Does a photon of visible light $(\lambda \approx 400 \text { to } 700 \mathrm{nm})$ 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?

The energy of a photon can be calculated using the formula: \[E = \dfrac{hc}{\lambda}\] where \(E\) is the energy of the photon, \(h\) is the Planck's constant \((h \approx 6.63 \times 10^{-34} \mathrm{Js})\), \(c\) is the speed of light \((c \approx 3 \times 10^8 \mathrm{m/s})\), and \(\lambda\) is the wavelength of the photon. We are given that the range of wavelengths for visible light is from \(400 nm\) to \(700 nm\). We must calculate the energies corresponding to these wavelengths. First let's convert the wavelengths to meters by multiplying them by \(1\times 10^{-9} m\): \[\lambda_1 = 400\ nm \times(1\times 10^{-9} m/nm) = 4 \times 10^{-7} m\] \[\lambda_2 = 700\ nm \times (1\times 10^{-9} m/nm) = 7 \times 10^{-7} m\] Now, calculate the energy corresponding to \(\lambda_1 = 4 \times 10^{-7} m\): \[E_1 = \dfrac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{m/s}}{4 \times 10^{-7} m} = 4.974 \times 10^{-19} J\] Calculate the energy corresponding to \(\lambda_2 = 7 \times 10^{-7} m\): \[E_2 = \dfrac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{m/s}}{7 \times 10^{-7} m} = 2.846 \times 10^{-19} J\] The energy of the photon varies between \(2.846\times 10^{-19} J\) and \(4.974\times 10^{-19} J\).

The energy levels of a hydrogen atom can be calculated using the formula: \[E_n = -\dfrac{13.6 \mathrm{eV}}{n^2}\] Thus, the energy required to excite an electron from \(n=1\) to \(n=5\) can be calculated as: \[E_{1 \rightarrow 5} = E_5 - E_1 = \left(\dfrac{-13.6 \mathrm{eV}}{5^2} \right) - \left(\dfrac{-13.6 \mathrm{eV}}{1^2}\right) = 12.09 \text{ eV}\] Now convert this energy to joules (using \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\)): \[E_{1 \rightarrow 5} = 12.09 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}) = 1.934 \times 10^{-18} J\]

Using the same formula as in Step 2, we calculate the energy required to excite an electron from \(n=2\) to \(n=6\): \[E_{2 \rightarrow 6} = E_6 - E_2 = \left(\dfrac{-13.6 \mathrm{eV}}{6^2} \right) - \left(\dfrac{-13.6 \mathrm{eV}}{2^2}\right) = 3.40 \text{ eV}\] Now convert this energy to joules: \[E_{2 \rightarrow 6} = 3.40 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}) = 5.44 \times 10^{-19} J\]

Compare the energy of the photon with the required energy for \(n=1\) to \(n=5\) transition: \[ \begin{cases} 2.846\times 10^{-19} J < 1.934\times 10^{-18} J \\ 4.974\times 10^{-19} J < 1.934\times 10^{-18} J \end{cases} \] Both photon energies are insufficient to excite an electron from the \(n=1\) to the \(n=5\) energy state. Now, compare the energy of the photon with the required energy for \(n=2\) to \(n=6\) transition: \[ \begin{cases} 2.846\times 10^{-19} J < 5.44\times 10^{-19} J \\ 4.974\times 10^{-19} J > 5.44\times 10^{-19} J \end{cases} \] The photon with energy \(4.974\times 10^{-19} J\) has sufficient energy to excite an electron from the \(n=2\) to the \(n=6\) energy state.