Question

For a hydrogen atom () in the fourth Bohr orbit (): (a) Calculate the time period of revolution of the electron. (b) Determine how many revolutions the electron completes in seconds (one microsecond).

Answer 1

a) 9.76 x 10^{-15} s, b) 1.02 x 10^8 revolutions

Answer 2

The problem asks us to calculate the time period of revolution of an electron in the fourth Bohr orbit of a hydrogen atom and determine the number of revolutions it completes in one microsecond.

Key Concepts and Formulas:

1. Radius of the nth Bohr orbit for a hydrogen atom: $r_n = n^2 a_0$ where $a_0$ is the Bohr radius, $a_0 = 0.529 \times 10^{-10}$ m.

2. Speed of the electron in the nth Bohr orbit for a hydrogen atom: $v_n = \frac{v_1}{n}$ where $v_1$ is the speed of the electron in the first Bohr orbit, $v_1 = 2.18 \times 10^6$ m/s.

3. Time period of revolution (T): The time period is the time taken to complete one revolution, which is the circumference of the orbit divided by the speed of the electron. $T_n = \frac{2\pi r_n}{v_n}$

Calculations:

(a) Calculate the time period of revolution of the electron in the fourth Bohr orbit ($n=4$).

First, calculate the radius of the fourth orbit ($r_4$): $r_4 = 4^2 \times a_0 = 16 \times 0.529 \times 10^{-10}$ m $r_4 = 8.464 \times 10^{-10}$ m

Next, calculate the speed of the electron in the fourth orbit ($v_4$): $v_4 = \frac{v_1}{4} = \frac{2.18 \times 10^6 \text{ m/s}}{4}$ $v_4 = 0.545 \times 10^6 \text{ m/s}$

Now, calculate the time period ($T_4$): $T_4 = \frac{2\pi r_4}{v_4} = \frac{2 \times 3.14159 \times 8.464 \times 10^{-10} \text{ m}}{0.545 \times 10^6 \text{ m/s}}$ $T_4 = \frac{53.18 \times 10^{-10}}{0.545 \times 10^6}$ s $T_4 = 97.58 \times 10^{-16}$ s $T_4 \approx 9.76 \times 10^{-15}$ s

(b) Determine how many revolutions the electron completes in 1 microsecond ($\mu s$).

1 microsecond = $1 \times 10^{-6}$ s. The number of revolutions is the total time divided by the time period of one revolution. Number of revolutions = $\frac{\text{Total Time}}{\text{Time Period}}$ Number of revolutions = $\frac{1 \times 10^{-6} \text{ s}}{9.76 \times 10^{-15} \text{ s/revolution}}$ Number of revolutions = $\frac{1}{9.76} \times 10^{(-6 - (-15))}$ Number of revolutions = $0.10245 \times 10^9$ Number of revolutions $\approx 1.02 \times 10^8$ revolutions

Explanation of the solution:

(a) The time period of an electron in a Bohr orbit is calculated using the formula $T_n = \frac{2\pi r_n}{v_n}$. For the fourth orbit ($n=4$), the radius $r_4$ is $16a_0$ and the speed $v_4$ is $v_1/4$. Substituting the values of $a_0 = 0.529 \times 10^{-10}$ m and $v_1 = 2.18 \times 10^6$ m/s, we get $r_4 = 8.464 \times 10^{-10}$ m and $v_4 = 0.545 \times 10^6$ m/s. Plugging these into the time period formula yields $T_4 \approx 9.76 \times 10^{-15}$ s.

(b) To find the number of revolutions in 1 microsecond ($10^{-6}$ s), we divide the total time by the time period per revolution: Number of revolutions = $\frac{10^{-6} \text{ s}}{9.76 \times 10^{-15} \text{ s/revolution}} \approx 1.02 \times 10^8$ revolutions.