An alpha particle (q=2e and m=4mp) is projected directly at a gold nucleus (q =

Question

An alpha particle (q=2e and m=4mp) is projected directly at a gold nucleus (q = 79e) which can be assumed to be fixed in place. Assume that the alpha particle starts from a point which is very far from the gold and that it has an initial speed of 4. 5 105 m/s. How close does it come to the gold nucleus, assuming that it moves in a straight line towards that nucleus. From comparison, the gold nucleus has a diameter of about 7. 5 fm. Are the assumptions of this problem consistent with this.

Please provide work and explanations.

Explanation / Answer

This can be solved by using the idea that PE final will equal KE initial

The KE of the aplha particle will be .5mv2

The potential Energy it will have when at its closest point will be kqq/r

so

(.5)(4)(1.67265 X 10-27)(4.5 X 105)2 = (9 X 109)(2)(79)(1.6 X 10-19)2/r

Solve for r

r = 5.37 X 10-11 m   which is 53.7 pm

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