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 use the formula derived from Bohr's atomic model. Here’s a step-by-step solution:

### Step 1: Identify the formula for the speed of an electron
The speed \( v \) of an electron in the \( n^{th} \) orbit of a hydrogen atom is given by the formula:
\[ v = 2.12 \times 10^6 \frac{Z}{n} \text{ m/s} \]
where:
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n \) is the principal quantum number (for the 4th orbit, \( n = 4 \)).

### Step 2: Substitute the values into the formula
For hydrogen (\( Z = 1 \)) and the 4th orbit (\( n = 4 \)):
\[ v = 2.12 \times 10^6 \frac{1}{4} \]

### Step 3: Calculate the speed
Now, perform the calculation:
\[ v = 2.12 \times 10^6 \times \frac{1}{4} = 2.12 \times 10^6 \times 0.25 = 0.53 \times 10^6 \text{ m/s} \]
This simplifies to:
\[ v = 0.53 \times 10^6 \text{ m/s} = 5.3 \times 10^5 \text{ m/s} \]

### Step 4: Convert the speed to a fraction of the speed of light
The speed of light \( c \) is approximately \( 3 \times 10^8 \text{ m/s} \). To express the speed of the electron as a fraction of the speed of light:
\[ \frac{v}{c} = \frac{5.3 \times 10^5}{3 \times 10^8} \]

### Step 5: Simplify the fraction
Calculating the fraction:
\[ \frac{5.3}{3} \times \frac{10^5}{10^8} = \frac{5.3}{3} \times 10^{-3} \]
This gives:
\[ \frac{v}{c} \approx 1.77 \times 10^{-3} \]

### Step 6: Conclusion
Thus, the speed of the electron in the 4th orbit of a hydrogen atom can be expressed as:
\[ v \approx \frac{c}{548} \]
This means the correct option is option 4.