Grade 11Inorganic Chemistry
Find out the value of electrostatic potential energy of 2 electrons separated by
3.0 angstrom in vacuum.
1 Answer
ABHISHEK JAIN AskiitiansExpert-IITD
Dear Shambhavi,
since P.E. = Kq1q2\r
charge on electron = 1.6 × 10^ -19
distance b/w two electrons = 3Å = 3 × 10^ -10 meter
coulamb's constant k = 9 × 10^9
put the values in above formula:
P.E. = 9 × 10^9 × 1.6 × 10^-19 × 1.6× 10^-19 ÷ 3 × 10^-10
= 7.676 × 10^-19 J
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