Alpha particle touching Gold nucleus

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To determine the minimum kinetic energy required for an alpha particle to touch a gold nucleus, the relevant equation is Ue = (k * q1 * q2) / r. Given the radius of the gold nucleus as 6.47 fm and the alpha particle as 2.36 fm, the total distance for the collision is the sum of their radii. The charges are q1 = 2 and q2 = 79, but the calculated kinetic energy of approximately 3E45 eV seems excessively high, indicating a potential error in the calculations. Clarification is sought regarding the units for charge (q) to resolve the issue.
kraigandrews
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Homework Statement


If the radius of the gold nucleus is 6.47 fm, and the radius of the α particle is 2.36 fm, then what is the minimum required kinetic energy for the α particle if we want the two nuclear surfaces to touch in a head-on collision?

Homework Equations


Ue=(kq1q2)/r


The Attempt at a Solution


so using Ue=KE
q1=2
q2=79

then it should be straightforward plug and chug, however the KE this yields is very high, around 3E45 eV this cannot be right, what am I doing wrong?
 
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The alpha-particle moves from infinity to a distance of ralpha + rau femtometres from the centre of the gold nucleus.
 
kraigandrews said:

The Attempt at a Solution


so using Ue=KE
q1=2
q2=79



what is the unit of q?

ehild
 
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