If `omega` the angular speed of electron in the nth orbit hydrogen atom, then

Options

A. `omega propn^(1//2)`
B. `omegaprop(1)/(n)`
C. `omega prop(1)/(n^(2))`
D. `omegaprop(1)/(n^(3))`

Explanation

To solve the problem regarding the angular speed (ω) of an electron in the nth orbit of a hydrogen atom, we can follow these steps:

### Step 1: Understand the relationship between linear speed and angular speed
The angular speed (ω) of an electron can be expressed in terms of its linear speed (v) and the radius (r) of its orbit:
\[ \omega = \frac{v}{r} \]

### Step 2: Determine the linear speed (v) of the electron
According to Bohr's model of the hydrogen atom, the linear speed (v) of the electron in the nth orbit is given by:
\[ v = \frac{2.2 \times 10^6 \cdot Z}{n} \]
where Z is the atomic number (for hydrogen, Z = 1).

### Step 3: Determine the radius (r) of the electron's orbit
The radius (r) of the electron in the nth orbit is given by:
\[ r = a_0 \cdot \frac{n^2}{Z} \]
where \( a_0 \) is the Bohr radius.

### Step 4: Substitute the expressions for v and r into the equation for ω
Now we can substitute the expressions for v and r into the equation for ω:
\[ \omega = \frac{v}{r} = \frac{\frac{2.2 \times 10^6 \cdot Z}{n}}{a_0 \cdot \frac{n^2}{Z}} \]

### Step 5: Simplify the expression for ω
Now, simplifying the expression:
\[ \omega = \frac{2.2 \times 10^6 \cdot Z^2}{a_0 \cdot n^3} \]
From this equation, we can see that ω is inversely proportional to \( n^3 \).

### Conclusion
Thus, we conclude that:
\[ \omega \propto \frac{1}{n^3} \]
This means that the correct answer is that ω is inversely proportional to \( n^3 \).

### Final Answer
The correct option is: **ω is inversely proportional to n³**.

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