The second line of the Balmer series occurs at a wavelength of 486.1 nm. What is the energy difference between the initial and final levels of the hydrogen atom in this emission process A) 2.44 * 10¹⁸ J B) 4.09 * 10^-19 J C) 4.09 * 10^-22 J D) 4.09 * 10^-28 J E) 1.07 * 10^-48 J

Question

Quizplus · Accepted Answer

The energy difference between the initial and final levels of the hydrogen atom can be calculated using the formula $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{J s}$), $c$ is the speed of light ($3.00 \times 10^8 \, \text{m/s}$), and $\lambda$ is the wavelength in meters ($486.1 \, \text{nm} = 486.1 \times 10^{-9} \, \text{m}$). Substituting these values gives an energy difference of approximately $4.09 \times 10^{-19} \, \text{J}$.

The Second Line of the Balmer Series Occurs at a Wavelength