The de-Broglie wavelength of an electron moving in the nth Bohr orbit of radius ris given by

Options

A. nπr
B. nr/2π
C. 2π r/n
D. π r/n

Answer

To find the de-Broglie wavelength of an electron moving in the nth Bohr orbit of radius \( r \), we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

### Step 2: Express momentum in terms of mass and velocity
The momentum \( p \) of the electron can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

### Step 3: Substitute momentum into the de-Broglie wavelength formula
Substituting the expression for momentum into the de-Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{mv} \]

### Step 4: Use the angular momentum quantization condition
According to Bohr's model, the angular momentum \( L \) of the electron in the nth orbit is quantized and given by:
\[ L = mvr = \frac{nh}{2\pi} \]
where \( n \) is the principal quantum number.

### Step 5: Solve for velocity
From the angular momentum equation, we can express the velocity \( v \) as:
\[ v = \frac{nh}{2\pi mr} \]

### Step 6: Substitute velocity back into the de-Broglie wavelength formula
Now, substituting this expression for \( v \) back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{m \left(\frac{nh}{2\pi mr}\right)} = \frac{h \cdot 2\pi mr}{nh} \]

### Step 7: Simplify the expression
This simplifies to:
\[ \lambda = \frac{2\pi r}{n} \]

### Final Answer
Thus, the de-Broglie wavelength of an electron moving in the nth Bohr orbit of radius \( r \) is given by:
\[ \lambda = \frac{2\pi r}{n} \]