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 $\frac{1}{4\pi\epsilon_{\mathrm{o}}}=9 imes10^9$ in SI units)

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 $\frac{1}{4\pi\epsilon_{\mathrm{o}}}=9	imes10^9$ in SI units)

Accepted Answer

B. 1. Consider an $\alpha$-particle moving directly toward the center of a gold nucleus.

At a very large distance, its electrostatic potential energy is nearly zero, so its energy is entirely kinetic:

$$
K_0=7.7\ \mathrm{MeV}.
$$

2. As the positively charged $\alpha$-particle approaches the positively charged gold nucleus, electrostatic repulsion slows it down.

At the **distance of closest approach** $r_0$, the particle momentarily comes to rest. Therefore, by the **law of conservation of energy**,

$$
	ext{Initial Kinetic Energy}
=
	ext{EPE}.
$$

Hence,

$$
\begin{aligned}
K_0
&=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_0}.
\end{aligned}
$$

3. An $\alpha$-particle has charge

$$
q_1=+2e,
$$

and a gold nucleus ($Z=79$) has charge

$$
q_2=+79e.
$$

Substituting these values,

$$
\begin{aligned}
K_0
&=\frac{1}{4\pi\epsilon_0}\frac{(2e)(79e)}{r_0},
\end{aligned}
$$

which gives

$$
\begin{aligned}
r_0
&=\frac{1}{4\pi\epsilon_0}\frac{158e^2}{K_0}.
\end{aligned}
$$

4. Convert the kinetic energy into SI units:

$$
\begin{aligned}
K_0
&=7.7	imes10^6\ \mathrm{eV} \
&=(7.7	imes10^6)(1.6	imes10^{-19})\ \mathrm{J} \
&=12.32	imes10^{-13}\ \mathrm{J}.
\end{aligned}
$$

5. Substitute the given values:

$$
\begin{aligned}
r_0
&=\frac{(9	imes10^9)	imes158	imes(1.6	imes10^{-19})^2}
{12.32	imes10^{-13}} \
&=\frac{(9	imes158	imes2.56)	imes10^{-29}}
{12.32	imes10^{-13}} \
&=295	imes10^{-16}\ \mathrm{m} \
&=2.95	imes10^{-14}\ \mathrm{m}.
\end{aligned}
$$

6. Therefore, the distance of closest approach is

$$
r_0=2.95	imes10^{-14}\ \mathrm{m}.
$$

Hence, the correct option is **(B)**.

If an alpha particle with energy 7.77.77.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold =79=79=79 and 14πϵo=9×109\frac{1}{4\pi\epsilon_{\mathrm{o}}}=9\times10^94πϵo​1​=9×109 in SI units)

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 $\frac{1}{4\pi\epsilon_{\mathrm{o}}}=9\times10^9$ in SI units)