The figure shows the potential energy of a proton (q =+e) and a lead nucles (q =

Question

The figure shows the potential energy of a proton (q =+e) and a lead nucles (q = +82e) The horizontal scale is in units of femtometers, where I Fm + 1 femtometers = 10_15 m. A proton is fired toward a lead nucleus from very far away. How much point 10 fm from the nucleus? Explain. How much Kinetic k energy does the proton of part a have when a have when it is 20 fm from the nucleus and moving toward it, before it, the collision? How much kinetic energy does the proton of part a have when it is 20fm and moving away from it, after the collision

Explanation / Answer

a) When proton reaches 10fm from necleus.

then PE = 4 x 10^-12 J   (from the graph)

Initally KE of proton will convert into PE completely and proton will stop there .

so KE needed = 4 x 10^-12 J

b) Using energy conservation,

PE + KE = constant

PE at 20 fm   = 1 x 10^-12

0 + 4 x 10^-12 = 1 x 10^-12   + KE

KE = 3 x 10^-12 J

c) at 20fm , PE is 1 x 10^-12 J

henc KE of at this point will be 3 x 10^-12 J.

irrespective of direction, it is moving in.

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