Question

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the electron is initially a great distance from the proton, at what distance from the proton is the speed of the electron instantaneously equal to twice the initial value?

Answer 1

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

         The initial speed of the  electron is $v_i = 3.2 *10^5 m/s$

         The mass of electron is $m = 9.11*10^{-31}kg$

         Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

         Let $KE_i$ be the initial kinetic energy of the electron \

          Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

  Considering that energy is conserved,

  The energy at the initial position of the electron = The energy at the final position of the electron

      i.e

             $KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial  position of the electron and at distance d of the electron to the proton

                Here $U_1 = 0$

So the equation becomes

                   $\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

 $k$ is electrostatic constant with value $8.99*10^9 N \cdot m^2 /C^2$

i.e $q = 1.602 *10^{-19}C$

           $v_d$ is the velocity at distance d from the proton = 2$v_i$

  So the equation becomes

             $\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}$

            $\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}$

           $3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}$

Making d the subject of the formula

           $d = \frac{2k(q)^2}{3mv_i^2}$

              $= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}$

              $=1.66*10^{-9}m$