An alpha particle, having charge +2e and mass 6.64 x 10^-27 kg, is a product of

Question

An alpha particle, having charge +2e and mass 6.64 x 10^-27 kg, is a product of certain radioactive decays. Assume that an alpha particle, initially very far from a gold nucleus, is fired with a velocity of 2.00 x 10^7 m/s directly toward the nucleus (charge +79e). How close does the alpha particle get to the nucleus before turning around? Assume that the gold nucleus remains stationary.

I get d= 2kq1q2/mv^2

d= 2(8.98x10^9)(2)(79)/6.64x10^-27((2.00x10^7)^2)

Am I putting it in my calculator wrong? What is going on? The answer should be 2.74x10^-14. I get numbers in the 10^24 Way off!

Explanation / Answer

Got it! I forgot to compensate for the charge of e=1.6x10^-19. The equation should be r= 2ke(2)(79)(1.6x10^-19)/mv^2

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