What is the wavelength of the photon emitted by a He⁺ when the electron jumps from the n = 7 to the n = 4 state? A)8.8 $\mu$m B)3.4 $\mu$m C)4.4 $\mu$m D)896 nm E)542 nm

Question

Quizplus · Accepted Answer

The wavelength of the photon emitted can be calculated using the Rydberg formula for hydrogen-like atoms: $ \frac{1}{\lambda} = RZ^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) $, where $ R $ is the Rydberg constant, $ Z $ is the atomic number (2 for He+), and $ n_1 $ and $ n_2 $ are the principal quantum numbers. For He+, the calculation gives a wavelength of approximately 542 nm.

What Is the Wavelength of the Photon Emitted by a He+

What Is the Wavelength of the Photon Emitted by a He⁺