrt{2\mathrm{k}/{\mathrm{m}}_{\mathrm{n}}}}({\mathrm{m}}_{\mathrm{n}}=\mathrm{massof}\mathrm{neutron}) \\ =\frac{\mathrm{r}}{\mathrm{c}}\sqrt{\frac{2{\mathrm{mc}}^2}{\mathrm{K}}} \\ =\frac{(1.2\times {10}^{-15}\mathrm{m}\times {\left(197\right)}^{1/3}}{\mathrm{c}}\sqrt{\frac{2(931.5\mathrm{MeV})}{5\mathrm{MeV}}}(\because \mathrm{from}\mathrm{equation}(3),\mathrm{r}={\mathrm{r}}_0{\mathrm{A}}^{1/3}) \\ =4\times {10}^{-22}\mathrm{s} \end{gathered}$$

Hence, the order of magnitude is ${10}^{-22}$.