Question

At what speed would a particles lave to be directed at gold foil if some are to contact gold nuclei?

Short answer

The speed is, $v=3.95·{10}^7\frac{\text{m}}{\text{s}}$

Step-by-step solution

Given data

Atomic number of gold$Z=79$

Radius of gold nucleus$r=7·{10}^{-15}\text{m}$

Mass of the alpha particle $m=6.644·{10}^{-27}\text{kg}$.

Concept of Isotopes

Kinetic energy of particle with mass mmoving at speedvis equal to:

$E_k=\frac{1}{2}·m·v^2$

Potential energy of charged projectile particle$q_p$ in electric field of nucleus with atomic numberZis given by:

$E_p=k·\frac{q_p·Z·e}{r}$

where $k=8.98·{10}^9\frac{{\text{Nm}}^2}{{\text{C}}^2}$is Coulomb's constant, $e=1.6·{10}^{-19}\text{C}$is charge of each proton in nucleus andris distance of charged projectile particle from center of observed nucleus.

Comparison of atomic mass and average mass of boron

We will obtain expression for required speed of the alpha particle from equation above. We will first multiply both sides of equation above by$\frac{2}{m}$:

$\begin{array}{rcl}\frac{1}{2}·m·v^2·\frac{2}{m}& =& k·\frac{2·e·Z·e}{r}·\frac{2}{m}{v}^2\\ & =& k·\frac{2·e·Z·e}{r}·\frac{2}{m}\\ & =& k·\frac{4·e·Z·e}{r·m}\\ & & \end{array}$

We will take square root of both sides of equation above:

$\begin{array}{rcl}\sqrt{v^2}& =& \sqrt{k·\frac{4·e·Z·e}{r·m}}\\ v& =& \sqrt{k·\frac{4·e·Z·e}{r·m}}\\ & =& \sqrt{k·\frac{4·e^2·Z}{r·m}}\\ & & \end{array}$

We will plug in numeric values to right side of equation above:

$\begin{array}{rcl}v& =& \sqrt{8.98·{10}^9\frac{{\text{Nm}}^2}{{\text{C}}^2}·\frac{4·{\left(1.6·{10}^{-19}\text{C}\right)}^2·79}{7·{10}^{-15}\text{m}·6.644·{10}^{-27}\text{kg}}}\\ & =& 3.95·{10}^7\frac{\text{m}}{\text{s}}\\ & & \end{array}$