Question

In hydrogen like system the ratio of coulombian force and gravitational force between an electron and a proton is in the order of:

• A. 10³⁹
• B. 10¹⁹
• C. 10²⁹
• D. 10³⁶

Answer 1

The Correct Option is A

The coulombian force (\(F_c\)) is given by:

\[ F_c = \frac{k Q_1 Q_2}{r^2} = \frac{9 \times 10^9 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{r^2} \]

The gravitational force (\(F_g\)) is given by:

\[ F_g = \frac{G m_1 m_2}{r^2} = \frac{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-27}}{r^2} \]

The ratio of the forces is:

\[ \frac{F_c}{F_g} \approx 0.23 \times 10^{40} \approx 2.3 \times 10^{39} \]

Therefore, the answer is approximately \(10^{39}\).

Answer 2

Step 1: Write expressions for both forces
Between an electron and a proton in a hydrogen-like system:
Coulomb (electrostatic) force: \[ F_e = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2}. \] Gravitational force: \[ F_g = G\frac{m_p m_e}{r^2}. \] Here, \( e \) = charge of electron = \( 1.6\times10^{-19} \, C \), \( m_e = 9.1\times10^{-31} \, kg \), \( m_p = 1.67\times10^{-27} \, kg \), and \( G = 6.67\times10^{-11} \, N\,m^2/kg^2. \)

Step 2: Compute their ratio
\[ \frac{F_e}{F_g} = \frac{\frac{1}{4\pi\varepsilon_0} e^2 / r^2}{G m_p m_e / r^2} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{G m_p m_e}. \] Since \( \frac{1}{4\pi\varepsilon_0} = 9\times10^9 \, N\,m^2/C^2 \), we get \[ \frac{F_e}{F_g} = \frac{9\times10^9 \times (1.6\times10^{-19})^2}{6.67\times10^{-11} \times 9.1\times10^{-31} \times 1.67\times10^{-27}}. \]

Step 3: Simplify numerically
Numerator: \[ 9\times10^9 \times (2.56\times10^{-38}) = 2.3\times10^{-28}. \] Denominator: \[ 6.67\times10^{-11} \times 1.52\times10^{-57} = 1.01\times10^{-67}. \] Therefore, \[ \frac{F_e}{F_g} = \frac{2.3\times10^{-28}}{1.01\times10^{-67}} \approx 2.3\times10^{39}. \] Hence, the ratio is of the order of \( 10^{39} \).

Final answer

10³⁹