Question

In hydrogen atom in its ground state, the first Bohr orbit has radius $r_1$. When the atom is raised to one of its excited states, the electron's orbital velocity becomes one-third. The radius of that orbit is \dots

• A. $2r_1$
• B. $3r_1$
• C. $4r_1$
• D. $9r_1$

Hint:

Since $v \propto 1/n$ and $r \propto n^2$, you can directly relate them: $r \propto 1/v^2$. If velocity decreases by a factor of 3, the radius must increase by a factor of $3^2 = 9$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In Bohr's model, the orbital velocity $v_n$ and radius $r_n$ of the $n^{th}$ orbit are related to the principal quantum number $n$.

Step 2: Formula Application:
$v_n \propto \frac{1}{n}$ and $r_n \propto n^2$.

Step 3: Explanation:
Given $v_{new} = \frac{1}{3} v_1 \implies \frac{1}{n} = \frac{1}{3} \implies n = 3$. Now, for $n = 3$, the radius $r_3 = r_1 \times n^2 = r_1 \times 3^2 = 9r_1$.

Step 4: Final Answer:
The radius of the orbit is $9r_1$.