Does a photon of visible light \((\lambda \approx 400\) to \(700 \mathrm{~nm})\) have sufficient energy to excite an electron in a hydrogen atom from the \(n=1\) to the \(n=5\) energy state? from the \(n=2\) to the \(n=6\) energy state?

Using the formula for the energy of an electron in a hydrogen atom, we can find the energy difference between the initial and final states: Energy of an electron in a hydrogen atom, \(E_n\), can be found using the formula: \[E_n = -\frac{13.6 \mathrm{~eV}}{n^2}\] For the first case, \(n_i = 1\) and \(n_f = 5\): \[E_1 = -\frac{13.6}{1^2} = -13.6 \mathrm{~eV}\] \[E_5 = -\frac{13.6}{5^2} = -0.544 \mathrm{~eV}\] \[\Delta E_{1 \rightarrow 5} = E_5 - E_1 = 13.056 \mathrm{~eV}\] For the second case, \(n_i = 2\) and \(n_f = 6\): \[E_2 = -\frac{13.6}{2^2} = -3.4 \mathrm{~eV}\] \[E_6 = -\frac{13.6}{6^2} = -0.378 \mathrm{~eV}\] \[\Delta E_{2 \rightarrow 6} = E_6 - E_2 = 3.022 \mathrm{~eV}\]

A photon's energy can be found using the formula: \[E_{photon} = \frac{hc}{\lambda}\] Where \(h\) is Planck's constant, \(6.626 \times 10^{-34} \mathrm{J\cdot s}\), \(c\) is the speed of light, \(3 \times 10^8 \mathrm{m/s}\), and \(\lambda\) is the wavelength of the photon. Since we're dealing with visible light and we are given limits on its wavelength, we can determine the energy range: \[E_{photon, min} = \frac{hc}{700 \times 10^{-9} \mathrm{m}} = 2.849 \times 10^{-19} \mathrm{J}\] \[E_{photon, max} = \frac{hc}{400 \times 10^{-9} \mathrm{m}} = 4.977 \times 10^{-19} \mathrm{J}\] We can convert these energies to electron volts (eV): \[E_{photon, min} = 1.77 \mathrm{~eV}\] \[E_{photon, max} = 3.10 \mathrm{~eV}\]

Now we simply compare the energy differences in each case with the energy range of visible light photons: For the first case, from \(n=1\) to \(n=5\), \(\Delta E_{1 \rightarrow 5} = 13.056 \mathrm{~eV}\). Since this value is larger than both \(E_{photon, min}\) and \(E_{photon, max}\), a photon of visible light does not have sufficient energy to excite an electron from the \(n=1\) to the \(n=5\) energy state. For the second case, from \(n=2\) to \(n=6\), \(\Delta E_{2 \rightarrow 6} = 3.022 \mathrm{~eV}\). This value is extremely close to the maximum energy of visible light photons, \(E_{photon, max} = 3.10 \mathrm{~eV}\). While it may not be possible for the majority of visible light photons to excite an electron from the \(n=2\) to \(n=6\) energy state, there could be a small fraction of high-energy visible light photons that might have sufficient energy to cause the excitation.