The De Broglie wavelenght associated with electron in n Bohr orbit is

Options

A. `(2pir)/(n) A^(0)`
B. `2pin A^(0)`
C. `(1)/(n)A^(0)`
D. `nA^(0)`

Explanation

To find the de Broglie wavelength associated with an electron in the n-th Bohr orbit, we can follow these steps:

### Step-by-Step Solution:

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

2. **Relate Angular Momentum to Velocity and Radius:**
The angular momentum can also be expressed in terms of the mass (m), velocity (v), and radius (r) of the orbit:
\[ L = mvr \]
Setting these two expressions for angular momentum equal gives:
\[ mvr = n \frac{h}{2\pi} \]

3. **Solve for Velocity (v):**
Rearranging the equation for velocity, we have:
\[ v = \frac{n h}{2\pi m r} \]

4. **Use de Broglie's Wavelength Formula:**
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
Substituting the expression for \( v \) from the previous step into this equation gives:
\[ \lambda = \frac{h}{m \left( \frac{n h}{2\pi m r} \right)} \]

5. **Simplify the Expression:**
Simplifying the above expression:
\[ \lambda = \frac{h \cdot 2\pi m r}{n h} \]
This simplifies to:
\[ \lambda = \frac{2\pi r}{n} \]

6. **Final Result:**
Thus, the de Broglie wavelength associated with an electron in the n-th Bohr orbit is:
\[ \lambda = \frac{2\pi r}{n} \]