If velocity of an electron in the first Bohr orbit is denoted by $v_0$, then the velocity ($v$) of the electron in other orbits (as a function of principle quantum number `n') is represented as:
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Remember: As an electron moves to higher energy levels (larger $n$), it is further from the nucleus and experiences a weaker electrostatic attraction, meaning it moves slower ($v \propto \frac{1}{n}$).
Step 1: Recall the Bohr velocity formula. In Bohr's model, the speed of the electron in the $n$-th orbit is $v_n = \frac{2\pi k Z e^2}{n h}$. Every term here is a constant except the $n$ in the bottom. So the speed depends only on $n$ in a simple way.
Step 2: Pull out the constants. Group all the fixed quantities together as one number. For the first orbit, $n = 1$, so $v_0 = \frac{2\pi k Z e^2}{1 \cdot h}$. That fixed group is exactly $v_0$.
Step 3: Write the general orbit in terms of $v_0$. Since the only change for a general $n$ is the $n$ in the denominator, we can write $v_n = \frac{v_0}{n}$. So the speed in orbit $n$ is just $v_0$ divided by $n$.
Step 4: Understand the relationship. This means velocity is inversely proportional to $n$. As $n$ gets bigger (outer orbits), the electron moves slower. As $n$ gets smaller, it moves faster.
Step 5: Picture the graph. A relation $v = \frac{v_0}{n}$ is a rectangular hyperbola shape when plotted against $n$. It starts high at $n = 1$ (where $v = v_0$) and keeps dropping, halving at $n = 2$, becoming a third at $n = 3$, and so on, but never reaching zero.
Step 6: Choose the curve. So the correct plot is the one falling off as $\frac{1}{n}$, passing through $v_0$ at $n = 1$ and $\frac{v_0}{2}$ at $n = 2$. That is option A.
