For the Bohr's first orbit of circumference ` 2pi r` , the de - Broglie wavelength of revolving electron will be

Options

A. `2pir`
B. `pir`
C. `(1)/(2pir)`
D. `(1)/(4pir)`

Explanation

To find the de Broglie wavelength of an electron revolving in Bohr's first orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the de Broglie Wavelength Formula**:
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is the Planck's constant,
- \( m \) is the mass of the electron,
- \( v \) is the velocity of the electron.

2. **Using Bohr's Angular Momentum Condition**:
According to Bohr's model, the angular momentum of the electron in the nth orbit is quantized and given by:
\[ mvr = \frac{nh}{2\pi} \]
where:
- \( n \) is the principal quantum number (1 for the first orbit),
- \( r \) is the radius of the orbit.

3. **Rearranging the Angular Momentum Equation**:
From the angular momentum equation, we can express \( mv \) in terms of \( n \), \( h \), and \( r \):
\[ mv = \frac{nh}{2\pi r} \]
This will be referred to as Equation (1).

4. **Substituting into the de Broglie Wavelength Formula**:
Now, we substitute \( mv \) from Equation (1) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{mv} = \frac{h}{\frac{nh}{2\pi r}} = \frac{h \cdot 2\pi r}{nh} \]
Simplifying this gives:
\[ \lambda = \frac{2\pi r}{n} \]

5. **Finding the Wavelength for the First Orbit**:
For the first orbit, we have \( n = 1 \):
\[ \lambda = \frac{2\pi r}{1} = 2\pi r \]

6. **Conclusion**:
Therefore, the de Broglie wavelength of the electron in Bohr's first orbit is:
\[ \lambda = 2\pi r \]

### Final Answer:
The de Broglie wavelength of the revolving electron in Bohr's first orbit is \( 2\pi r \).