Answer :
Answer:
Explanation:
F = 2.12 x 10⁻²⁶ / x²
Work done by electric field of nucleus
W = ∫ Fdx
= ∫2.12 x 10⁻²⁶ / x² dx
= 2.12 x 10⁻²⁶ ( - 1 / x )
= - 2.12 x 10⁻²⁶ ( 1/5 - 1 / 8 x 10⁻¹⁰ )
= - .265 x 10⁻¹⁶ J
1/ 2 x mv² = .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ - .265 x 10⁻¹⁶
= 7.515 x 10⁻¹⁷ - .265 x 10⁻¹⁶
=( .7515 - .265 )x 10⁻¹⁶
= .4865 x 10⁻¹⁶
.5 x 1.67 x 10⁻²⁷ x v² = .4865 x 10⁻¹⁶
v² = .5826 x 10¹¹
v² = 5.826 x 10¹⁰
v = 2.41 x 10⁵ m /s
b )
Let r be the closest distance
Potential at this point
2.12 x 10⁻²⁶ ( 1 / r )
Kinetic energy
= 0
Total energy = 2.12 x 10⁻²⁶ ( 1 / r )
Total energy at 5 m
= .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ + 0 ( potential energy at 5 m will be negligible as compared with that near the center )
= 7.515 x 10⁻¹⁷ J
So ,
2.12 x 10⁻²⁶ ( 1 / r ) = 7.515 x 10⁻¹⁷
r = 2.12 x 10⁻²⁶ / 7.515 x 10⁻¹⁷
= .282 x 10⁻⁹
= 2.82 x 10⁻¹⁰ m
c ) As electric field is conservative , no dissipation of energy takes place . Hence its speed at 5m on returning back to this point will be same as
3.00 x 10⁵ m /s