According to Bohr's theory of hydrogen atom , the product of the binding energy of the electron in the nth orbit and its radius in the nth orbit

Options

A. is proportional to `n^(2)`
B. is inversely proportional to `n^(3)`n
C. has a constant value of 10.2 eV-Å`
D. has a constant value of `7.2 eV-Å`

Answer

To solve the problem, we need to find the product of the binding energy of the electron in the nth orbit and its radius in the nth orbit according to Bohr's theory of the hydrogen atom.

### Step-by-Step Solution:

1. **Understand the Binding Energy and Radius:**
According to Bohr's theory:
- The binding energy \( E_n \) of the electron in the nth orbit is given by:
\[ E_n \propto -\frac{1}{n^2} \]
- The radius \( R_n \) of the nth orbit is given by:
\[ R_n \propto n^2 \]

2. **Calculate the Product:**
We need to find the product \( E_n \times R_n \):
\[ E_n \times R_n \propto \left(-\frac{1}{n^2}\right) \times (n^2) \]
Simplifying this gives:
\[ E_n \times R_n \propto -1 \]
This shows that the product is independent of \( n \).

3. **Substituting Values:**
For the first orbit (n=1), we can use known values:
- The binding energy for the first orbit \( E_1 = 13.6 \, \text{eV} \)
- The radius for the first orbit \( R_1 = 0.53 \, \text{Å} \)

Now, we calculate the product:
\[ E_1 \times R_1 = 13.6 \, \text{eV} \times 0.53 \, \text{Å} \]

4. **Perform the Calculation:**
\[ E_1 \times R_1 = 13.6 \times 0.53 = 7.228 \, \text{eV} \cdot \text{Å} \]
Rounding this gives approximately:
\[ E_1 \times R_1 \approx 7.2 \, \text{eV} \cdot \text{Å} \]

5. **Conclusion:**
The product of the binding energy of the electron in the nth orbit and its radius in the nth orbit is a constant value of approximately \( 7.2 \, \text{eV} \cdot \text{Å} \).

### Final Answer:
The product of the binding energy of the electron in the nth orbit and its radius in the nth orbit is \( 7.2 \, \text{eV} \cdot \text{Å} \).

---