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: Understanding the Question:
The question asks us to identify the correct plot of the velocity ($v$) of an electron in a hydrogen-like atom's Bohr orbits as a function of the principal quantum number ($n$).
Step 2: Key Formula or Approach:
According to Bohr's model of the hydrogen atom, the velocity of an electron in the $n$-th orbit is given by:
\[ v_n = \frac{2\pi k Z e^2}{n h} = v_0 \frac{Z}{n} \]
Step 3: Detailed Explanation:
• Let us look at the relationship between velocity ($v$) and the principal quantum number ($n$) for a given hydrogen-like atom ($Z$ is constant):
\[ v \propto \frac{1}{n} \]
• When the principal quantum number is $n=1$ (the first Bohr orbit), the velocity is given as $v_0$.
• Mathematically, the function $y = \frac{C}{x}$ represents a rectangular hyperbola.
• As $n$ increases, the velocity of the electron decreases asymptotically toward zero.
• Let us evaluate the options:
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• Option A shows a hyperbolic curve where the velocity at $n=3$ is exactly equal to $\frac{v_0}{3}$, which perfectly matches our equation $v_n = \frac{v_0}{n}$.
• Option B shows a curve where the velocity at $n=3$ is $\frac{v_0}{9}$, which corresponds to a $1/n^2$ relationship (incorrect).
• Options C and D show the velocity increasing as $n$ increases, which is physically incorrect.
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Step 4: Final Answer:
Therefore, the correct graph is the one shown in Option A.
