91Ó°ÊÓ

Atomic number of gold$Z=79$

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

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

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.

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: