The velocity of an e in excited state of H-atom is `1.093 xx 10^(6)m//s`, what is the circumference of this orbit?

Options

A. (a) `3.32xx10^(-10)` m
B. (b) `6.64xx10^(-10)` m
C. (c) `13.30xx10^(-10)` m
D. (d) `13.28xx10^(-8)` m

Answer

To find the circumference of the orbit of an electron in an excited state of a hydrogen atom, we can follow these steps:

### Step 1: Identify the given data
- The velocity of the electron \( v = 1.093 \times 10^6 \, \text{m/s} \)
- The atomic number of hydrogen \( Z = 1 \)

### Step 2: Use Bohr's theory to find the principal quantum number \( n \)
According to Bohr's theory, the velocity of an electron in an orbit is given by the formula:
\[ v = \frac{2.18 \times 10^6 \cdot Z}{n} \]
Substituting the known values:
\[ 1.093 \times 10^6 = \frac{2.18 \times 10^6 \cdot 1}{n} \]

### Step 3: Solve for \( n \)
Rearranging the equation to solve for \( n \):
\[ n = \frac{2.18 \times 10^6}{1.093 \times 10^6} \]
Calculating \( n \):
\[ n = 2 \]

### Step 4: Calculate the radius of the orbit for \( n = 2 \)
The radius of the orbit in Bohr's model is given by:
\[ R_n = \frac{0.529 \, \text{Å} \cdot n^2}{Z} \]
Substituting \( n = 2 \) and \( Z = 1 \):
\[ R_2 = \frac{0.529 \cdot 2^2}{1} = 0.529 \cdot 4 = 2.116 \, \text{Å} \]

### Step 5: Calculate the circumference of the orbit
The circumference \( C \) of the orbit is given by:
\[ C = 2 \pi R_n \]
Substituting the value of \( R_2 \):
\[ C = 2 \pi \cdot 2.116 \, \text{Å} \]
Calculating \( C \):
\[ C = 2 \cdot 3.14 \cdot 2.116 \approx 13.30 \, \text{Å} \]

### Step 6: Convert the circumference from Ångströms to meters
Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):
\[ C = 13.30 \, \text{Å} = 13.30 \times 10^{-10} \, \text{m} \]

### Final Answer
The circumference of the orbit is:
\[ C \approx 1.33 \times 10^{-9} \, \text{m} \]

---