Question

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When the proton is a distance R from the nucleus its velocity has decreased to 1/2vo. How far from the nucleus will the proton be when its velocity has dropped to 1/4vo

Answer 1

The value is
`R_f = (4)/(5) R`

Step-by-step explanation:

From the question we are told that

The initial velocity of the proton is
`v_o`

At a distance R from the nucleus the velocity is
`v_1 = (1)/(2) v_o`

The velocity considered is
`v_2 = (1)/(4) v_o`

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

`\Delta K = \Delta P`

Here
`\Delta K` is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

`\Delta K = K__(R)} - K_i`

=>
`\Delta K = (1)/(2) * m * v_1^2 - (1)/(2) * m * v_o^2`

=>
`\Delta K = (1)/(2) * m * ((1)/(2) * v_o )^2 - (1)/(2) * m * v_o^2`

=>
`\Delta K = (1)/(2) * m * (1)/(4) * v_o ^2 - (1)/(2) * m * v_o^2`

And
`\Delta P` is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

`\Delta P = P_f - P_i`

Here
`P_i` is zero because the electric potential energy at the initial stage is zero so

`\Delta P = k * (q_1 * q_2 )/(R) - 0`

So

`(1)/(2) * m * (1)/(4) * v_o ^2 - (1)/(2) * m * v_o^2 = k * (q_1 * q_2 )/(R) - 0`

=>
`(1)/(2) * m *v_0^2 [ (1)/(4) -1 ] = k * (q_1 * q_2 )/(R)`

=>
`- (3)/(8) * m *v_0^2 = k * (q_1 * q_2 )/(R) ---(1 )`

Generally considering from initial position to a position of distance
`R_f` from the nucleus

Here
`R_f` represented the distance of the proton from the nucleus where the velocity is
`(1)/(4) v_o`

Generally from the law of energy conservation we have that

`\Delta K_f = \Delta P_f`

Here
`\Delta K` is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

`\Delta K_f = K_f - K_i`

=>
`\Delta K_f = (1)/(2) * m * v_2^2 - (1)/(2) * m * v_o^2`

=>
`\Delta K_f = (1)/(2) * m * ((1)/(4) * v_o )^2 - (1)/(2) * m * v_o^2`

=>
`\Delta K_f = (1)/(2) * m * (1)/(16) * v_o ^2 - (1)/(2) * m * v_o^2`

And
`\Delta P` is the change in electric potential energy from initial position to a position of distance
`R_f` from the nucleus , this is mathematically represented as

`\Delta P_f = P_f - P_i`

Here
`P_i` is zero because the electric potential energy at the initial stage is zero so

`\Delta P_f = k * (q_1 * q_2 )/(R_f ) - 0`

So

`(1)/(2) * m * (1)/(8) * v_o ^2 - (1)/(2) * m * v_o^2 = k * (q_1 * q_2 )/(R_f )`

=>
`(1)/(2) * m *v_o^2 [-(15)/(16) ] = k * (q_1 * q_2 )/(R_f )`

=>
`- (15)/(32) * m *v_o^2 = k * (q_1 * q_2 )/(R_f ) ---(2)`

Divide equation 2 by equation 1

`(- (15)/(32) * m *v_o^2 )/(- (3)/(8) * m *v_0^2 ) } = (k * (q_1 * q_2 )/(R_f ) )/(k * (q_1 * q_2 )/(R ) )}`

=>
`-(15)/(32 ) * -(8)/(3) = (R)/(R_f)`

=>
`(5)/(4) = (R)/(R_f)`

=>
`R_f = (4)/(5) R`