Question:

Suppose in a hypothetical world the angular momentum is quantized to be even integral multiples of $\frac{h}{2}$. The largest possible wavelength emitted by hydrogen atoms in the visible range in a world according to Bohr's model will be (consider $h_e = 1224$ MeV-fm)

Updated On: Apr 15, 2025
  • 153 nm
  • 409 nm
  • 121 nm
  • 487 nm
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The Correct Option is D

Solution and Explanation

In this hypothetical world, we assume that the angular momentum of electrons in a hydrogen atom is quantized to even integral multiples of \( \frac{h}{2} \), instead of integral multiples of \( \frac{h}{2\pi} \) as in Bohr's model. This affects the allowed energy levels and, consequently, the wavelengths of light emitted. Using Bohr's model, the energy levels of the hydrogen atom are given by the formula: \[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \] In the hypothetical world, this quantization would result in modified energy levels. However, the largest wavelength emitted corresponds to the transition from the highest possible energy level to the second-highest level. This transition will give the longest wavelength, which lies in the visible range. Given the value \( h_e = 1224 \, \text{MeV-fm} \), and the conditions stated, the largest possible wavelength for hydrogen atoms in the visible range is: \[ \boxed{487 \, \text{nm}} \]

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