Problem

An electron is in the first Bohr orbit of hydrogen. Find:
(a) The speed of the electron.
(b) The time required for the electron to circle the nucleus.

Data

• Energy level ($n$): $1$
• Planck’s constant ($h$): $6.626 \times 10^{-34} \, \mathrm{Js}$
• Mass of the electron ($m$): $9.109 \times 10^{-31} \, \mathrm{kg}$
• Radius of the first Bohr orbit ($r_1$): $0.53 \times 10^{-10} \, \mathrm{m}$
• $\pi$: $3.14$

To Find:

1. Speed of the electron ($v$)
2. Time period ($T$)

Prerequisite Concepts

1. Bohr’s Second Postulate:
   The angular momentum of the electron is quantized:

$$m v r_n = n \frac{h}{2 \pi}$$

Solving for $v$:

$$v = n \frac{h}{2 \pi m r_1}$$

2. Time Period ($T$):
   The time required for the electron to circle the nucleus is:

$$T = \frac{\text{Circumference of the orbit}}{\text{Speed of the electron}}$$

Substituting:

$$T = \frac{2 \pi r_1}{v}$$

Solution

Part (a): Speed of the Electron ($v$)

Using Bohr’s second postulate:

$$v = n \frac{h}{2 \pi m r_1}$$

Substitute $n = 1$, $h = 6.626 \times 10^{-34} \, \mathrm{Js}$, $m = 9.109 \times 10^{-31} \, \mathrm{kg}$, and $r_1 = 0.53 \times 10^{-10} \, \mathrm{m}$:

$$v = \frac{6.626 \times 10^{-34}}{2 \cdot 3.14 \cdot 9.109 \times 10^{-31} \cdot 0.53 \times 10^{-10}}$$

Simplify:

$$v = \frac{6.626 \times 10^{-34}}{3.038 \times 10^{-30}}$$ $$v = 2.19 \times 10^{6} \, \mathrm{m/s}$$

Part (b): Time Period ($T$)

Using the formula for time period:

$$T = \frac{2 \pi r_1}{v}$$

Substitute $r_1 = 0.53 \times 10^{-10} \, \mathrm{m}$ and $v = 2.19 \times 10^{6} \, \mathrm{m/s}$:

$$T = \frac{2 \cdot 3.14 \cdot 0.53 \times 10^{-10}}{2.19 \times 10^{6}}$$

Simplify:

$$T = \frac{3.3284 \times 10^{-10}}{2.19 \times 10^{6}}$$ $$T = 1.52 \times 10^{-16} \, \mathrm{s}$$

Answer

1. Speed of the electron:

$$v = 2.19 \times 10^{6} \, \mathrm{m/s}$$

2. Time period:

$$T = 1.52 \times 10^{-16} \, \mathrm{s}$$

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