Consider a transition in which the electron of a hydrogen atom is excited from \(n=1\) to \(n=\infty\). (a) What is the end result of this transition? (b) What is the wavelength of light that must be absorbed to accomplish this process? (c) What will occur if light with a shorter wavelength than that in part (b) is used to excite the hydrogen atom? (d) How are the results of parts \((\mathrm{b})\) and \((\mathrm{c})\) related to the plot shown in Exercise \(6.88 ?\)

When the electron of a hydrogen atom is excited from its ground state (n=1) to n=∞, it means that the electron has gained enough energy to completely ionize and leave the hydrogen atom. The end result of this transition is a free electron and the hydrogen nucleus, or a hydrogen ion.

To calculate the wavelength of the absorbed light, we'll use the Rydberg formula for hydrogen: \[\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\] Here, \(R_H\) is the Rydberg constant for hydrogen, approximately \(1.097\times10^7 \, m^{-1}\), \(n_1\) and \(n_2\) are the initial and final energy levels respectively, and \(\lambda\) is the wavelength of the absorbed light. For this transition, \(n_1 = 1\) and \(n_2 = \infty\). Substitute these values into the formula: \[\frac{1}{\lambda} = R_H \left(\frac{1}{1^2} - \frac{1}{\infty^2}\right)\] Simplify the equation: \[\frac{1}{\lambda} = R_H\] Now, solve for the wavelength (\(\lambda\)): \[\lambda = \frac{1}{R_H}\] Plug in the value of \(R_H\): \[\lambda = \frac{1}{1.097\times10^7 \, m^{-1}}\] \[\lambda \approx 9.11\times10^{-8} \, m = 91.1 \, nm\] The wavelength of light needed to excite the electron from n=1 to n=∞ is approximately 91.1 nm.

If light with a shorter wavelength than 91.1 nm is used to excite the hydrogen atom, it would provide more energy than needed to ionize the hydrogen atom. As a result, this excess energy would be imparted to the ejected electron as kinetic energy.

The plot in Exercise 6.88 shows the energy levels of a hydrogen atom as a function of the principal quantum number n. The energy levels converge as n approaches infinity, with the energy difference between consecutive levels decreasing. The wavelength of light needed to ionize an electron in the hydrogen atom corresponds to the energy difference between the ground state (n=1) and n=∞. The result found in part (b), 91.1 nm, is the minimum wavelength needed to ionize the hydrogen atom. Any shorter wavelength light would provide more energy, as discussed in part (c), causing the ejected electron to have kinetic energy.