MP Board · Class 12 · Physics · AtomsUsing the Bohr model, calculate the speed of the electron in the first, second, and third orbits of a hydrogen atom. Also, find the period of revolution in the innermost orbit. (Given: $e = 1.6 \times 10^{-19}\text{ C}$, $\varepsilon0 = 8.85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2}$, $h = 6.63 \times 10^{-34}\text{ J s}$, $m = 9.1 \times 10^{-31}\text{ kg}$)

Step-by-Step Numerical Working

1. Formula for Velocity of Electron in the $n$-th Orbit:\nFrom Bohr's theory, the velocity $v_n$ of an electron in the $n$-th orbit of a hydrogen atom is given by: $$v_n = \frac{e^2}{2 \varepsilon_0 h n}$$ \nAlternatively, substituting the standard values of constants: $$v_n = \frac{c \cdot \alpha}{n} \quad \text{or using the basic formula directly:} $$v_n = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2\pi ze^2}{nh}$$ \nLet us use the direct fundamental formula derivation parameters: $$v_n = \frac{nh}{2\pi m r_n}$$\nWhere radius $r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2}$. Substituting $r_n$ into the velocity equation: $$v_n = \frac{nh}{2\pi m} \cdot \frac{\pi m e^2}{n^2 h^2 \varepsilon_0} = \frac{e^2}{2 \varepsilon_0 h n}$$

2. Calculation for First Orbit ($n = 1$): $$v_1 = \frac{(1.6 \times 10^{-19})^2}{2 \times (8.85 \times 10^{-12}) \times (6.63 \times 10^{-34}) \times 1}$$ $$v_1 = \frac{2.56 \times 10^{-38}}{2 \times 8.85 \times 6.63 \times 10^{-46}}$$ $$v_1 = \frac{2.56 \times 10^{-38}}{117.349 \times 10^{-46}} = \frac{2.56}{117.349} \times 10^{8}$$ $$v_1 \approx 2.18 \times 10^6 \text{ m/s}$$

3. Calculation for Second Orbit ($n = 2$):\nSince $v_n \propto \frac{1}{n}$: $$v_2 = \frac{v_1}{2} = \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6 \text{ m/s}$$

4. Calculation for Third Orbit ($n = 3$): $$v_3 = \frac{v_1}{3} = \frac{2.18 \times 10^6}{3} = 7.27 \times 10^5 \text{ m/s}$$

5. Calculation of Period of Revolution in Innermost Orbit ($n = 1$):\nFirst, find the radius of the first orbit ($r_1$): $$r_1 = \frac{(1)^2 \times (6.63 \times 10^{-34})^2 \times (8.85 \times 10^{-12})}{\pi \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-19})^2}$$ $$r_1 = \frac{1 \times 4.3956 \times 10^{-68} \times 8.85 \times 10^{-12}}{3.1416 \times 9.1 \times 10^{-31} \times 2.56 \times 10^{-38}}$$ $$r_1 = \frac{38.901 \times 10^{-80}}{73.214 \times 10^{-69}} \approx 5.31 \times 10^{-11} \text{ m} = 0.531 \text{ Å}$$ \nNow, the time period $T$ is given by the circumference of the orbit divided by the velocity: $$T = \frac{2\pi r_1}{v_1}$$ $$T = \frac{2 \times 3.1416 \times 5.31 \times 10^{-11}}{2.18 \times 10^6}$$ $$T = \frac{33.336 \times 10^{-11}}{2.18 \times 10^6} \approx 1.53 \times 10^{-16} \text{ s}$$

Final Answer:

• Speed in 1st orbit = $2.18 \times 10^6 \text{ m/s}$
• Speed in 2nd orbit = $1.09 \times 10^6 \text{ m/s}$
• Speed in 3rd orbit = $7.27 \times 10^5 \text{ m/s}$
• Time period in innermost orbit = $1.53 \times 10^{-16} \text{ s}$