Closest distance of approach physics electric field help!

https://isaacphysics.org/questions/closest_approach_nucleus?board=6a3a18e1-24e8-4c2c-a09e-675e4202d4e4&stage=a_level
I got 8.89*10^-13 m by using conservation of energy, but they said I have to also conserve momentum which I did to get (4*1.67*10^-27*3.5*10^6)/(1.67*10^-27(4+197))=V
I then plugged the new value of velocity in to get 2.24*10^-9 m which is incorrect
Please could you help me understand where I went wrong ? Thank you

I think they have given you a hint that you need to apply both the conservation of energy and linear momentum to solve the problem.
If you apply the conservation of energy, linear momentum is not conserved for both the alpha particle and nucleus.
Why?
I believe you assume when the alpha particle is at the closest possible approach to the nucleus, both the nucleus and alpha particle are stationary. However, when the alpha particle is moving toward the nucleus, the system has initial momentum and your approach definitely violates the conservation of linear momentum.

So i tried to conserve momentum by saying m_1*v= m_1*v_2+m_2*v_2 to calculate the velocity at the end (they are both moving at the same velocity so they arent moving relative to each other and got the value of v above (69,651.74) then used that in the conservation of energy one
is that wrong?

Are you finding v or v_2? In fact, you are not required to find the common velocity.

So i tried to conserve momentum by saying m_1*v= m_1*v_2+m_2*v_2 to calculate the velocity at the end (they are both moving at the same velocity so they arent moving relative to each other and got the value of v above (69,651.74) then used that in the conservation of energy one
is that wrong?

I was using v_2 to put into the equation for conservation of energy to try to link conservation of momentum and energy, cause that was in the hints in the question
Then used it to find the distance