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 Question:
An electron in a hydrogen atom jumps from the ground state to an excited state. We are given how its velocity changes and must determine how its orbital radius changes.

Step 2: Key Formula or Approach:
From Bohr's model of the hydrogen atom, we must know the dependencies of velocity and radius on the principal quantum number $n$:
1. Velocity: $v_n \propto \frac{1}{n}$
2. Radius: $r_n \propto n^2$

Step 3: Detailed Explanation:
Let the ground state be $n = 1$. The velocity is $v_1$ and radius is $r_1$.
The electron is excited to a new state $n$. Its new velocity is $v_n = v_1 / 3$.
Using the velocity proportionality:
$$\frac{v_n}{v_1} = \frac{1/n}{1/1} = \frac{1}{n}$$
Given $\frac{v_n}{v_1} = \frac{1}{3}$, we can conclude that the principal quantum number of the excited state is $n = 3$.
Now, calculate the new radius using the radius proportionality:
$$\frac{r_n}{r_1} = \frac{n^2}{1^2} = n^2$$
Substitute $n = 3$:
$$r_3 = r_1 \times (3)^2 = 9r_1$$

Step 4: Final Answer:
The radius of that orbit is $9r_1$, matching option (d).