According to the Bohr theory of Hydrogen atom, the speed of the electron, its energy and the radius of its orbit varies with the principal quantum number n, respectively, as

Options

A. `(1)/(n), (1)/(n^(2)), n^(2)`
B. `(1)/(n), n^(2), (1)/(n^(2))`
C. `n^(2), (1)/(n^(2)), n^(2)`
D. `n, (1)/(n^(2)), (1)/(n^(2))`

Answer

To solve the question regarding how the speed of the electron, its energy, and the radius of its orbit vary with the principal quantum number \( n \) according to Bohr's theory of the hydrogen atom, we can follow these steps:

### Step 1: Understanding the relationship of speed with quantum number \( n \)

According to Bohr's model, the speed of the electron in the nth orbit is given by the formula:

\[ v_n = \frac{Z e^2}{2 \epsilon_0 h} \cdot \frac{1}{n} \]

For a hydrogen atom where \( Z = 1 \), this simplifies to:

\[ v_n \propto \frac{1}{n} \]

**Hint:** The speed of the electron decreases as the principal quantum number \( n \) increases.

### Step 2: Understanding the relationship of energy with quantum number \( n \)

The total energy of the electron in the nth orbit is given by:

\[ E_n = -\frac{Z^2 e^4 m}{2 (4 \pi \epsilon_0)^2 h^2 n^2} \]

For hydrogen (\( Z = 1 \)), this simplifies to:

\[ E_n \propto -\frac{1}{n^2} \]

**Hint:** The energy becomes less negative (increases) as \( n \) increases, indicating that the electron is less bound.

### Step 3: Understanding the relationship of radius with quantum number \( n \)

The radius of the nth orbit is given by:

\[ r_n = \frac{n^2 h^2}{4 \pi^2 m e^2} \cdot Z^{-1} \]

For hydrogen, this simplifies to:

\[ r_n \propto n^2 \]

**Hint:** The radius of the orbit increases with the square of the principal quantum number \( n \).

### Conclusion

Based on the above relationships, we can summarize the variations as follows:

- The speed of the electron \( v_n \) varies as \( \frac{1}{n} \)
- The energy of the electron \( E_n \) varies as \( -\frac{1}{n^2} \)
- The radius of the orbit \( r_n \) varies as \( n^2 \)

Thus, the final answer is:

- Speed: \( \frac{1}{n} \)
- Energy: \( -\frac{1}{n^2} \)
- Radius: \( n^2 \)

### Final Answer:
- Speed of electron: \( \frac{1}{n} \)
- Energy of electron: \( -\frac{1}{n^2} \)
- Radius of orbit: \( n^2 \)