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)`?

Explanation

To solve the problem of estimating how many revolutions an electron makes when it is in the state \( n = 2 \) before transitioning to state \( n = 1 \), we can follow these steps:

### Step 1: Calculate the velocity of the electron in the \( n = 2 \) state.
The velocity \( V_n \) of an electron in a hydrogen atom can be calculated using the formula:
\[ V_n = \frac{2.18 \times 10^6 \cdot Z}{n} \text{ m/s} \]
For hydrogen, \( Z = 1 \) and \( n = 2 \):
\[ V_2 = \frac{2.18 \times 10^6 \cdot 1}{2} = 1.09 \times 10^6 \text{ m/s} \]

### Step 2: Calculate the radius of the \( n = 2 \) orbit.
The radius \( R_n \) of the \( n \)-th orbit is given by:
\[ R_n = n^2 \cdot R_1 \]
Where \( R_1 \) (the Bohr radius) is approximately \( 5.3 \times 10^{-11} \text{ m} \). For \( n = 2 \):
\[ R_2 = 2^2 \cdot R_1 = 4 \cdot 5.3 \times 10^{-11} = 2.12 \times 10^{-10} \text{ m} \]

### Step 3: Calculate the number of revolutions in the lifetime of the excited state.
The average lifetime of the excited state is given as \( 10^{-8} \text{ s} \). The distance traveled by the electron in this time can be calculated as:
\[ \text{Distance} = V_2 \cdot \text{time} = 1.09 \times 10^6 \text{ m/s} \cdot 10^{-8} \text{ s} = 1.09 \times 10^{-2} \text{ m} \]

### Step 4: Calculate the circumference of the \( n = 2 \) orbit.
The circumference \( C \) of the orbit is given by:
\[ C = 2 \pi R_2 = 2 \pi (2.12 \times 10^{-10}) \approx 1.33 \times 10^{-9} \text{ m} \]

### Step 5: Calculate the number of revolutions made before the transition.
The number of revolutions \( N \) can be calculated by dividing the distance traveled by the circumference of the orbit:
\[ N = \frac{\text{Distance}}{C} = \frac{1.09 \times 10^{-2}}{1.33 \times 10^{-9}} \approx 8.2 \times 10^6 \]

Thus, the electron makes approximately \( 8.2 \times 10^6 \) revolutions before transitioning from state \( n = 2 \) to state \( n = 1 \).