Avarage lifetime of a hydrogen atom excited to `n =2` state `10^(-6)s` find the number of revolutions made by the electron on the avarage before it jump to the ground state

Explanation

To solve the problem of finding the number of revolutions made by an electron in a hydrogen atom excited to the n=2 state before it jumps to the ground state, we can follow these steps:

### Step 1: Understand the Given Information
We are given:
- Average lifetime of the hydrogen atom in the n=2 state, \( \tau = 10^{-6} \) seconds.
- We need to find the number of revolutions made by the electron before it transitions to the ground state.

### Step 2: Calculate the Time Period of the Electron's Revolution
The time period \( T \) of the electron in the n=2 state can be calculated using the formula:

\[ T = \frac{4 \pi \epsilon_0 n^3 h^2}{m_e e^4} \]

Where:
- \( \epsilon_0 \) = permittivity of free space = \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- \( n \) = principal quantum number (for n=2, \( n^3 = 2^3 = 8 \))
- \( h \) = Planck's constant = \( 6.63 \times 10^{-34} \, \text{Js} \)
- \( m_e \) = mass of the electron = \( 9.1 \times 10^{-31} \, \text{kg} \)
- \( e \) = charge of the electron = \( 1.6 \times 10^{-19} \, \text{C} \)

Substituting these values into the formula:

\[ T = \frac{4 \pi (8.85 \times 10^{-12}) (8) (6.63 \times 10^{-34})^2}{(9.1 \times 10^{-31}) (1.6 \times 10^{-19})^4} \]

### Step 3: Perform the Calculation
Calculating the numerator and denominator separately:

1. **Numerator:**
\[ 4 \pi (8.85 \times 10^{-12}) (8) (6.63 \times 10^{-34})^2 \]

2. **Denominator:**
\[ (9.1 \times 10^{-31}) (1.6 \times 10^{-19})^4 \]

After performing the calculations, we find:

\[ T \approx 1.2247735 \times 10^{-19} \text{ seconds} \]

### Step 4: Calculate the Number of Revolutions
The number of revolutions \( N \) made by the electron before it jumps to the ground state can be calculated using the formula:

\[ N = \frac{\tau}{T} \]

Substituting the values we have:

\[ N = \frac{10^{-6}}{1.2247735 \times 10^{-19}} \]

### Step 5: Perform the Final Calculation
Calculating \( N \):

\[ N \approx 8.2 \times 10^8 \text{ revolutions} \]

### Final Answer
The number of revolutions made by the electron before it jumps to the ground state is approximately \( 8.2 \times 10^8 \).

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