Question 135: Atoms & Nuclei

Question

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is $\_\_\_\_$ m. (Atomic number of gold $=79$ and $\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 	imes 10^9$ in SI units)

Accepted Answer

Let an $\alpha$-particle fired directly at the centre of a target nucleus, gold foil.
The $\alpha$-particle has maximum kinetic energy $(\mathrm{K})$ and practically zero electrostatic potential energy at very far away from target material.
The initial Kinetic Energy ( $\mathrm{K}_0$ ) of the alpha particle is :
$$ \mathrm{K}_0=7.7 \mathrm{MeV} $$
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.
The electrostatic potential energy $(U)$ between two point charges $q_1$ and $q_2$ separated by a distance $r_0$ is :
$$ \mathrm{U}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}_0} $$
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,
$$ \mathrm{K}_0=7.7 \mathrm{MeV}=7.7 	imes 10^6 \mathrm{eV} $$
Since $1 \mathrm{eV}=1.6 	imes 10^{-19} \mathrm{~J}$, we get :
$$ K_0=\left(7.7 	imes 10^6ight) 	imes\left(1.6 	imes 10^{-19}ight) \mathrm{J} $$
$$\Rightarrow $$ $$ \mathrm{K}_0=12.32 	imes 10^{-13} \mathrm{~J} $$
Putting in the closest approach formula,
$$ r_0=\frac{\left(9 	imes 10^9ight) 	imes 158 	imes\left(1.6 	imes 10^{-19}ight)^2}{12.32 	imes 10^{-13}} $$
$$\Rightarrow $$ $r_0=\frac{(9 	imes 158 	imes 2.56) 	imes 10^{-29}}{12.32 	imes 10^{-13}}$
$$\Rightarrow $$ $\mathrm{r}_0=295 	imes 10^{-16} \mathrm{~m}$
$$\Rightarrow $$ $$ \mathrm{r}_0=2.95 	imes 10^{-14} \mathrm{~m} $$
Therefore, the distance of closest approach is $\mathrm{r}_0=2.95 	imes 10^{-14} \mathrm{~m}$.
Hence, the correct option is (C).

Answer

$2.95 	imes 10^{-16}$

Answer

$3.85 	imes 10^{-14}$

Answer

$2.95 	imes 10^{-14}$

Answer

$3.85 	imes 10^{-16}$

Atoms & Nuclei

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is \_\_\_\_ m. (Atomic number of gold =79 and \frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^9 in SI units)

Choose the correct answer: