Question 133: Atoms & Nuclei

Question

$7.9 \mathrm{MeV} \alpha$-particle scatters from a target material of atomic number 79 . From the given data the estimated diameter of nuclei of the target material is (approximately) $\_\_\_\_$ m.
$$ \left[\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 	imes 10^9 \mathrm{Nm}^2 / \mathrm{C}^2 	ext { and electron charge }=1.6 	imes 10^{-19} \mathrm{C}ight] $$

Accepted Answer

Let an $\alpha$-particle fired directly at the center of a target nucleus.
The $\alpha$-particle has maximum kinetic energy $(\mathrm{K})$ and practically zero electrostatic potential energy at very far away from target material.
As it gets closer, the positive charge of the $\alpha$-particle and the positive charge of the target nucleus repel each other. The $\alpha$-particle slows down, converting its kinetic energy into electrostatic potential energy.
At the closest approach ( $\mathrm{r}_0$ ) it momentarily stops before reversing direction. At this exact point, all its initial kinetic energy has been converted into potential energy.
By the Law of Conservation of Energy:
Initial Kinetic Energy $\left(\mathrm{K}_0ight)=$ Final Potential Energy $(\mathrm{U})$
The electrostatic potential energy between two point charges ( $\mathrm{q}_1$ and $\mathrm{q}_2$ ) separated by a distance r is :
$$ U=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r} $$
An $\alpha$-particle is a Helium nucleus, so it has 2 protons. So, the charge of $\alpha$-particle $\left(\mathrm{q}_1ight)$ is $\mathrm{q}_1=+2 \mathrm{e}$
The atomic number of the target is $\mathrm{Z}=79$, so it has 79 protons, So, the charge of target nucleus $\left(\mathrm{q}_2ight)$ is $\mathrm{q}_2= +79 \mathrm{e}$
Substituting into our energy conservation equation :
$$ \mathrm{K}_0=\frac{1}{4 \pi \epsilon_0} \frac{(2 \mathrm{e})(79 \mathrm{e})}{\mathrm{r}_0} $$
$$\Rightarrow $$ $$ \mathrm{r}_0=\frac{1}{4 \pi \epsilon_0} \frac{158 \mathrm{e}^2}{\mathrm{~K}_0} $$
The kinetic energy of alpha particle is,
$$ \mathrm{K}_0=7.9 \mathrm{MeV}=7.9 	imes 10^6 \mathrm{eV} $$
Since $1 \mathrm{eV}=1.6 	imes 10^{-19} \mathrm{~J}$, we get:
$$ K_0=\left(7.9 	imes 10^6ight) 	imes\left(1.6 	imes 10^{-19}ight) \mathrm{J} $$
$$\Rightarrow $$ $$ \mathrm{K}_0=12.64 	imes 10^{-13} \mathrm{~J} $$
Putting in the closest approach formula,
$$ r_0=\frac{\left(9 	imes 10^9ight) 	imes 158 	imes\left(1.6 	imes 10^{-19}ight)^2}{12.64 	imes 10^{-13}} $$
$$\Rightarrow $$ $r_0=\frac{(9 	imes 158 	imes 2.56) 	imes 10^{-29}}{12.64 	imes 10^{-13}}$
$$\Rightarrow $$ $\mathrm{r}_0=288 	imes 10^{-16} \mathrm{~m}$
$$\Rightarrow $$ $$ \mathrm{r}_0=2.88 	imes 10^{-14} \mathrm{~m} $$
Therefore, the distance of closest approach is $\mathrm{r}_0=2.88 	imes 10^{-14} \mathrm{~m}$.
The distance from the centre of the nucleus to this stopping point is the distance of closest approach. Because the $\alpha$-particle cannot physically penetrate the nucleus using just the electrostatic force in this classical model, $r_0$ gives us an upper estimate for the radius of the target nucleus.
Diameter $=2 	imes$ Radius
Diameter $=2 	imes \mathrm{r}_0$
Diameter $=2 	imes\left(2.88 	imes 10^{-14} \mathrm{~m}ight)$
Diameter $=5.76 	imes 10^{-14} \mathrm{~m}$
Therefore, the diameter of nuclei of the target material is approximately $5.76 	imes 10^{-14} \mathrm{~m}$.
Hence, the correct option is (B).

Answer

$2.88 	imes 10^{-14}$

Answer

$5.76 	imes 10^{-14}$

Answer

$1.44 	imes 10^{-13}$

Answer

$1.69 	imes 10^{-12}$

Atoms & Nuclei

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