1. किसी दूरी पर स्थित एक इलेक्ट्रॉन एवं एक प्रोटॉन के बीच लगने वाले कूलॉम के स्थैतिक वैद्युत एवं गुरूत्वीय बल का अनुपात 2.4 × 1039, अनुक्रमानुपाती स्थिरांक। K = 1 / 4πε₀ का गुरूत्व स्थिरांक G से अनुपात का मान लगभग होगाः

Step-by-step solution

1. 1

   Given Data

   Ratio of electrostatic force to gravitational force: $\frac{F_e}{F_g} = 2.4 \times 10^{39}$

   $$\frac{F_e}{F_g} = 2.4 \times 10^{39}$$
2. 2

   Force Formulas

   Electrostatic force $F_e = K \frac{q_1 q_2}{r^2}$ and Gravitational force $F_g = G \frac{m_1 m_2}{r^2}$

   $$F_e = K \frac{q_e q_p}{r^2}, F_g = G \frac{m_e m_p}{r^2}$$
3. 3

   Ratio Expression

   Divide $F_e$ by $F_g$ to get $\frac{F_e}{F_g} = \frac{K q_e q_p}{G m_e m_p}$

   $$\frac{F_e}{F_g} = \frac{K q_e q_p / r^2}{G m_e m_p / r^2} = \frac{K q_e q_p}{G m_e m_p}$$
4. 4

   Substitute Values

   Given $q_e = q_p = 1.6 \times 10^{-19} \text{C}$, $m_p = 1.67 \times 10^{-27} \text{kg}$, $m_e = 9.1 \times 10^{-31} \text{kg}$

   $$2.4 \times 10^{39} = \frac{K}{G} \frac{(1.6 \times 10^{-19})^2}{(9.1 \times 10^{-31})(1.67 \times 10^{-27})}$$
5. 5

   Calculate $q^2 / (m_e m_p)$

   Calculate the numerical part: $\frac{(1.6 \times 10^{-19})^2}{(9.1 \times 10^{-31})(1.67 \times 10^{-27})} \approx \frac{2.56 \times 10^{-38}}{15.197 \times 10^{-58}}$

   $$\frac{2.56 \times 10^{-38}}{15.197 \times 10^{-58}} \approx 0.168 \times 10^{20} \approx 1.68 \times 10^{19}$$
6. 6

   Solve for $K/G$

   Rearrange the equation: $\frac{K}{G} = \frac{2.4 \times 10^{39}}{1.68 \times 10^{19}}$

   $$\frac{K}{G} = \frac{2.4 \times 10^{39}}{1.68 \times 10^{19}}$$
7. 7

   Final Ratio

   The ratio $\frac{K}{G} \approx 1.42 \times 10^{20}$. This is closest to $10^{40}$ in the options.

   $$\frac{K}{G} \approx 1.42 \times 10^{20}$$
8. 8

   Re-evaluating for $10^{40}$

   If $\frac{K}{G} = 10^{40}$, then $\frac{F_e}{F_g} = \frac{K}{G} \frac{q^2}{m_e m_p} = 10^{40} \times 1.68 \times 10^{19} = 1.68 \times 10^{59}$

   $$\frac{F_e}{F_g} = 10^{40} \times 1.68 \times 10^{19} = 1.68 \times 10^{59}$$