the electrostatic force of repulsion between two positively charged Ion carried equal charges is 3.7 into 10 to the power minus 19 and when they separated by distance of 5 angstrom how many electrons are missing from each other
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Electrostatic force F = q1q2/(4πϵor2)
Let q1 = q2 = ne where n is the number of electrons missing from each atom and e the charge on an electron = 1.6 x 10-19C.
1/(4πϵo) = 9 x 109 N.m2/C2.
r = 5 x 10-10 m
F = 3.7 x 10-9 N = n2. (1.6 x 10-19 C)2 x 9 x 109 N.m2/C2/(5 x 10-10 m)2.
⇒ n2 = 3.7 x 10-9 N x (5 x 10-10 m)2 / (1.6 x 10-19 C)2 x 9 x 109 N.m2/C2.
= 4
⇒ n = 2
The number of electrons missing from each atom = 2
Let q1 = q2 = ne where n is the number of electrons missing from each atom and e the charge on an electron = 1.6 x 10-19C.
1/(4πϵo) = 9 x 109 N.m2/C2.
r = 5 x 10-10 m
F = 3.7 x 10-9 N = n2. (1.6 x 10-19 C)2 x 9 x 109 N.m2/C2/(5 x 10-10 m)2.
⇒ n2 = 3.7 x 10-9 N x (5 x 10-10 m)2 / (1.6 x 10-19 C)2 x 9 x 109 N.m2/C2.
= 4
⇒ n = 2
The number of electrons missing from each atom = 2
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