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. **Understand the Concept of de Broglie Wavelength**:
The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( v \) is its velocity.

2. **Relate Velocity to Orbit**:
For an electron in a circular orbit, the angular momentum is quantized and given by:
\[ mvr = \frac{n h}{2\pi} \]
where \( n \) is the principal quantum number (for the third orbit, \( n = 3 \)) and \( r \) is the radius of the orbit.

3. **Substituting for Velocity**:
Rearranging the angular momentum equation to find \( v \):
\[ v = \frac{n h}{2\pi m r} \]

4. **Substituting into the de Broglie Wavelength Formula**:
Substitute the expression for \( v \) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{m \left( \frac{n h}{2\pi m r} \right)} = \frac{2\pi r}{n} \]

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

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

7. **Final Result**:
Therefore, the de Broglie wavelength of the electron in the third orbit is:
\[ \lambda = 6\pi a_0 \]

### Final Answer:
The de Broglie wavelength of the electron when it moves in the third orbit after absorbing some definite amount of energy is \( 6\pi a_0 \).

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