An electron confined to a one-dimensional box has a ground-state energy of \(40.0 \mathrm{eV} .\) (a) If the electron makes a transition from its first excited state to the ground state, what is the wavelength of the emitted photon? (b) If the box were somehow made twice as long, how would the photon's energy change for the same transition (first excited state to ground state)?

In a one-dimensional box electron energy is quantized, and each energy level is given by the following formula: \(E_n = \frac{n^2 \cdot h^2}{8 \cdot L^2 \cdot m_e}\), where \(E_n\) is the energy of the nth level, \(n\) is the state number, \(L\) is the length of the box, and \(m_e\) is the mass of the electron.

Given the ground state energy, E1, which is \(40.0\:\text{eV}\). We can find the length of the box by finding the value of \(L\) from ground state energy formula: \(E_1 = \frac{h^2}{8 \cdot L^2 \cdot m_e}\). Now, rearrange the formula to find \(L\): \(L = \sqrt{\frac{h^2}{8 \cdot E_1 \cdot m_e}}\). Convert the energy in electron volts to joules by multiplying by the electron charge, \(e = 1.602 \times 10^{-19} \text{C}\): \(E_1 = 40.0\:\text{eV} \times 1.602 \times 10^{-19} \text{C/eV} = 6.408 \times 10^{-18} \text{J}\). Now, using the values of Planck's constant, \(h = 6.626 \times 10^{-34} \text{Js}\), and electron mass, \(m_e = 9.11 \times 10^{-31} \text{kg}\), we calculate the length of the box: \(L = \sqrt{\frac{(6.626 \times 10^{-34} \text{Js})^2}{8 \cdot (6.408 \times 10^{-18} \text{J}) \cdot (9.11 \times 10^{-31} \text{kg})}} = 3.682\times 10^{-10}\text{m}\).

The first excited state corresponds to \(n = 2\). Using the formula for energy levels: \(E_2 = \frac{2^2 \cdot h^2}{8 \cdot L^2 \cdot m_e} = 4 \cdot E_1 = 4 \times 40.0\:\text{eV} = 160.0\:\text{eV}\).

The energy difference between the first excited state and ground state is: \(\Delta E = E_2 - E_1 = 160.0\:\text{eV} - 40.0\:\text{eV} = 120.0\:\text{eV}\). Convert the energy difference to joules: \(\Delta E = 120.0\:\text{eV} \times 1.602 \times 10^{-19} \text{C/eV} = 1.922 \times 10^{-17} \text{J}\). Now, using the energy-wavelength relation for a photon, \(E_{\text{photon}} = h \cdot c / \lambda\), we can find the wavelength of the emitted photon: \(\lambda = \frac{h \cdot c}{\Delta E}\). Using the speed of light, \(c = 3.00 \times 10^8 \text{m/s}\), we get: \(\lambda = \frac{6.626 \times 10^{-34} \text{Js} \cdot 3.00 \times 10^8 \text{m/s}}{1.922 \times 10^{-17}\text{J}} = 1.034\times 10^{-7}\text{m}\). So the wavelength of the emitted photon is \(1.034\times 10^{-7}\text{m}\).

If the box were made twice as long, the new length of the box would be \(2L\). We can now find the new ground state energy, \(E'_1\), and the new first excited state energy, \(E'_2\), using the energy levels formula: \(E'_1 = \frac{h^2}{8 \cdot (2L)^2 \cdot m_e} = \frac{E_1}{4}\). \(E'_2 = \frac{2^2 \cdot h^2}{8 \cdot (2L)^2 \cdot m_e} = E_2 / 4\). The energy difference, \(\Delta E'\), between the first excited state and the new ground state is: \(\Delta E' = E'_2 - E'_1 = \frac{3}{4} \cdot E_1\). The new energy difference in electron volts would be: \(\Delta E' = \frac{3}{4} \cdot 40.0\:\text{eV} = 30.0\:\text{eV}\). When the box is made twice as long, the photon's energy for the same transition (first excited state to ground state) changes from \(120.0\:\text{eV}\) to \(30.0\: \text{eV}\).