Question

In a hydrogen atom an electron makes a transition from \(n_1 \rightarrow n_2\), where \(n_1\) and \(n_2\) are the principal quantum numbers of these two states. Assuming Bohr's model to be valid, if the time period of electron in the initial state is eight times of that in the final state, the possible values of \(n_1\) and \(n_2\) are:

• A. \(n_1=8,\ n_2=4\)
• B. \(n_1=4,\ n_2=8\)
• C. \(n_1=8,\ n_2=2\)
• D. \(n_1=2,\ n_2=8\)

Hint:

In Bohr model: \[ T_n \propto n^3 \] \[ r_n \propto n^2 \] \[ v_n \propto \frac{1}{n} \]

Answer 1

The Correct Option is A

Step 1: Write relation for time period in Bohr orbit.
In Bohr model: \[ T_n \propto n^3 \] where: \[ T_n=\text{time period of electron in }n^{th}\text{ orbit} \]

Step 2: Use the given condition.
Given: \[ T_1=8T_2 \] Using: \[ T \propto n^3 \] we get: \[ \frac{T_1}{T_2}=\frac{n_1^3}{n_2^3} \] Thus: \[ \frac{n_1^3}{n_2^3}=8 \] \[ \left(\frac{n_1}{n_2}\right)^3=2^3 \] \[ \frac{n_1}{n_2}=2 \]

Step 3: Check the options.
Option (A): \[ \frac{8}{4}=2 \] satisfies the condition. Other options do not satisfy: \[ \frac{n_1}{n_2}=2 \]

Step 4: Identify the correct option.
Therefore: \[ \boxed{n_1=8,\ n_2=4} \] Hence, the correct answer is: \[ \boxed{\mathrm{(A)}} \]