Question

Item 4 part a if a proton and an electron are released when they are 4.50×10−10 m apart (typical atomic distances), find the initial acceleration of each of them.

Answer 1

Acceleration is defined as the rate of change of velocity with respect to time.

Formulas of force are given by:

`F=ma` (1)

where,

F = force

m = mass

a = acceleration

`F=(kq_(1)q_(2))/(r^(2))` (2)

where,

F = force

k = coulomb constant (
`9* 10^(9)Nm^(2)C^(-2)`)

r = distance between the charged particles

`q_(1)` and
`q_(2)` are the signed magnitude of charges

Use the formula (2) for calculating the value of force, we get:

Substitute the value of
`q_(1)` and
`q_(2)`, k and r to find the value of force.

F=
`(9* 10^(9) Nm^(2)C^(-2)* (1.6* 10^(-19) C)^(2))/((4.50* 10^(-10))^(2))`

=
`1.1378* 10^(-9) N`

Now, put the above value force in formula (1) to identify the initial acceleration.

`1.1378* 10^(-9) N=ma` (1) (mass of electron =
`9.1 * 10^(-31) kg`)

acceleration of electron =
`(1.1378* 10^(-9) N)/(9.1* 10^(-31)kg)`

=
`1.25* 10^(21)N/kg`

And,

acceleration of proton =
`(1.1378* 10^(-9) N)/(mass of proton)`

mass of proton =
`1.67 *  10^(-27)`

Thus,

acceleration of proton =
`(1.1378* 10^(-9) N)/(1.67 *  10^(-27) kg)`

=
`6.81 *  10^(17) N/kg`

Now, initial acceleration of electron and proton is
`1.25* 10^(21)m/s^(2)` and
`6.81 *  10^(17) m/s^(2)` as 1 N =
`kgm/s^(2)`.