If the average life time of an excited state of hydrogen is of the order of `10^(-8) s`, estimate how many whits an alectron makes when it is in the state `n = 2` and before it suffers a transition to state` n = 1 (Bohrredius a_(0) = 5.3 xx 10^(-11)m)`?

Options

A. `10^(7)`
B. `8 xx 10^(6)`
C. `2 xx 10^(5)`
D. none of these

Explanation

To solve the problem step by step, we will follow the concepts of Bohr's model of the hydrogen atom.

### Step 1: Understanding the Problem
We need to estimate how many orbits an electron makes in the excited state \( n = 2 \) before transitioning to the ground state \( n = 1 \). The average lifetime of the excited state is given as \( \tau = 10^{-8} \) seconds.

### Step 2: Calculate the Radius for \( n = 2 \)
According to Bohr's model, the radius of the electron's orbit in the \( n \)-th state is given by:
\[ r_n = n^2 \cdot a_0 \]
where \( a_0 = 5.3 \times 10^{-11} \, \text{m} \) is the Bohr radius.

For \( n = 2 \):
\[ r_2 = 2^2 \cdot a_0 = 4 \cdot (5.3 \times 10^{-11}) = 2.12 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the Velocity of the Electron in \( n = 2 \)
The velocity of the electron in the \( n \)-th state is given by:
\[ v_n = \frac{e^2}{2 \epsilon_0 h} \cdot \frac{1}{n} \]
However, we can also use the angular momentum quantization condition:
\[ m v_n r_n = \frac{n h}{2 \pi} \]
Rearranging gives:
\[ v_n = \frac{n h}{2 \pi m r_n} \]

Substituting \( r_n \):
\[ v_2 = \frac{2 h}{2 \pi m (2^2 a_0)} = \frac{h}{2 \pi m (4 a_0)} \]

### Step 4: Calculate the Time Period \( T \)
The time period \( T \) for one complete orbit is given by:
\[ T = \frac{2 \pi r_n}{v_n} \]
Substituting \( r_n \) and \( v_n \):
\[ T = \frac{2 \pi (4 a_0)}{v_2} \]

### Step 5: Calculate the Number of Orbits
The number of orbits \( N \) made by the electron before transitioning is given by:
\[ N = \frac{\tau}{T} \]

### Step 6: Substitute Values and Calculate
Using the known values:
- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Mass of the electron \( m = 9.11 \times 10^{-31} \, \text{kg} \)

We can now calculate \( T \) and subsequently \( N \).

1. Calculate \( v_2 \):
\[ v_2 = \frac{h}{2 \pi m (4 a_0)} \]

2. Calculate \( T \):
\[ T = \frac{2 \pi (4 a_0)}{v_2} \]

3. Finally, calculate \( N \):
\[ N = \frac{10^{-8}}{T} \]

### Final Calculation
After performing the calculations, we find that:
\[ N \approx 8 \times 10^6 \]

### Conclusion
The number of orbits the electron makes in the \( n = 2 \) state before transitioning to \( n = 1 \) is approximately \( 8 \times 10^6 \).

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