Answer:
The value is 
Explanation:
From the question we are told that
The initial velocity of the proton is
At a distance R from the nucleus the velocity is
The velocity considered is 
Generally considering from initial position to a position of distance R from the nucleus
Generally from the law of energy conservation we have that

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

=> 
=> 
=> 
And
is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

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

So

=> ![\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20m%20%20%2Av_0%5E2%20%5B%20%5Cfrac%7B1%7D%7B4%7D%20-1%20%5D%20%20%3D%20%20%20k%20%20%2A%20%20%5Cfrac%7Bq_1%20%2A%20q_2%20%7D%7BR%7D)
=> 
Generally considering from initial position to a position of distance
from the nucleus
Here
represented the distance of the proton from the nucleus where the velocity is 
Generally from the law of energy conservation we have that

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

=> 
=> 
=> 
And
is the change in electric potential energy from initial position to a position of distance
from the nucleus , this is mathematically represented as

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

=> ![\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20m%20%20%2Av_o%5E2%20%5B-%5Cfrac%7B15%7D%7B16%7D%20%5D%20%20%3D%20%20%20k%20%20%2A%20%20%5Cfrac%7Bq_1%20%2A%20q_2%20%7D%7BR_f%20%7D)
=> 
Divide equation 2 by equation 1

=> 
=> 
=> 