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

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

[tex]\Delta K = K__{R}} - K_i[/tex]

=> [tex]\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2[/tex]

=> [tex]\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2[/tex]

=> [tex]\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2[/tex]

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

[tex]\Delta P = P_f - P_i[/tex]

Here [tex]P_i[/tex] is zero because the electric potential energy at the initial stage is zero so

[tex]\Delta P = k * \frac{q_1 * q_2 }{R} - 0[/tex]

So

[tex]\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0[/tex]

=> [tex]\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}[/tex]

=> [tex]- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )[/tex]

Generally considering from initial position to a position of distance [tex]R_f[/tex] from the nucleus

Here [tex]R_f[/tex] represented the distance of the proton from the nucleus where the velocity is [tex]\frac{1}{4} v_o[/tex]

Generally from the law of energy conservation we have that

[tex]\Delta K_f = \Delta P_f[/tex]

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

[tex]\Delta K_f = K_f - K_i[/tex]

=> [tex]\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2[/tex]

=> [tex]\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2[/tex]

=> [tex]\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2[/tex]

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

[tex]\Delta P_f = P_f - P_i[/tex]

Here [tex]P_i[/tex] is zero because the electric potential energy at the initial stage is zero so

[tex]\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0[/tex]

So

[tex]\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }[/tex]

=> [tex]\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }[/tex]

=> [tex]- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)[/tex]

Divide equation 2 by equation 1

[tex]\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}[/tex]

=> [tex]-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}[/tex]

=> [tex]\frac{5}{4} = \frac{R}{R_f}[/tex]

=> [tex]R_f = \frac{4}{5} R[/tex]