The circumference of the `4^(th)` bohr orbit of hydrogen is 5.32 nm .The wavelenght of electron revolving in the first bohr orbit will be

Options

A. 0.133 nm
B. 0.3325 nm
C. 2.65 nm
D. 133 nm

Answer

To find the wavelength of the electron revolving in the first Bohr orbit of hydrogen, we can follow these steps:

### Step 1: Understand the relationship between circumference and radius
The circumference of a circular orbit is given by the formula:
\[ C = 2\pi r \]
where \( r \) is the radius of the orbit.

### Step 2: Use the given circumference of the 4th Bohr orbit
We are given that the circumference of the 4th Bohr orbit is 5.32 nm. Therefore, we can write:
\[ C_4 = 2\pi r_4 = 5.32 \, \text{nm} \]

### Step 3: Calculate the radius of the 4th Bohr orbit
From the circumference formula, we can solve for \( r_4 \):
\[ r_4 = \frac{C_4}{2\pi} = \frac{5.32 \, \text{nm}}{2\pi} \]

Calculating this gives:
\[ r_4 \approx \frac{5.32}{6.2832} \approx 0.847 \, \text{nm} \]

### Step 4: Use the Bohr radius formula to find the radius of the 1st Bohr orbit
The radius of the nth Bohr orbit is given by:
\[ r_n = r_0 n^2 \]
where \( r_0 \) is the radius of the first Bohr orbit (approximately 0.529 Å or 0.0529 nm) and \( n \) is the principal quantum number.

For the 4th orbit (n=4):
\[ r_4 = r_0 \cdot 4^2 = r_0 \cdot 16 \]

### Step 5: Relate the radius of the 4th orbit to the radius of the 1st orbit
We can express \( r_0 \) in terms of \( r_4 \):
\[ r_0 = \frac{r_4}{16} \]

Substituting \( r_4 \):
\[ r_0 = \frac{0.847 \, \text{nm}}{16} \approx 0.0529 \, \text{nm} \]

### Step 6: Calculate the wavelength of the electron in the first Bohr orbit
The wavelength \( \lambda \) of the electron can be calculated using the de Broglie wavelength formula:
\[ \lambda = \frac{h}{mv} \]
However, for Bohr orbits, we can use the relationship:
\[ \lambda = \frac{C_1}{n} \]
where \( C_1 \) is the circumference of the first orbit.

The circumference of the first orbit is:
\[ C_1 = 2\pi r_0 \]

Substituting \( r_0 \):
\[ C_1 = 2\pi (0.0529 \, \text{nm}) \approx 0.331 \, \text{nm} \]

Thus, the wavelength of the electron in the first Bohr orbit is:
\[ \lambda = C_1 = 0.331 \, \text{nm} \]

### Final Answer
The wavelength of the electron revolving in the first Bohr orbit is approximately **0.331 nm**.

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