The radius of the second orbit of an electron in hydrogen atom is `2.116A`. The de Broglie wavelength associated with this electron in this orbit would be

Options

A. `6.64 Å`
B. `1.058 Å`
C. `2.116 Å`
D. `13.28 Å`

Answer

To find the de Broglie wavelength associated with the electron in the second orbit of a hydrogen atom, we can follow these steps:

### Step 1: Understand the relationship between de Broglie wavelength and angular momentum
The de Broglie wavelength (λ) 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. The momentum \( p \) can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

### Step 2: Use the angular momentum quantization condition
For an electron in a hydrogen atom, the angular momentum \( L \) is quantized and given by:
\[ L = mvr = n\frac{h}{2\pi} \]
where \( n \) is the principal quantum number (for the second orbit, \( n = 2 \)) and \( r \) is the radius of the orbit.

### Step 3: Express momentum in terms of known quantities
From the angular momentum equation, we can express \( mv \) as:
\[ mv = \frac{n h}{2\pi r} \]

### Step 4: Substitute \( mv \) into the de Broglie wavelength formula
Substituting \( mv \) into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{mv} = \frac{h}{\frac{n h}{2\pi r}} = \frac{2\pi r}{n} \]

### Step 5: Substitute the known values
We know:
- The radius of the second orbit \( r = 2.116 \, \text{Å} \)
- The principal quantum number for the second orbit \( n = 2 \)

Substituting these values into the equation:
\[ \lambda = \frac{2\pi (2.116 \, \text{Å})}{2} \]

### Step 6: Calculate the wavelength
Calculating this gives:
\[ \lambda = \pi (2.116 \, \text{Å}) \approx 3.14 \times 2.116 \, \text{Å} \approx 6.64 \, \text{Å} \]

### Final Answer
The de Broglie wavelength associated with the electron in the second orbit of the hydrogen atom is approximately:
\[ \lambda \approx 6.64 \, \text{Å} \]

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