The de-Broglie wavelength of an electron in the first Bohr orbit is

Options

A. equal to one- fourth the circumference of the first orbit
B. equal to half the circumference of first orbit
C. equal to twice the circumference of first orbit.
D. equal to the circumference of the first orbit.

Explanation

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

### Step-by-Step Solution:

1. **Understand Bohr's Model**: According to Bohr's model of the atom, electrons revolve around the nucleus in specific orbits without radiating energy. The angular momentum of the electron in these orbits is quantized.

2. **Angular Momentum Quantization**: The angular momentum (L) of the electron is given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number (for the first orbit, \( n = 1 \)) and \( h \) is Planck's constant.

3. **Relate Angular Momentum to Linear Momentum**: The angular momentum can also be expressed in terms of linear momentum (p) and radius (r) of the orbit:
\[ L = p \cdot r \]
where \( p = mv \) (mass times velocity).

4. **Substituting Linear Momentum**: We can express the linear momentum in terms of the de-Broglie wavelength (\( \lambda \)):
\[ p = \frac{h}{\lambda} \]
Therefore, substituting this into the angular momentum equation gives:
\[ \frac{h}{\lambda} \cdot r = n \frac{h}{2\pi} \]

5. **Canceling \( h \)**: We can cancel \( h \) from both sides:
\[ \frac{r}{\lambda} = \frac{n}{2\pi} \]

6. **Solving for Wavelength**: Rearranging the equation for \( \lambda \), we get:
\[ \lambda = \frac{2\pi r}{n} \]

7. **Substituting for the First Orbit**: For the first Bohr orbit, \( n = 1 \):
\[ \lambda = 2\pi r \]

8. **Relating Wavelength to Circumference**: The circumference of the first orbit is given by:
\[ C = 2\pi r \]
Therefore, we can conclude that:
\[ \lambda = C \]

9. **Final Answer**: Thus, the de-Broglie wavelength of an electron in the first Bohr orbit is equal to the circumference of the first orbit.

### Conclusion:
The correct option is that the de-Broglie wavelength of an electron in the first Bohr orbit is equal to the circumference of the first orbit.