An electron in a hydrogen atom makes a transition from `n_(1) to n_(2)`. If the time period of electron in the initial state is eight times that in the final state then Find the ratio `n_(1)/n_(2)`

Options

A. 2
B. 3
C. 4
D. 8

Answer

To solve the problem step by step, we will use Bohr's model of the hydrogen atom, which provides a relationship between the time period of an electron in an orbit and the principal quantum number \( n \).

### 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 \( n \):
\[ T \propto n^3 \]
This can be expressed mathematically as:
\[ T \propto n^3 \implies T = k n^3 \]
where \( k \) is a constant.

### Step 2: Set up the relationship for the initial and final states
Let \( T_1 \) be the time period of the electron in the initial state (with quantum number \( n_1 \)) and \( T_2 \) be the time period in the final state (with quantum number \( n_2 \)). According to the problem, we have:
\[ T_1 = 8 T_2 \]

### Step 3: Express the time periods in terms of \( n_1 \) and \( n_2 \)
Using the proportionality established in Step 1, we can write:
\[ T_1 = k n_1^3 \quad \text{and} \quad T_2 = k n_2^3 \]

### Step 4: Substitute the expressions into the equation
Substituting these expressions into the equation \( T_1 = 8 T_2 \):
\[ k n_1^3 = 8 (k n_2^3) \]
We can cancel \( k \) from both sides (assuming \( k \neq 0 \)):
\[ n_1^3 = 8 n_2^3 \]

### Step 5: Solve for the ratio \( \frac{n_1}{n_2} \)
Taking the cube root of both sides gives:
\[ \frac{n_1}{n_2} = \sqrt[3]{8} = 2 \]

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

### Final Answer
The ratio \( \frac{n_1}{n_2} \) is \( 2 \).

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