If `a_(0)` represents the bOhr radius, the de Broglie wavelength of the electron when it moves in the third orbit after absorbing some definite, amount of energy will be

Options

A. `a_(0)//3`
B. `9a_(0)`
C. `2pia_(0)`
D. `6pia_(0)`

Explanation

To find the de Broglie wavelength of an electron moving in the third orbit after absorbing energy, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to determine the de Broglie wavelength of an electron in the third orbit of a hydrogen atom, where the Bohr radius is denoted as \( a_0 \).

2. **Using the Angular Momentum Quantization**:
The angular momentum of an electron in a Bohr orbit is given by:
\[ mvr = \frac{nh}{2\pi} \]
where \( n \) is the principal quantum number (for the third orbit, \( n = 3 \)), \( m \) is the mass of the electron, \( v \) is its velocity, and \( r \) is the radius of the orbit.

3. **Using the de Broglie Wavelength Relation**:
The de Broglie wavelength \( \lambda \) is related to momentum \( p \) by:
\[ p = \frac{h}{\lambda} \]
where \( p = mv \) (momentum of the electron).

4. **Relating Angular Momentum and de Broglie Wavelength**:
We can equate the two expressions for momentum:
\[ mv \cdot r = \frac{nh}{2\pi} \]
Substituting \( mv \) from the de Broglie relation:
\[ \frac{h}{\lambda} \cdot r = \frac{nh}{2\pi} \]

5. **Solving for Wavelength**:
Rearranging the equation gives:
\[ \frac{r}{\lambda} = \frac{n}{2\pi} \]
Therefore, the de Broglie wavelength can be expressed as:
\[ \lambda = \frac{2\pi r}{n} \]

6. **Finding the Radius for the Third Orbit**:
The radius for the \( n \)-th orbit in the Bohr model is given by:
\[ r_n = n^2 a_0 \]
For the third orbit (\( n = 3 \)):
\[ r_3 = 3^2 a_0 = 9 a_0 \]

7. **Substituting the Radius into the Wavelength Equation**:
Now substituting \( r_3 \) into the wavelength equation:
\[ \lambda = \frac{2\pi (9 a_0)}{3} = 6\pi a_0 \]

8. **Final Answer**:
Thus, the de Broglie wavelength of the electron when it moves in the third orbit is:
\[ \lambda = 6\pi a_0 \]

### Summary:
The de Broglie wavelength of the electron in the third orbit after absorbing energy is \( 6\pi a_0 \).