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Q.

How far would the electron of hydrogen atom have to be removed from the nucleus for the force of electrostatic attraction to equal the weight of the atom ?
(Avogadro's number =6 .02 x 10²³ gm-atom)

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An Intiative by Sri Chaitanya

a

1.19 m

b

0.25 m

c

1.5 m

d

12 m

answer is A.

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Detailed Solution

$\frac{1}{4 {πε}_{0}} \times \frac{e^{2}}{r^{2}} = mg or r^{2} = \frac{e^{2}}{4 {πε}_{0} \times mg} ∴ r^{2} = \frac{(9 \times 10^{9}) {(1 \cdot 6 \times 10^{- 19})}^{2}}{\{\frac{1 \times 10^{- 3}}{(6 \cdot 02 \times 10^{23})}\} \times 9 \cdot 8} Solving, we get r = 0.25 m$

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