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

F = 91.27 N

Step-by-step 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 =
`(kqQ)/(r^2)`

where k = Coulomb's constant

Therefore:

`F = (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.