If `a_(0)` be the radius of first Bohr's orbit of H-atom, the de-Broglie's wavelength of an electron revolving in the second Bohr's orbit will be:

Options

A. `6pia_(0)`
B. `4pia_(0)`
C. `2pia_(0)`
D. None of these

Answer

To find the de Broglie's wavelength of an electron revolving in the second Bohr's orbit of a hydrogen atom, we can follow these steps:

### Step 1: Understand the relationship between angular momentum and the Bohr model
The angular momentum \( L \) of an electron in the nth orbit is quantized and given by the formula:
\[ L = mvr_n = \frac{nh}{2\pi} \]
where:
- \( m \) is the mass of the electron,
- \( v \) is the velocity of the electron,
- \( r_n \) is the radius of the nth orbit,
- \( n \) is the principal quantum number,
- \( h \) is Planck's constant.

### Step 2: Write the expression for the momentum
The momentum \( p \) of the electron can be expressed as:
\[ p = mv \]
Substituting this into the angular momentum equation gives:
\[ p \cdot r_n = \frac{nh}{2\pi} \]

### Step 3: Calculate for the second Bohr's orbit
For the second Bohr's orbit (\( n = 2 \)):
\[ p \cdot r_2 = \frac{2h}{2\pi} = \frac{h}{\pi} \]
Thus, we can express the momentum as:
\[ p = \frac{h}{\pi r_2} \]

### Step 4: Use the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
Substituting the expression for momentum:
\[ \lambda = \frac{h}{\left(\frac{h}{\pi r_2}\right)} = \pi r_2 \]

### Step 5: Find the radius of the second Bohr's orbit
The radius of the nth Bohr orbit is given by:
\[ r_n = \frac{0.529 \, n^2}{Z} \]
For hydrogen (\( Z = 1 \)) and \( n = 2 \):
\[ r_2 = 0.529 \cdot 2^2 = 0.529 \cdot 4 = 2.116 \, \text{Å} \]
Since \( a_0 = 0.529 \, \text{Å} \), we can express \( r_2 \) in terms of \( a_0 \):
\[ r_2 = 4a_0 \]

### Step 6: Substitute \( r_2 \) back into the wavelength equation
Now substituting \( r_2 \) into the wavelength equation:
\[ \lambda = \pi r_2 = \pi (4a_0) = 4\pi a_0 \]

### Final Answer
Thus, the de Broglie's wavelength of an electron revolving in the second Bohr's orbit is:
\[ \lambda = 4\pi a_0 \]

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