An electron is moving in Bohr's fourth orbit. Its de Broglie wavelength is `lambda`. What is the circumference of the fourth orbit?

Options

A. `2//lambda`
B. `2lambda`
C. `4lambda`
D. `3//lambda`

Explanation

To find the circumference of the fourth orbit of an electron moving in Bohr's model, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between angular momentum and orbits**: According to Bohr's model, the angular momentum (L) of an electron in an orbit is quantized and given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number (orbit number), and \( h \) is Planck's constant.

2. **Relate angular momentum to the circumference of the orbit**: The angular momentum can also be expressed in terms of mass (m), velocity (v), and radius (r) of the orbit:
\[ L = mvr \]
Therefore, we can set the two expressions for angular momentum equal to each other:
\[ mvr = n \frac{h}{2\pi} \]

3. **Express the circumference in terms of radius**: The circumference (C) of the orbit is given by:
\[ C = 2\pi r \]
We need to express \( r \) in terms of \( n \) and \( \lambda \) (de Broglie wavelength).

4. **Use the de Broglie wavelength relation**: The de Broglie wavelength \( \lambda \) is related to the momentum of the electron:
\[ \lambda = \frac{h}{mv} \]
Rearranging this gives:
\[ mv = \frac{h}{\lambda} \]

5. **Substitute \( mv \) into the angular momentum equation**: Substitute \( mv \) from the de Broglie wavelength into the angular momentum equation:
\[ mvr = n \frac{h}{2\pi} \implies r = \frac{n h}{2\pi mv} \]
Now substituting \( mv \) gives:
\[ r = \frac{n h}{2\pi \left(\frac{h}{\lambda}\right)} = \frac{n \lambda}{2\pi} \]

6. **Calculate the circumference**: Now substitute \( r \) back into the circumference formula:
\[ C = 2\pi r = 2\pi \left(\frac{n \lambda}{2\pi}\right) = n \lambda \]
For the fourth orbit (\( n = 4 \)):
\[ C = 4 \lambda \]

### Final Answer:
The circumference of the fourth orbit is:
\[ C = 4\lambda \]