Use the Balmer equation to determine n for the spectral line of wavelength 656 nm.The constant for Balmer's equation is 3.2881 × 10¹⁵ s^-1. A)3 B)4 C)5 D)6 E)7

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The Balmer equation is used to find the wavelength of light emitted when an electron moves between energy levels in the hydrogen atom. It is given by $1/\lambda = R_H (1/2^2 - 1/n^2)$, where $\lambda$ is the wavelength of the emitted light, $R_H$ is the Rydberg constant for hydrogen (3.2881 × 10^15 s^-1 in this case), and $n$ is the principal quantum number of the higher energy level. Rearranging the equation to solve for $n$, and substituting $\lambda = 656 \, nm = 656 \times 10^{-9} \, m$ and $R_H$, we find that $n$ corresponds to the principal quantum number 3 for the transition that results in the emission of light with a wavelength of 656 nm. This is characteristic of the H-alpha line, one of the Balmer series of lines in the hydrogen spectrum, which specifically involves a transition from the n=3 energy level to n=2.

Use the Balmer Equation to Determine N for the Spectral