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Found in: Page 1307

### Fundamentals Of Physics

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

# A projectile alpha particle is headed directly toward a target aluminum nucleus. Both objects are assumed to be spheres. What energy is required of the alpha particle if it is to momentarily stop just as its “surface” touches the “surface” of the aluminum nucleus? Assume that the target nucleus remains stationary.

The energy required of the alpha particle is 6.79 MeV.

See the step by step solution

## Step 1: Given data

An alpha particle projected to the aluminum nucleus stops momentarily at a distance.

## Step 2: Understanding the concept of Rutherford scattering experiment

Rutherford scattering experiment shows a beam of high-energy streams of α-particles from a radioactive source directed at a thin sheet (100 nm thickness) of gold. To study the deflection caused to the α-particles, he placed a fluorescent zinc sulfide screen around the thin gold foil.

The distance of the closest approach is defined as the minimum distance of the charged particle from the nucleus at which initial kinetic energy is the same as the potential energy of the nucleus.

Formulae:

The potential energy of the system of charged particles, $P.E=\frac{k{q}_{1}{q}_{2}}{r}\dots ....\dots \left(1\right)$

The radius of an atom of mass number, A in a nucleus, $r={r}_{0}{A}^{1/3}·······\left(2\right)where{r}_{0}=1.2fm$

## Step 3: Calculation of the kinetic energy of alpha particle

Using equation (2), we estimate the nuclear radii of the alpha particle and aluminum as follows:

${r}_{\alpha }=\left(1.2×{10}^{-15}m\right){\left(4\right)}^{1/3}\phantom{\rule{0ex}{0ex}}=1.90×{10}^{-15}m$

${r}_{Al}=\left(1.2×{10}^{-15}m\right){\left(27\right)}^{1/3}\phantom{\rule{0ex}{0ex}}=3.69×{10}^{-15}m$

Thus, the distance between the centers of the nuclei when their surfaces touch is given by:

$\mathrm{r}={\mathrm{r}}_{\mathrm{\alpha }}+{\mathrm{r}}_{\mathrm{Al}}\phantom{\rule{0ex}{0ex}}=1.90×{10}^{-15}\mathrm{m}+3.60×{10}^{-15}\mathrm{m}\phantom{\rule{0ex}{0ex}}=5.50×{10}^{-15}\mathrm{m}$

Now, from the condition of closest distance of approach, the kinetic energy is equal to potential energy of the system and thus is given using equation (i) as follows;

$K.E=\frac{\left(9×{10}^{9}N.{m}^{2}/{C}^{2}\right)\left(2×1.6×{10}^{-19}C\right)\left(13×1.6×{10}^{-19}C\right)}{5.50×{10}^{-15}m}\phantom{\rule{0ex}{0ex}}=1.09×{10}^{-12}J\phantom{\rule{0ex}{0ex}}=6.79×{10}^{6}eV\phantom{\rule{0ex}{0ex}}=6.79MeV$

Hence, the kinetic energy value is 6.79 MeV.