The circumference of the second orbit of an atom or ion having single electron,`4times10^(-9)`m.The de-Broglie wavelength of electron revolving in this orbit should be

Options

A. `2xx10^(-9)m`
B. `4xx10^(-9)m`
C. `8xx10^(-9)m`
D. `1xx10^(-9)m`

Answer

To find the de Broglie wavelength of an electron revolving in the second orbit of an atom or ion with a given circumference, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- The circumference of the second orbit, \( C = 4 \times 10^{-9} \, \text{m} \).
- The principal quantum number for the second orbit, \( n = 2 \).

2. **Use the de Broglie Wavelength Formula**:
The de Broglie wavelength \( \lambda \) is related to the circumference of the orbit and the principal quantum number by the formula:
\[ n \lambda = C \]
where \( n \) is the principal quantum number.

3. **Rearrange the Formula to Find Wavelength**:
To find the de Broglie wavelength, we can rearrange the formula:
\[ \lambda = \frac{C}{n} \]

4. **Substitute the Values**:
Substitute the known values into the equation:
\[ \lambda = \frac{4 \times 10^{-9} \, \text{m}}{2} \]

5. **Calculate the Wavelength**:
Performing the division:
\[ \lambda = 2 \times 10^{-9} \, \text{m} \]

### Final Answer:
The de Broglie wavelength of the electron revolving in the second orbit is:
\[ \lambda = 2 \times 10^{-9} \, \text{m} \]

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