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

Answer:

The value is $R_f = \frac{4}{5} R$

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  $v_1 = \frac{1}{2} v_o$

    The  velocity considered is  $v_2 = \frac{1}{4} v_o$

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

      $\Delta K = K__{R}} - K_i$

=>    $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=>    $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=>    $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

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

             $\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

So

           $\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

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

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

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  $\frac{1}{4} v_o$

     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

      $\Delta K_f = K_f - K_i$

=>    $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=>    $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=>    $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

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

             $\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0$      

So

          $\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }$

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

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

Divide equation 2  by equation 1

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

=>           $-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}$

=>           $\frac{5}{4} = \frac{R}{R_f}$

=>             $R_f = \frac{4}{5} R$