An electron in a hydrogen atom makes a transition `n_(1) rarr n_(2)`, where `n_(1) and n_(2)` are principal quantum numbers of the states. Assume the Bohr's model to be valid. The time period of the electron in the initial state is eight times to that of final state. What is ratio of `n_(2)//n_(1)`

Options

A. `8 : 1`
B. `4 : 1`
C. `2 : 1`
D. `1 : 2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between time period and principal quantum number
According to Bohr's model, the time period \( T \) of an electron in an orbit is directly proportional to the cube of the principal quantum number \( n \):
\[ T \propto n^3 \]
Since the atomic number \( Z \) for hydrogen is 1, we can express this relationship as:
\[ T \propto n^3 \]

### Step 2: Set up the relationship based on the given information
We are given that the time period of the electron in the initial state \( T_1 \) is 8 times that of the final state \( T_2 \):
\[ T_1 = 8 T_2 \]

### Step 3: Relate the time periods to the principal quantum numbers
Using the proportionality established in Step 1, we can write:
\[ T_1 = k n_1^3 \quad \text{and} \quad T_2 = k n_2^3 \]
where \( k \) is a constant. Substituting these into the equation from Step 2 gives:
\[ k n_1^3 = 8 (k n_2^3) \]
We can cancel \( k \) from both sides (assuming \( k \neq 0 \)):
\[ n_1^3 = 8 n_2^3 \]

### Step 4: Solve for the ratio of \( n_2 \) to \( n_1 \)
Now, we can express \( n_1 \) in terms of \( n_2 \):
\[ n_1^3 = 8 n_2^3 \implies n_1 = 2 n_2 \]
To find the ratio \( \frac{n_2}{n_1} \):
\[ \frac{n_2}{n_1} = \frac{n_2}{2 n_2} = \frac{1}{2} \]

### Conclusion
Thus, the ratio of \( n_2 \) to \( n_1 \) is:
\[ \frac{n_2}{n_1} = \frac{1}{2} \]