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 the circumference of the Bohr orbits and the radius.
The circumference of a Bohr orbit is given by the formula:
\[ C_n = 2\pi R_n \]
where \( R_n \) is the radius of the nth orbit.

### Step 2: Use the given circumference of the 4th Bohr orbit.
We are given that the circumference of the 4th Bohr orbit (\( C_4 \)) is 5.32 nm. We can express this in meters:
\[ C_4 = 5.32 \, \text{nm} = 5.32 \times 10^{-9} \, \text{m} \]

### Step 3: Calculate the radius of the 4th Bohr orbit.
Using the circumference formula:
\[ C_4 = 2\pi R_4 \]
We can rearrange this to find \( R_4 \):
\[ R_4 = \frac{C_4}{2\pi} = \frac{5.32 \times 10^{-9}}{2\pi} \]

### Step 4: Calculate the value of \( R_4 \).
Calculating \( R_4 \):
\[ R_4 = \frac{5.32 \times 10^{-9}}{6.2832} \approx 8.47 \times 10^{-10} \, \text{m} \]

### Step 5: Use the formula for the radius of the nth Bohr orbit.
The radius of the nth Bohr orbit is given by:
\[ R_n = R_0 \frac{n^2}{Z} \]
For hydrogen (\( Z = 1 \)):
\[ R_n = R_0 n^2 \]
where \( R_0 \) is the Bohr radius, approximately \( 5.29 \times 10^{-11} \, \text{m} \).

### Step 6: Calculate the radius of the 1st Bohr orbit.
For the first Bohr orbit (\( n = 1 \)):
\[ R_1 = R_0 \cdot 1^2 = R_0 = 5.29 \times 10^{-11} \, \text{m} \]

### Step 7: 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, we can also relate the radius of the orbit to the wavelength:
\[ \lambda = \frac{C_1}{n} \]
where \( C_1 = 2\pi R_1 \).

### Step 8: Calculate the circumference of the 1st Bohr orbit.
Using the radius we calculated:
\[ C_1 = 2\pi R_1 = 2\pi (5.29 \times 10^{-11}) \]

### Step 9: Calculate \( C_1 \).
Calculating \( C_1 \):
\[ C_1 = 2\pi (5.29 \times 10^{-11}) \approx 3.32 \times 10^{-10} \, \text{m} \]

### Step 10: Find the wavelength of the electron in the first Bohr orbit.
Since \( n = 1 \):
\[ \lambda = C_1 \approx 3.32 \times 10^{-10} \, \text{m} \]

### Final Answer:
The wavelength of the electron revolving in the first Bohr orbit is approximately:
\[ \lambda \approx 3.32 \times 10^{-10} \, \text{m} \text{ or } 0.332 \, \text{nm} \]

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