Question

The de Broglie wavelength of the electron in the ground state is \(λ_1\) and that in the \(n = 3\) level is \(λ_3\) then \(λ_3\) is given by

• A. \(\frac{λ_1}{3}\)
• B. \(\frac{λ_1}{2}\)
• C. \(2λ_1\)
• D. \(3λ_1\)

Hint:

Bohr's quantisation says 2 pi r = n lambda, and r is proportional to n^2, so lambda is proportional to n.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In Bohr's model, the electron's orbit holds a whole number of de Broglie wavelengths, so \(2\pi r_n = n\lambda_n\).

Step 2: Key Formula or Approach:
\[ \lambda_n = \frac{2\pi r_n}{n},\qquad r_n\propto n^2 \]

Step 3: Detailed Explanation:
\[ \lambda_n\propto\frac{n^2}{n} = n \]
So the wavelength is proportional to the orbit number:
\[ \frac{\lambda_3}{\lambda_1} = \frac31 \Rightarrow \lambda_3 = 3\lambda_1 \]
Alternative check: the speed in orbit \(n\) is \(v_n\propto\dfrac1n\), so the momentum \(mv\) falls as \(1/n\), and \(\lambda = \dfrac{h}{mv}\) rises as \(n\). The result is the same.

Final Answer:
\(\lambda_3 = 3\lambda_1\), option (D). \[ \boxed{3\lambda_1 \text{ (D)}} \]