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Q86P

Expert-verifiedFound in: Page 1307

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.

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

**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, $r={r}_{0}{A}^{1/3}\xb7\xb7\xb7\xb7\xb7\xb7\xb7\left(2\right)where{r}_{0}=1.2fm$

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

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

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

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

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

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;

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

Hence, the kinetic energy value is 6.79 MeV.

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