Question

An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge = +79e). What is the electric force acting on the alpha particle when the alpha particle is 2.0 × 10−14 m from the gold nucleus?​

Answer 1

Answer:

F = 91.27 N

Explanation:

Parameters given:

Charge of alpha particle, q = +2.0e = $2.0 * 1.60217662 * 10^{-19}$ = $3.204 * 10^{-19} C$

Charge of gold nucleus, Q = +79e = $79 * 1.60217662 * 10^{-19}$ = $1.266 * 10^{-17} C$

Distance between the alpha particle and gold nucleus, r = $2.0 * 10^{-14} m$

Electric force is given as:

F = $\frac{kqQ}{r^2}$

where k = Coulomb's constant

Therefore:

$F = \frac{9 * 10^9 * 3.204 * 10^{-19} * 1.266 * 10^{-17}}{(2.0 * 10^{-14})^2} \\\\\\F = 91.27 N$

The electric force acting on the alpha particle when the alpha particle is $2.0 * 10^{-14} m$ from the gold nucleus is 91.27 N.