The speed of an electron in the `4^"th"` orbit of hydrogen atom is

Options

A. c
B. `c/137`
C. `c/2192`
D. `c/548`

Answer

To find the speed of an electron in the 4th orbit of a hydrogen atom, we can follow these steps:

### Step 1: Understand the Forces Acting on the Electron
In a hydrogen atom, the electron orbits the nucleus due to the electrostatic force of attraction between the positively charged nucleus (proton) and the negatively charged electron. This electrostatic force provides the necessary centripetal force for the electron's circular motion.

### Step 2: Set Up the Equations
The electrostatic force \( F_e \) can be expressed using Coulomb's Law:
\[ F_e = \frac{k \cdot e^2}{R^2} \]
where:
- \( k \) is Coulomb's constant,
- \( e \) is the charge of the electron,
- \( R \) is the radius of the orbit.

The centripetal force \( F_c \) required for circular motion is given by:
\[ F_c = \frac{m v^2}{R} \]
where:
- \( m \) is the mass of the electron,
- \( v \) is the speed of the electron.

### Step 3: Equate the Forces
Since the electrostatic force provides the centripetal force, we can set these two expressions equal to each other:
\[ \frac{k \cdot e^2}{R^2} = \frac{m v^2}{R} \]

### Step 4: Rearranging the Equation
Multiplying both sides by \( R \) gives us:
\[ \frac{k \cdot e^2}{R} = m v^2 \]
From this, we can express the speed \( v \):
\[ v^2 = \frac{k \cdot e^2}{m R} \]

### Step 5: Use the Angular Momentum Quantization Condition
According to the Bohr model, the angular momentum \( L \) of the electron is quantized:
\[ L = m v R = n \frac{h}{2\pi} \]
where:
- \( n \) is the principal quantum number (for the 4th orbit, \( n = 4 \)),
- \( h \) is Planck's constant.

From this, we can express \( R \):
\[ R = \frac{n h}{2 \pi m v} \]

### Step 6: Substitute R in the Speed Equation
Substituting \( R \) from the angular momentum equation into the speed equation:
\[ v^2 = \frac{k \cdot e^2 \cdot 2 \pi m v}{n h} \]
Rearranging gives:
\[ v = \frac{k \cdot e^2 \cdot 2 \pi m}{n h} \]

### Step 7: Plug in the Constants
Using the known values:
- \( k \) (Coulomb's constant) = \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( e \) (charge of electron) = \( 1.6 \times 10^{-19} \, \text{C} \),
- \( m \) (mass of electron) = \( 9.11 \times 10^{-31} \, \text{kg} \),
- \( h \) (Planck's constant) = \( 6.63 \times 10^{-34} \, \text{Js} \),
- \( n = 4 \).

Substituting these values into the equation gives us the speed of the electron in the 4th orbit.

### Final Calculation
After substituting the constants and simplifying, we find:
\[ v = \frac{c}{4 \cdot 137} \]
where \( c \) is the speed of light.

### Conclusion
Thus, the speed of the electron in the 4th orbit of a hydrogen atom is:
\[ v = \frac{c}{548} \]