Chapter 27, Problem 73GP

To determine

The closest distance between $\text{α}$ particles and gold nucleus.

Answer to Problem 73GP

  $4.7\times {10}^{-14}\text{m}$ .

Explanation of Solution

Given:

Kinetic energy of $\text{α}$ particles $=4.8Mev$

Gold nucleus charge $=+79\text{e}$

Formula used:

It is known that: $r_{\min }=\frac{k{Z}_{\alpha }{Z}_{Au}{e}^2}{KE}$

Where,

  $k$ is the constant factor.

  ${\text{Z}}_{\text{α}}$ is the number of positive charge on $\text{α}$ particles.

  ${\text{Z}}_{\text{Au}}$ is the number of positive charge for the gold atom.

  $e$ is charge of electron.

  $r_{\min }$ is the radius.

  $KE$ is the kinetic energy.

Calculation:

At the closest approach kinetic energy of both particles is zero when we ignore the recoil motion.

Now, substitute the data into the formula $r_{\min }=\frac{k{Z}_{\alpha }{Z}_{Au}{e}^2}{KE}$ ,

  $\therefore r_{\min }=\frac{\left(9.00\times {10}^9\text{}\text{N}\cdot {\text{m}}^2/C^2\right)\left(2\right)\left(79\right){\left(1.60\times {10}^{-19}\text{C}\right)}^2}{\left(4.8\text{Mev}\right)\left(1.60\times {10}^{-13}\text{J/Mev}\right)}$

  $\therefore r_{\min }=\frac{\left(9.00\times {10}^9\text{}\text{N}\cdot {\text{m}}^2/C^2\right)\times 158\times 2.56\times {10}^{-38}C}{\left(4.8\text{Mev}\right)\left(1.60\times {10}^{-13}\text{J/Mev}\right)}$

  $\therefore r_{\min }=4.7\times {10}^{-14}m$

Conclusion: The closest distance is $4.7\times {10}^{-14}m$ that $\text{α}$ particle can get to the gold nucleus.