What is the wavelength,in nm,of the spectral line resulting from the transition of an electron from n = 3 to n = 2 in a Bohr hydrogen atom? The constant for Balmer's equation is 3.2881 × 10¹⁵ s^-1. A)18.3 nm B)656 nm C)547 nm D)152 nm E)252 nm

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Quizplus · Accepted Answer

The wavelength of the spectral line resulting from the transition of an electron from n = 3 to n = 2 in a Bohr hydrogen atom can be calculated using the Rydberg formula for hydrogen spectral lines: $ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) $, where $R$ is the Rydberg constant for hydrogen. Given the constant for Balmer's equation as $3.2881 \times 10^{15} \, \text{s}^{-1}$, we can calculate the wavelength in meters and then convert to nanometers (nm). The correct calculation leads to a wavelength in the visible spectrum, specifically around 656 nm, which corresponds to the red part of the hydrogen spectrum and is a well-known value for the H-alpha line in the Balmer series.

What Is the Wavelength,in Nm,of the Spectral Line Resulting from the Transition