Answer:
The distance is [tex]d =1.66*10^{-9}m[/tex]
Explanation:
From the question we are told that
The initial speed of the electron is [tex]v_i = 3.2 *10^5 m/s[/tex]
The mass of electron is [tex]m = 9.11*10^{-31}kg[/tex]
Let [tex]d[/tex] be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value
Let [tex]KE_i[/tex] be the initial kinetic energy of the electron \
Let [tex]KE_d[/tex] be the kinetic energy of the electron at the distance [tex]d[/tex] 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
[tex]KE_i +U_1 = KE_d + U_2[/tex]
[tex]U_1 \ and \ U_2[/tex] are the potential energy at the initial position of the electron and at distance d of the electron to the proton
Here [tex]U_1 = 0[/tex]
So the equation becomes
[tex]\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}[/tex]
Here [tex]q_1 \ and \ q_2[/tex] 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
[tex]k[/tex] is electrostatic constant with value [tex]8.99*10^9 N \cdot m^2 /C^2[/tex]
i.e [tex]q = 1.602 *10^{-19}C[/tex]
[tex]v_d[/tex] is the velocity at distance d from the proton = 2[tex]v_i[/tex]
So the equation becomes
[tex]\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}[/tex]
[tex]\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}[/tex]
[tex]3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}[/tex]
Making d the subject of the formula
[tex]d = \frac{2k(q)^2}{3mv_i^2}[/tex]
[tex]= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}[/tex]
[tex]=1.66*10^{-9}m[/tex]
