The de Broglie wavelength of an electron in the ` n^(th)` Bohr orbit is related to the radius R of the orbit as

Options

A. `n lambda = pi R `
B. `n lambda = (3)/(2) pi R `
C. ` n lambda = 2 pi R `
D. ` n lambda = 4 pi R `

Answer

To find the relationship between the de Broglie wavelength of an electron in the \( n^{th} \) Bohr orbit and the radius \( R \) of that orbit, we can follow these steps:

### Step 1: Understand de Broglie Wavelength
The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle. For an electron, the momentum \( p \) can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

### Step 2: Relate Momentum to Angular Momentum
According to Bohr's model, the angular momentum (\( L \)) of the electron in the \( n^{th} \) orbit is quantized and given by:
\[ L = mvr = \frac{nh}{2\pi} \]
where \( r \) is the radius of the orbit and \( n \) is the principal quantum number.

### Step 3: Substitute Momentum in de Broglie Equation
From the de Broglie wavelength equation, substituting for momentum, we have:
\[ \lambda = \frac{h}{mv} \]
Now, we can express \( mv \) in terms of angular momentum:
\[ mv = \frac{nh}{2\pi r} \]
Substituting this into the de Broglie wavelength equation gives:
\[ \lambda = \frac{h}{\frac{nh}{2\pi r}} = \frac{2\pi r}{n} \]

### Step 4: Rearranging the Equation
From the above equation, we can rearrange it to express the relationship between the de Broglie wavelength and the radius:
\[ n\lambda = 2\pi r \]

### Conclusion
Thus, the relationship between the de Broglie wavelength of an electron in the \( n^{th} \) Bohr orbit and the radius \( R \) of that orbit is:
\[ n\lambda = 2\pi R \]