The radius of Bohr's first orbit is `a_(0)` . The electron in `n^(th) `orbit has a radius `:`

Options

A. `na_(0)`
B. `n^(2)//a_(0)`
C. `n^(2)a_(0)`
D. `a_(0)//n^(2)`

Answer

To find the radius of the electron in the nth orbit according to Bohr's model, we can follow these steps:

### Step 1: Understand the radius of the first orbit
The radius of Bohr's first orbit is given as \( a_0 \). This is a known constant in Bohr's model of the hydrogen atom.

### Step 2: Use Bohr's quantization condition
According to Bohr's model, the angular momentum of the electron in the nth orbit is quantized and given by the equation:
\[ L = m v_n R_n = \frac{n h}{2 \pi} \]
where:
- \( L \) is the angular momentum,
- \( m \) is the mass of the electron,
- \( v_n \) is the velocity of the electron in the nth orbit,
- \( R_n \) is the radius of the nth orbit,
- \( n \) is the principal quantum number (1, 2, 3,...),
- \( h \) is Planck's constant.

### Step 3: Set up the centripetal force equation
The centripetal force required to keep the electron in circular motion is provided by the electrostatic (Coulomb) force between the electron and the nucleus. This can be expressed as:
\[ \frac{m v_n^2}{R_n} = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{R_n^2} \]
where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( e \) is the charge of the electron.

### Step 4: Combine the two equations
From the angular momentum equation, we can express \( v_n \) in terms of \( R_n \):
\[ v_n = \frac{n h}{2 \pi m R_n} \]
Substituting this expression for \( v_n \) into the centripetal force equation gives:
\[ \frac{m \left(\frac{n h}{2 \pi m R_n}\right)^2}{R_n} = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{R_n^2} \]
Simplifying this equation leads to:
\[ \frac{n^2 h^2}{4 \pi^2 m R_n^3} = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{R_n^2} \]

### Step 5: Solve for \( R_n \)
Rearranging the equation to solve for \( R_n \) gives:
\[ R_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2} \]
This shows that the radius of the nth orbit is proportional to \( n^2 \).

### Step 6: Relate \( R_n \) to \( a_0 \)
Since \( a_0 = \frac{h^2 \epsilon_0}{\pi m e^2} \), we can express \( R_n \) in terms of \( a_0 \):
\[ R_n = n^2 a_0 \]

### Final Answer
Thus, the radius of the electron in the nth orbit is:
\[ R_n = n^2 a_0 \]

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