🤓 Based on our data, we think this question is relevant for Professor Roychowdhury's class at RUTGERS.

$\overline{)\mathbf{E}\mathbf{}\mathbf{=}\mathbf{}\mathbf{hv}}\phantom{\rule{0ex}{0ex}}\mathbf{E}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{hc}}{\mathbf{\lambda}}\phantom{\rule{0ex}{0ex}}\mathbf{E}\mathbf{}\mathbf{=}\mathbf{}\frac{(6.626\times {10}^{-34}J\xb7\overline{)s})(3.00\times {10}^{8}{\displaystyle \frac{\overline{)m}}{\overline{)s}}})}{\mathbf{94}\mathbf{.}\mathbf{91}\mathbf{}\overline{)\mathbf{nm}}\mathbf{}\mathbf{\times}{\displaystyle \frac{{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\overline{)\mathbf{m}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{nm}}}}}$

**E = 2.0944x10 ^{-18} J**

A ground-state H atom absorbs a photon of wavelength 94.91 nm, and its electron attains a higher energy level. The atom then emits two photons: one of wavelength 1281 nm to reach an intermediate energy level, and a second to return to the ground state. What higher level did the electron reach?