Animated Solution for Physics - Atoms and Nuclei: The electron in a hydrogen atom makes a transition $n_1\to n_2$, where $n_1$ and $n_2$ are the principal quantum numbers of two states. Assume the Bohr's model to be valid. The time period of the electron in the initial state is eight times that in the final state. The possible values of $n_1$ and $n_2$ are

Imagine an electron orbiting the nucleus of a hydrogen atom, much like a planet orbiting the sun. According to Bohr's model, these orbits are quantized, meaning the electron can only exist in certain allowed paths, each defined by a principal quantum number $n$. The problem tells us about an electron making a transition from an initial state $n_1$ to a final state $n_2$. We are given a crucial piece of information: the time period of the electron in the initial state is exactly eight times the time period in the final state. Our goal is to find the possible pairs of $(n_1,n_2)$ that satisfy this condition.

To relate the time period to the principal quantum number, we need to recall the basic kinematics of circular motion. The time period $T_n$ of an electron in the $n^{\text{th}}$ orbit is simply the circumference of the orbit divided by the electron's orbital velocity:

Now, we need to know how the radius $r_n$ and velocity $v_n$ depend on $n$. From Bohr's postulates, we know that the radius of the $n^{\text{th}}$ orbit scales with the square of the principal quantum number:

Similarly, the orbital velocity is inversely proportional to $n$:

Substituting these proportionalities back into our time period equation, we get:

This is a beautiful and powerful result! It tells us that the time it takes for an electron to complete one full orbit scales with the cube of its principal quantum number.

Armed with this relationship, let's return to the condition given in the problem:

Since $T_n\propto n^3$, we can rewrite this as:

Taking the cube root of both sides, we arrive at a simple, elegant linear relationship:

This means the initial quantum number must be exactly twice the final quantum number. Now, all we have to do is check the given options to see which ones fit this rule.

- Option (a): $n_1=4,n_2=2$. Here, $4=2(2)$, which is True. - Option (b): $n_1=8,n_2=2$. Here, $8 eq 2(2)$, which is False. - Option (c): $n_1=8,n_2=1$. Here, $8 eq 2(1)$, which is False. - Option (d): $n_1=6,n_2=3$. Here, $6=2(3)$, which is True.

Therefore, the possible values are given by options (a) and (d).