Question

An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the electron be moving when it is 3.00 cm from the proton?

Answer 1

The speed of the electron is 91.86 m/s.

Step-by-step explanation:

Given that,

Distance of electron from proton = 6.00 cm

Distance of proton = 3.00 cm

We need to calculate the initial potential energy

`U_(1)=(kq_(1)q_(2))/(r)`

Put the value into the formula

`U_(i)=(9*10^(9)*(1.6*10^(-19))^2)/(6.00*10^(-2))`

`U_(i)=3.84*10^(-27)\ J`

The final potential energy

`U_(f)=(9*10^(9)*(1.6*10^(-19))^2)/(3.00*10^(-2))`

`U_(f)=7.68*10^(-27)\ J`

We need to calculate the speed of the electron

Using conservation of energy

`(1)/(2)mv^2=U_(f)-U_(i)`

Put the value into the formula

`(1)/(2)*9.1*10^(-31)* v^2=7.68*10^(-27)-3.84*10^(-27)`

`v=\sqrt{(2*3.84*10^(-27))/(9.1*10^(-31))}`

`v=91.86\ m/s`

Hence, The speed of the electron is 91.86 m/s.

Answer 2

electron moving at 91.82 m/s

Step-by-step explanation:

given data

distance d1 = 6 cm

distance d2 = 3 cm

to find out

how fast will the electron be moving

solution

we know potential energy formula that is

potential energy =
`k(q1q2)/(r)`

here k is 9×
`10^(9)` N-m²/C²

m mass of electron is 9.11 ×
`10^(-31)` kg

and q = 1.60 ×
`10^(-19)` C

we consider here initial potential energy = U1

U1 =
`9*10^(9)(1.60*10^(-19)*1.60*10^(-19))/(0.06^2)`

U1 = 3.84 ×
`10^(-27)` J

and final potential energy = U2

U2 =
`9*10^(9)(1.60*10^(-19)*1.60*10^(-19))/(0.03)`

U2 = 7.68 ×
`10^(-27)` J

and speed of electron = v

so we will apply here conservation of energy

0.5×m×v² = U2 - U1 ................1

so

0.5×9.11×
`10^(-31)` ×v² = 3.84 ×
`10^(-27)`

v = 91.82 m/s

so electron moving at 91.82 m/s