Question

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) \(10^{19}\)

Short answer

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) \(10^{15}\).

Step-by-step solution

Write down the known values

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

Calculate the velocity of the electron

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: \[v_1 = \frac{e^2}{2 \epsilon_0 h}\]

Calculate the radius of the first Bohr orbit

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: \[r_1 = \frac{ε_0 h^2}{π m e^2}\]

Calculate the time period of the electron's revolution

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: \[T_1 = \frac{2π (\frac{ε_0 h^2}{π m e^2})}{\frac{e^2}{2 \epsilon_0 h}} = \frac{4π^2 m ε_0^2 h^3}{m^2 e^4}\]

Calculate the frequency of the electron's revolution

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) \(10^{15}\).