The de Broglie wavelength of an electron in the nth Bohr orbit is related to the radius R of the orbit as:

Options

A. `nlambda=nR`
B. `nlambda=3/2piR`
C. `nlambda=2piR`
D. `nlambda=4piR`

Explanation

To solve the problem, we need to establish the relationship between the de Broglie wavelength of an electron in the nth Bohr orbit and the radius of that orbit.

### Step-by-Step Solution:

1. **Understand de Broglie Wavelength**:
The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle. For an electron, momentum \( p \) can be expressed as \( mv \), where \( m \) is the mass and \( v \) is the velocity of the electron.

2. **Relate Momentum to the Bohr Model**:
According to the Bohr model, the angular momentum (\( L \)) of an electron in the nth orbit is quantized and given by:
\[ L = mvr = \frac{nh}{2\pi} \]
where \( n \) is the principal quantum number and \( r \) is the radius of the orbit.

3. **Express Velocity in Terms of Angular Momentum**:
From the angular momentum equation, we can express the velocity \( v \) as:
\[ mv = \frac{nh}{2\pi r} \]

4. **Substitute into the de Broglie Wavelength Formula**:
Now, substituting \( mv \) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{mv} = \frac{h}{\frac{nh}{2\pi r}} = \frac{2\pi r}{n} \]

5. **Rearranging the Equation**:
Multiplying both sides by \( n \):
\[ n\lambda = 2\pi r \]

6. **Conclusion**:
Thus, we find the relationship between the de Broglie wavelength and the radius of the orbit:
\[ n\lambda = 2\pi r \]
This corresponds to option 3.

### Final Answer:
The correct option is:
\[ n\lambda = 2\pi r \]