The possible value of radius of orbit of H-atom is (`lambda` is the de-Broglie wavelength of electron)

Options

A. `(10lambda)/(11π)`
B. `(13lambda)/(7π)`
C. `(3lambda)/(2π)`
D. `(17lambda)/(9π)`

Answer

To find the possible value of the radius of the orbit of a hydrogen atom in terms of the de Broglie wavelength (\(\lambda\)) of the electron, we can follow these steps:

### Step 1: Understand the quantization of angular momentum
According to Bohr's model of the hydrogen atom, the angular momentum \(L\) of the electron is quantized and is given by the formula:
\[ L = mvr = n \frac{h}{2\pi} \]
where:
- \(m\) is the mass of the electron,
- \(v\) is the velocity of the electron,
- \(r\) is the radius of the orbit,
- \(n\) is a positive integer (quantum number),
- \(h\) is Planck's constant.

### Step 2: Relate angular momentum to the radius
From the angular momentum equation, we can express the radius \(r\) in terms of \(n\):
\[ r = \frac{n h}{2 \pi mv} \]

### Step 3: Use the de Broglie wavelength
According to de Broglie's hypothesis, the wavelength \(\lambda\) associated with a particle is given by:
\[ \lambda = \frac{h}{mv} \]
From this equation, we can express \(mv\) in terms of \(\lambda\):
\[ mv = \frac{h}{\lambda} \]

### Step 4: Substitute \(mv\) into the radius equation
Now, substitute \(mv\) back into the radius formula:
\[ r = \frac{n h}{2 \pi \left(\frac{h}{\lambda}\right)} \]
This simplifies to:
\[ r = \frac{n \lambda}{2 \pi} \]

### Conclusion
Thus, the possible value of the radius of the orbit of the hydrogen atom in terms of the de Broglie wavelength \(\lambda\) is:
\[ r = \frac{n \lambda}{2 \pi} \]