Calculate the angular frequency of revolution of an electron occupying the second Bohr orbit of `He^(+)` ion.

Answer

To calculate the angular frequency of revolution of an electron occupying the second Bohr orbit of the \( He^+ \) ion, we can follow these steps:

### Step 1: Understand the formulas
The angular frequency \( \omega \) can be expressed in terms of the velocity \( v \) of the electron and the radius \( r_n \) of the orbit:
\[ \omega = \frac{v}{r_n} \]

### Step 2: Determine the velocity of the electron
The velocity \( v \) of an electron in the \( n \)-th orbit of a hydrogen-like atom is given by:
\[ v = \frac{2 \pi z e^2}{n h} \]
where:
- \( z \) is the atomic number (for \( He^+ \), \( z = 2 \)),
- \( e \) is the charge of the electron,
- \( n \) is the principal quantum number (for the second orbit, \( n = 2 \)),
- \( h \) is Planck's constant.

### Step 3: Determine the radius of the orbit
The radius \( r_n \) of the \( n \)-th orbit is given by:
\[ r_n = \frac{n^2 h^2}{4 \pi^2 m z e^2} \]
where:
- \( m \) is the mass of the electron.

### Step 4: Substitute \( v \) and \( r_n \) into the equation for \( \omega \)
Substituting the expressions for \( v \) and \( r_n \) into the equation for \( \omega \):
\[ \omega = \frac{\frac{2 \pi z e^2}{n h}}{\frac{n^2 h^2}{4 \pi^2 m z e^2}} = \frac{2 \pi z e^2 \cdot 4 \pi^2 m z e^2}{n h \cdot n^2 h^2} \]
This simplifies to:
\[ \omega = \frac{8 \pi m z^2 e^4}{n^3 h^3} \]

### Step 5: Substitute known values
Now we can substitute the known values:
- \( z = 2 \) (for \( He^+ \)),
- \( n = 2 \),
- \( m = 9.1 \times 10^{-28} \) grams,
- \( e = 4.8 \times 10^{-10} \) ESU,
- \( h = 6.63 \times 10^{-27} \) erg·s.

### Step 6: Calculate \( \omega \)
Substituting these values into the equation:
\[ \omega = \frac{8 \pi (9.1 \times 10^{-28}) (2^2) (4.8 \times 10^{-10})^4}{(2^3) (6.63 \times 10^{-27})^3} \]

### Step 7: Perform the calculation
After performing the calculation, we find:
\[ \omega \approx 2.067 \times 10^{16} \text{ s}^{-1} \]

### Final Answer
Thus, the angular frequency of revolution of the electron in the second Bohr orbit of the \( He^+ \) ion is:
\[ \omega \approx 2.067 \times 10^{16} \text{ s}^{-1} \]

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