The circumference of second orbit of hydrogen atom, if the wavelength of electron is `5 xx 10^(-9)m` will be

Options

A. `5 xx 10^(-8)m`
B. `10^(-8)m`
C. `10^(-10)m`
D. `10^(-18)m`

Answer

To solve the problem of finding the circumference of the second orbit of a hydrogen atom given the wavelength of the electron, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Formula**: The circumference of the nth orbit of a hydrogen atom is given by the formula:
\[ 2\pi r = n \lambda \]
where \( r \) is the radius of the orbit, \( n \) is the principal quantum number (which indicates the orbit number), and \( \lambda \) is the wavelength of the electron.

2. **Identify Given Values**: From the problem, we know:
- The wavelength of the electron, \( \lambda = 5 \times 10^{-9} \, m \)
- The orbit number, \( n = 2 \) (since we are looking for the second orbit)

3. **Substitute Values into the Formula**: Plugging the known values into the formula:
\[ 2\pi r = n \lambda \]
becomes:
\[ 2\pi r = 2 \times (5 \times 10^{-9} \, m) \]

4. **Calculate the Right Side**: Now, calculate the right side:
\[ 2 \times (5 \times 10^{-9}) = 10 \times 10^{-9} \, m = 1 \times 10^{-8} \, m \]

5. **Final Expression for Circumference**: Thus, we have:
\[ 2\pi r = 1 \times 10^{-8} \, m \]
This means that the circumference of the second orbit of the hydrogen atom is:
\[ C = 1 \times 10^{-8} \, m \]

### Conclusion
The circumference of the second orbit of the hydrogen atom, given the wavelength of the electron is \( 5 \times 10^{-9} \, m \), is \( 1 \times 10^{-8} \, m \).