The Rutherford revolution Per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of (A) \(10^{15}\) (B) \(10^{20}\) (C) \(10^{10}\) (D) \(10^{19}\)

From the given information we know: - We are talking about the first Bohr orbit of the hydrogen atom.

According to Bohr's model, the velocity of an electron in the nth orbit is given by: \[v_n = \frac{e^2}{2 \epsilon_0 n h}\] For the first Bohr orbit (n = 1), the velocity is: \[v_1 = \frac{e^2}{2 \epsilon_0 h}\]

Bohr's model also provides a formula for the radius of the nth orbit, which is: \[r_n = \frac{ε_0 n^2 h^2}{π m e^2}\] For the first Bohr orbit (n = 1), the radius is: \[r_1 = \frac{ε_0 h^2}{π m e^2}\]

To find the time period (T) of an electron's revolution, we can use the formula for the time period of a circular orbit: \[T = \frac{2πr}{v}\] Substituting the values of radius and velocity for the first Bohr orbit, we get: \[T_1 = \frac{2π (\frac{ε_0 h^2}{π m e^2})}{\frac{e^2}{2 \epsilon_0 h}} = \frac{4π^2 m ε_0^2 h^3}{m^2 e^4}\]

The frequency (f) of an electron's revolution is the reciprocal of the time period: \[f_1 = \frac{1}{T_1} = \frac{m^2 e^4}{4π^2 m ε_0^2 h^3}\] Now, substitute the appropriate values for the constants: - e = elementary charge = \(1.6 \times 10^{-19} C\) - ε_0 = vacuum permittivity = \(8.854 \times 10^{-12} C^2/(N \cdot m^2)\) - m = mass of electron = \(9.1 \times 10^{-31} kg\) - h = Planck's constant = \(6.626 \times 10^{-34} Js\) The expression for the frequency becomes: \[f_1 = \frac{(9.1 \times 10^{-31})^2 (1.6 \times 10^{-19})^4}{4π^2(9.1 \times 10^{-31}) (8.854 \times 10^{-12})^2 (6.626 \times 10^{-34})^3}\] After calculating the expression, we get \(f_1 \approx 6.58 \times 10^{15} s^{-1}\) #Conclusion# The Rutherford revolution per second made by an electron in the first Bohr orbit of a hydrogen atom is of the order of \(10^{15}\). So, the correct answer is (A) \(10^{15}\).