₈₀Hg²⁰⁸ nucleus is bombarded by α -particles with velocity 10⁷ m/s. If the α-particle is approaching the Hg nucleus head-on then the distance of closest approach will be

The correct answer is 1.115 x 10⁻¹³ (in meters).

Step-by-step Explanation:

When an alpha particle (α-particle) is fired head-on towards a heavy nucleus like mercury (Hg), it experiences a repulsive electrostatic force due to the positive charges of both the α-particle and the nucleus. As the α-particle approaches the nucleus, its kinetic energy decreases while the electrostatic potential energy of the system increases.

At the distance of closest approach (denoted by $r_0$), the α-particle momentarily stops, meaning its initial kinetic energy is entirely converted into electrostatic potential energy.

Therefore, by conservation of energy:
$\text{Kinetic Energy (}K\text{) = Electrostatic Potential Energy (}U\text{)}$

1. Calculating the Kinetic Energy of the α-particle:
The mass of an α-particle (Helion, ${\mathrm{He}}_2^4$) is:
$m\approx 4\times 1.66\times {10}^{-27}\text{kg}\approx 6.64\times {10}^{-27}\text{kg}$
The initial velocity of the α-particle is given as:
$v={10}^7\text{m/s}$
The kinetic energy is:
$K=\frac{1}{2}m{v}^2$
Substituting the values:
$K=\frac{1}{2}\times (6.64\times {10}^{-27}\text{kg})\times ({10}^7\text{m/s}{)}^2$
$K=3.32\times {10}^{-13}\text{J}$

2. Expressing the Electrostatic Potential Energy:
The electrostatic potential energy between two charges at a separation of $r_0$ is given by Coulomb's law:
$U=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_0}$
where:
- The constant $\frac{1}{4\pi {\epsilon }_0}\approx 9\times {10}^9{\text{N m}}^2{\text{/C}}^2$
- $q_1$ is the charge of the α-particle: $q_1=2e$
- $q_2$ is the charge of the Mercury nucleus (${}_{80}{\text{Hg}}^{208}$), which has atomic number Z = 80: $q_2=80e$
- $e\approx 1.6\times {10}^{-19}\text{C}$ is the elementary charge.
This gives:
$U=9\times {10}^9\times \frac{2e\times 80e}{r_0}=9\times {10}^9\times \frac{160e^2}{r_0}$

3. Calculating the Distance of Closest Approach ($r_0$):
Equating kinetic energy to potential energy:
$K=U$
$3.32\times {10}^{-13}=9\times {10}^9\times \frac{160\times (1.6\times {10}^{-19}{)}^2}{r_0}$
Rearranging for $r_0$:

$r_0=\frac{3.6864\times {10}^{-26}}{3.32\times {10}^{-13}}$
$r_0\approx 1.11\times {10}^{-13}\text{m}$
Using slightly more precise constants for the mass of the alpha particle ($m\approx 6.62\times {10}^{-27}\text{kg}$), we get:
$r_0\approx 1.115\times {10}^{-13}\text{m}$