The product of angular speed and tangential speed of electron in `n^(th)` orbit of hydrogen atom is

Options

A. Inversly proportional to `n^2`
B. Directly proportional to `n^2`
C. inversly proportional to` n^4`
D. Independent of `n`

Answer

To find the product of angular speed and tangential speed of an electron in the \( n^{th} \) orbit of a hydrogen atom, we can follow these steps:

### Step 1: Define Angular Speed and Tangential Speed
- The angular speed (\( \omega \)) of an electron in the \( n^{th} \) orbit can be defined as:
\[ \omega = \frac{v_t}{r} \]
where \( v_t \) is the tangential speed and \( r \) is the radius of the orbit.

### Step 2: Tangential Speed in the \( n^{th} \) Orbit
- The tangential speed (\( v_t \)) of an electron in the \( n^{th} \) orbit is given by:
\[ v_t = \frac{2\pi r}{T} \]
where \( T \) is the time period of revolution.

### Step 3: Relationship of Radius and Speed
- For the hydrogen atom, the radius of the \( n^{th} \) orbit is given by:
\[ r_n = n^2 a_0 \]
where \( a_0 \) is the Bohr radius.

### Step 4: Expressing Angular Speed
- The angular speed can be expressed as:
\[ \omega = \frac{v_t}{r} \]

### Step 5: Product of Angular Speed and Tangential Speed
- Now, we can find the product of angular speed and tangential speed:
\[ \text{Product} = \omega \cdot v_t \]
Substituting \( \omega \) from Step 1:
\[ \text{Product} = \left(\frac{v_t}{r}\right) \cdot v_t = \frac{v_t^2}{r} \]

### Step 6: Using the Relation of \( v_t \) and \( r \)
- From the Bohr model, we know that:
\[ v_t^2 \propto \frac{1}{n^4} \]
and
\[ r \propto n^2 \]
Therefore, we can write:
\[ \frac{v_t^2}{r} \propto \frac{1/n^4}{n^2} = \frac{1}{n^6} \]

### Conclusion
- Thus, the product of angular speed and tangential speed of an electron in the \( n^{th} \) orbit of a hydrogen atom is inversely proportional to \( n^6 \).

### Final Answer
The product of angular speed and tangential speed of an electron in the \( n^{th} \) orbit of hydrogen atom is \( \frac{1}{n^6} \).

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