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

Only even angular-momentum states survive, so the ladder of levels is $n = 2, 4, 6, \ldots$ with $E_n = -13.6/n^2$ eV. A visible photon needs energy between $1240/700 \approx 1.77$ eV and $1240/400 \approx 3.10$ eV.

List the candidate downward jumps and their energies. A jump ending on $n=2$ from $n=4$ releases $E_2 - E_4 = -3.40 + 0.85 = 2.55$ eV; from $n=6$ it releases $3.40 - 0.378 = 3.02$ eV; from $n=8$ it releases $3.19$ eV. Jumps ending on $n=4$ (for example $6 \to 4$) release only about $0.47$ eV, which is infrared and therefore not counted.

Among the visible jumps, the least energetic is $4 \to 2$ at $2.55$ eV, and least energy means longest wavelength:
\[ \lambda = \frac{hc}{\Delta E} = \frac{1240\ \text{eV nm}}{2.55\ \text{eV}} \approx 486\ \text{nm}. \]
Rounding gives the listed value of about 487 nm, which is the longest wavelength hydrogen can emit in the visible band in this world.
\[\boxed{\lambda \approx 487\ \text{nm}}\]