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) About the author Ella
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 Reply
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