Question

Suppose the angular momentum is quantized as even integral multiples of \(h/2\pi\) in an imaginary world. According to Bohr's model, what is the longest possible wavelength that hydrogen atoms emit in the visible range in such a world?

• A. 387 nm
• B. 487 nm
• C. 510 nm
• D. 760 nm

Hint:

Allowed levels are only even n (2, 4, 6, ...). Longest visible wavelength = smallest visible energy gap, which is the 4 to 2 transition.

Answer 1

The Correct Option is B

Step 1: In this modified world the quantization rule is \(mvr = (2k)\dfrac{h}{2\pi}\), with \(k = 1, 2, 3, \ldots\) So the only allowed principal quantum numbers are the even integers \(n = 2, 4, 6, 8, \ldots\)

Step 2: The energy expression of the Bohr model is unchanged in form:
\[ E_n = -\frac{13.6}{n^2}\ \text{eV}. \]
So the accessible levels have energies \(E_2 = -3.40\) eV, \(E_4 = -0.85\) eV, \(E_6 = -0.378\) eV, \(E_8 = -0.2125\) eV.

Step 3: Visible light spans roughly \(400\ \text{nm}\) to \(700\ \text{nm}\). The longest visible wavelength corresponds to the smallest photon energy that still lies in this band. Transitions ending on \(n = 4\) (like \(6 \to 4\)) give \(\Delta E \approx 0.47\) eV, i.e. infrared, so they are excluded. The smallest visible transition is \(4 \to 2\):
\[ \Delta E = E_2 - E_4 = -3.40 - (-0.85) = 2.55\ \text{eV}. \]

Step 4: Convert to wavelength using \(\lambda = \dfrac{1240\ \text{eV nm}}{\Delta E}\):
\[ \lambda = \frac{1240}{2.55} \approx 486\ \text{nm}. \]
This lies in the visible range, and every other allowed transition to \(n=2\) gives a shorter wavelength, so this is the longest visible emission.
\[\boxed{\lambda \approx 487\ \text{nm}}\]