Question

In hydrogen atom, an electron jumps from higher orbit to lower orbit so that radius of the lower orbit is \(\left(\frac{1}{4}\right)^{\text{th}}\) of radius of higher orbit. If speed of electron in higher orbit was V, then speed in lower orbit is

• A. \(\frac{V}{4}\)
• B. \(\frac{V}{2}\)
• C. \(V\)
• D. \(2V\)

Hint:

In Bohr's model r is proportional to n squared and v is proportional to 1/n, so v is proportional to 1 over the square root of r.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
In Bohr's model, \(r \propto n^2\) and \(v \propto \dfrac{1}{n}\).
So \(v \propto \dfrac{1}{\sqrt{r}}\).

Step 2: Radius Ratio
The lower orbit has radius \(r_2 = \dfrac{r_1}{4}\), so \(\dfrac{r_1}{r_2} = 4\).

Step 3: Speed Ratio
\(\dfrac{v_2}{v_1} = \sqrt{\dfrac{r_1}{r_2}} = \sqrt{4} = 2\)

Step 4: Result
\(v_2 = 2v_1 = 2V\)

Final Answer:
The speed of the electron in the lower orbit is 2V. \[ \boxed{\text{(D) }2V} \]