Question

The ratio of the electrostatic force of attraction to the gravitational force between the proton and electron of the hydrogen atom is of the order of

• A. ${10}^{39}$
• B. ${10}^{-39}$
• C. ${10}^8$
• D. ${10}^{-8}$

Answer 1

Answer: (A)

Electrostatic force of attraction $=\frac{k{q}_1{q}_2}{r^2}$

$=\frac{9\times {10}^9\times (1\cdot 6\times {10}^{-19}{)}^2}{r^2}$
Gravitational force of attraction $=\frac{G{m}_e{m}_p}{r^2}$

$=\frac{(6\cdot 67\times {10}^{-11})(9\cdot 1\times {10}^{-31})(1\cdot 67\times {10}^{-27})}{r^2}$

$Ratio\frac{F_e}{F_g}=\frac{9\times {10}^9\times (1\cdot 6\times {10}^{-19}{)}^2}{(6\cdot 67\times {10}^{-11})(9\cdot 1\times {10}^{-31})(1\cdot 67\times {10}^{-27})}$

$=\frac{23\cdot 04\times {10}^{-29}}{101\cdot 36\times {10}^{-69}}$

$=2\cdot 27\times {10}^{39}$