Which is the correct relation between de- Brogile wavelength of an electron in the `n^(th)` Bohr orbit and radius of the orbit R?

Options

A. `lambda = n2piR`
B. `lambda = (2piR)/n`
C. `lambda = (4 pi R)/n`
D. `lambda = (2 pi R)/nh`

Answer

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

### Step 1: Understand de Broglie's Hypothesis
De Broglie proposed that particles, like electrons, exhibit both particle and wave properties. The wavelength associated with a particle is given by the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 2: Determine the Momentum of the Electron
In the \( n^{th} \) Bohr orbit, the momentum \( p \) of the electron can be expressed in terms of its mass \( m \) and velocity \( v \):
\[ p = mv \]

### Step 3: Relate the Wavelength to the Orbit
The circumference of the \( n^{th} \) orbit is given by:
\[ C = 2\pi R_n \]
where \( R_n \) is the radius of the \( n^{th} \) orbit. According to de Broglie's hypothesis, this circumference must be an integral multiple of the de Broglie wavelength \( \lambda \):
\[ C = n\lambda \]

### Step 4: Substitute the Wavelength
Substituting the expression for the wavelength into the equation gives:
\[ 2\pi R_n = n\lambda \]
This implies:
\[ \lambda = \frac{2\pi R_n}{n} \]

### Step 5: Conclusion
Thus, the relation between the de Broglie wavelength of an electron in the \( n^{th} \) Bohr orbit and the radius of the orbit \( R_n \) is:
\[ \lambda = \frac{2\pi R_n}{n} \]

### Final Relation
This indicates that the de Broglie wavelength is directly proportional to the radius of the orbit, scaled by the factor of \( \frac{2\pi}{n} \).