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Q.

An electron is revolving around the nucleus with a radius of $0.53 A°$ of a Hydrogen atom. The force of attraction between electron and nucleus is

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a

8 .6 \times 10^{- 9} N

b

8 .2 \times 10^{- 8} N

c

4 .5 \times 10^{- 8} N

d

8 .1 \times 10^{- 6} N

answer is A.

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Detailed Solution

F = \frac{1}{4 π ε_{0}} . \frac{q^{2}}{r^{2}}

$= \frac{9 \times 10^{9} \times 2.56 \times 10^{- 38}}{53 \times 53 \times 10^{- 24}} = 8.2 \times 10^{- 8} N$

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