The angular speed of the electron in the `n^(th)` Bohr orbit of the hydrogen atom is proportional to

Options

A. directly proportional to n
B. inversely proportional to `sqrt(n)`
C. inversely proportional to `n^2`
D. inversely proportional to `n^3`

Explanation

To determine the relationship between the angular speed of the electron in the `n^(th)` Bohr orbit of the hydrogen atom, we can follow these steps:

### Step 1: Understand Angular Momentum in Bohr's Model
In Bohr's model of the hydrogen atom, the angular momentum \( L \) of the electron in the \( n^{th} \) orbit is quantized and given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number and \( h \) is Planck's constant.

### Step 2: Relate Angular Momentum to Angular Speed
The angular momentum \( L \) can also be expressed in terms of the moment of inertia \( I \) and angular speed \( \omega \):
\[ L = I \omega \]
For an electron moving in a circular orbit, the moment of inertia \( I \) is given by:
\[ I = m r_n^2 \]
where \( m \) is the mass of the electron and \( r_n \) is the radius of the \( n^{th} \) orbit.

### Step 3: Substitute Moment of Inertia into Angular Momentum Equation
Substituting the expression for moment of inertia into the angular momentum equation gives:
\[ n \frac{h}{2\pi} = m r_n^2 \omega \]
From this, we can solve for \( \omega \):
\[ \omega = \frac{n h}{2\pi m r_n^2} \]

### Step 4: Determine the Relationship of \( r_n \)
From Bohr's model, the radius \( r_n \) of the \( n^{th} \) orbit is given by:
\[ r_n \propto \frac{n^2}{Z} \]
where \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)). Thus, we can write:
\[ r_n \propto n^2 \]
This means:
\[ r_n^2 \propto n^4 \]

### Step 5: Substitute \( r_n^2 \) into the Angular Speed Equation
Substituting \( r_n^2 \) back into the equation for \( \omega \):
\[ \omega = \frac{n h}{2\pi m (n^4)} \]
This simplifies to:
\[ \omega \propto \frac{n}{n^4} = \frac{1}{n^3} \]

### Conclusion
Thus, the angular speed \( \omega \) of the electron in the \( n^{th} \) Bohr orbit is inversely proportional to the cube of the principal quantum number \( n \):
\[ \omega \propto \frac{1}{n^3} \]

### Final Answer
The angular speed of the electron in the \( n^{th} \) Bohr orbit of the hydrogen atom is proportional to \( \frac{1}{n^3} \).

---