Question

A proton moves from large distance, with a

speed of v m/s directly towards a free proton

initially at rest. The distance of closest

approach of two protons is (symbol have

usual meanings)

​

Answer 1

Explanation:

Speed=7.45×1−

5

m/s let mass=m

Let x is closest distance.

K.E loss=gain in potential energy

21 mv 2 = rK(9 p) 2,q p =e,=1.6×10 −19 =m p

=1.675×10 −27

=[ 1.675×10 −279×10 9×(1.6×10 −19 ) 2×2 ]21=5×10 −13 m