Consider a \(1 \mathrm{~s} \rightarrow 2 \mathrm{~s}\) electron transition. For which atom would this require shorter wavelength light, \(\mathrm{H}\) or He? Justify your answer.

To calculate the energy difference required for the electron transition, we will use the Rydberg formula for hydrogen-like atoms: \[E_n = -Z^2\frac{E_0}{n^2}\] where \(E_n\) is the energy of a particular energy level, \(Z\) is the atomic number, \(E_0\) is the Rydberg constant, and \(n\) is the principal quantum number. The relationship between energy and wavelength is given by: \[E = \frac{hc}{\lambda}\] where \(E\) is the energy, \(h\) is the Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength.

First, find the energy difference for the hydrogen (H) atom. The atomic number of hydrogen is 1, so \(Z = 1\). The initial energy level is 1, and the final energy level is 2. Using the Rydberg formula, find the initial and final energy levels: \[E_1 = -\frac{E_0}{1^2} = -E_0\] \[E_2 = -\frac{E_0}{2^2} = -\frac{E_0}{4}\] The energy difference for hydrogen is: \[\Delta E_H = E_2 - E_1 = -\frac{E_0}{4} - (-E_0) = \frac{3E_0}{4}\] Next, find the energy difference for the helium (He) atom. The atomic number of helium is 2, so \(Z = 2\). The initial and final energy levels remain the same (1 and 2). Using the Rydberg formula, find the initial and final energy levels: \[E_1 = -\left(2^2\right)\frac{E_0}{1^2} = -4E_0\] \[E_2 = -\left(2^2\right)\frac{E_0}{2^2} = -E_0\] The energy difference for helium is: \[\Delta E_{He} = E_2 - E_1 = -E_0 - (-4E_0) = 3E_0\]

Now we can use the energy-wavelength relationship to compare the wavelengths required for the electron transitions in H and He atoms. Using the relationship \(E = \frac{hc}{\lambda}\), find the wavelengths: For hydrogen: \[\lambda_H = \frac{hc}{\Delta E_H} = \frac{hc}{\frac{3E_0}{4}} = \frac{4hc}{3E_0}\] For helium: \[\lambda_{He} = \frac{hc}{\Delta E_{He}} = \frac{hc}{3E_0} = \frac{hc}{3E_0}\] Now compare the wavelengths: \[\frac{4hc}{3E_0} > \frac{hc}{3E_0}\] \[\lambda_H > \lambda_{He}\] Since the wavelength for the transition in the hydrogen atom is greater than the wavelength for the transition in the helium atom, the helium atom requires shorter wavelength light for the electron transition.

The \(1 \mathrm{~s} \rightarrow 2 \mathrm{~s}\) electron transition in a helium (He) atom requires shorter wavelength light compared to the hydrogen (H) atom.