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(a) How much work would it take to push two protons very slowly from a separation of2.00×10-19C(a typical atomic distance) to3.00×10-10m(a typical nuclear distance)? (b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation?

Short Answer

Expert verified

(a) The amount done is7.68×10-14J

(b) The velocity is 6.78×106ms.

Step by step solution

01

Determine the formulas for the work done.

Consider the formula for the work required to move a charge from the point a to the point b is:

Wab=Ub-Ua

Here,UaandUbare the initial and final electric potential respectively.

Consider the expression for the initial potential as:

Ua=14πε0q1q2ra

Consider the expression for the final potential as:

Ub=14πε0q2q2rb

Consider the formula for the change in the kinetic energy when the mass of proton and electron are equal and are projected in the same direction.

K=mv222+mv222=mv22

Then, the change in the work done is.

U=mv22

Rewrite the expression in terms of the velocity as,

v2=Um ….. (1)

02

Determine the work done.

(a)

Consider the equation for the work done as:

W=Ub-UaW=14πε01rb-1ra

Put given values

W=9×1091.6×10-19213.0×10-15-12×10-10W=-7.68×10-14J

But work done is never negative so7.68×10-14J.

Hence, the amount done is 7.68×10-14J.

03

Determine the velocity.

(b)

Substitute the values in the equation (1) and solve as:

v=7.68×10-14J1.67×10-27kgv=6.78×106ms

Hence, the velocity is 6.78×106ms.

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