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: Use the speed of the electron:
In orbit $n$, the speed is $v_n = \dfrac{v_1}{n}$ (since $v\propto Z/n$).

Step 2: Use $\lambda = h/mv$:
$\lambda_n = \dfrac{h}{mv_n} = \dfrac{nh}{mv_1} = n\lambda_1$.

Step 3: Evaluate:
For $n = 3$, $\lambda_3 = 3\lambda_1$.

Step 4: Link to standing waves:
Bohr's rule says the orbit holds exactly $n$ de Broglie waves around its circumference. For $n = 3$ the orbit is nine times larger than for $n = 1$, but it holds three waves, so each wave is $9/3 = 3$ times longer.

Final Answer:
Option (D). \[ \boxed{3\lambda_1 \text{ (D)}} \]