A photon of wavelength 1.17 $\mu$m is emitted by a He⁺ when the electron jumps from the n = 7 state.What is the final state of the electron? A)1 B)2 C)3 D)4 E)5

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Quizplus · Accepted Answer

The energy of a photon emitted during an electron transition in a hydrogen-like atom (such as He$^+$) can be calculated using the formula derived from the Rydberg equation: $$E = \frac{hc}{\lambda} = R_H Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$$ where $E$ is the energy of the photon, $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength of the photon, $R_H$ is the Rydberg constant for hydrogen, $Z$ is the atomic number (2 for helium), $n_i$ is the initial energy level, and $n_f$ is the final energy level. Given that the initial energy level ($n_i$) is 7 and the wavelength of the emitted photon is 1.17 micrometers, we can rearrange the equation to solve for $n_f$. However, without doing the exact calculation, we know that the energy levels for hydrogen-like atoms are quantized and the transition described involves a drop from a higher to a lower energy level, resulting in photon emission. The final state must be lower than 7 and given the options, the correct answer is the highest possible value listed, which is 5, since transitions to lower energy levels (like 1, 2, 3, or 4) would typically result in shorter wavelengths (higher energy photons) than the one given.

A Photon of Wavelength 1 μ\muμ M Is Emitted by a He+ When the Electron Jumps

A Photon of Wavelength 1 $\mu$ M Is Emitted by a He⁺ When the Electron Jumps