The de-Broglie wavelength of an electron in 4th orbit is (where, r=radius of 1st orbit)

Options

A. `2pir`
B. `4pir`
C. `8pir`
D. `16pir`

Explanation

To find the de-Broglie wavelength of an electron in the 4th orbit, we will follow these steps:

### Step 1: Understand the relationship between the circumference of an orbit and the de-Broglie wavelength.
According to the Bohr model, the circumference of the nth orbit is an integral multiple of the de-Broglie wavelength (λ). Mathematically, this can be expressed as:
\[ 2\pi r_n = n \lambda \]
where \( r_n \) is the radius of the nth orbit and \( n \) is the principal quantum number.

### Step 2: Write the equation for the 4th orbit.
For the 4th orbit (n = 4), the equation becomes:
\[ 2\pi r_4 = 4 \lambda \]
From this, we can express the wavelength as:
\[ \lambda = \frac{2\pi r_4}{4} = \frac{\pi r_4}{2} \]

### Step 3: Determine the radius of the 4th orbit.
The radius of the nth orbit in the Bohr model is given by the formula:
\[ r_n = a_0 n^2 \]
where \( a_0 \) is the radius of the first orbit. Given that \( r = a_0 \), we can write:
\[ r_4 = a_0 \cdot 4^2 = 16 a_0 \]
Since \( a_0 \) is given as \( r \), we substitute:
\[ r_4 = 16r \]

### Step 4: Substitute \( r_4 \) back into the wavelength equation.
Now, substituting \( r_4 \) into the wavelength equation:
\[ \lambda = \frac{\pi (16r)}{2} = 8\pi r \]

### Conclusion
Thus, the de-Broglie wavelength of an electron in the 4th orbit is:
\[ \lambda = 8\pi r \]

### Final Answer
The correct option is **8πr**.

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