An electron in a box is in the ground state with energy 2.0 eV. (a) Find the width of the box. (b) How much energy is needed to excite the electron to its first excited state? (c) If the electron makes a transition from an excited state to the ground state with the simultaneous emission of \(30.0-\mathrm{eV}\) photon, find the quantum number of the excited state?

Given the ground state energy, \(E_1 = 2.0 eV\), we have to convert this into Joules: \(E_1 = 2.0 eV \times 1.602 \times 10^{-19} J/eV = 3.204 \times 10^{-19} J\) Now, we plug in the values into the energy level formula for n = 1: \[3.204 \times 10^{-19} J = \frac{1^2 \cdot (6.626 \times 10^{-34} Js)^2}{8 \cdot (9.109 \times 10^{-31} kg) \cdot L^2}\] Solving for width L, we get: \[L = \sqrt{\frac{1^2 \cdot (6.626 \times 10^{-34} Js)^2}{8 \cdot (9.109 \times 10^{-31} kg) \cdot 3.204 \times 10^{-19} J}} = 2.61 \times 10^{-9} m\]

To find the energy required to excite the electron to its first excited state (n = 2), we first calculate the energy level for n = 2, using the width of the box (L) from part (a): \[E_2 = \frac{2^2 \cdot (6.626 \times 10^{-34} Js)^2}{8 \cdot (9.109 \times 10^{-31} kg) \cdot (2.61 \times 10^{-9} m)^2} = 8.0 \times 10^{-19} J\] Now, we find the energy difference for the transition: \(\Delta E = E_2 - E_1 = 8.0 \times 10^{-19} J - 3.204 \times 10^{-19} J = 4.796 \times 10^{-19} J\) Converting this energy difference to eV, we have: \(\Delta E = 4.796 \times 10^{-19} J \times \frac{1eV}{1.602 \times 10^{-19} J} = 3.0 eV\)

We are given the energy difference for the transition as a 30 eV photon. First, convert this into Joules: \(\Delta E = 30.0 eV \times 1.602 \times 10^{-19} J/eV = 4.806 \times 10^{-18} J\) Since the energy difference corresponds to a transition from an excited state to the ground state, we can write: \(\Delta E = E_n - E_1\), where n is the quantum number of the excited state. Thus, rearrange this formula to find the energy of the excited state: \(E_n = \Delta E + E_1 = 4.806 \times 10^{-18} J + 3.204 \times 10^{-19} J = 5.126 \times 10^{-18} J\) Now, use the energy level formula for n (unknown), using the width of the box (L): \[5.126 \times 10^{-18} J = \frac{n^2 \cdot (6.626 \times 10^{-34} Js)^2}{8 \cdot (9.109 \times 10^{-31} kg) \cdot (2.61 \times 10^{-9} m)^2}\] Solving for n, we get approximately: \[n = \sqrt{\frac{8 \cdot (9.109 \times 10^{-31} kg) \cdot (2.61 \times 10^{-9} m)^2 \cdot 5.126 \times 10^{-18} J}{(6.626 \times 10^{-34} Js)^2}} \approx 4\] So, the quantum number of the excited state is approximately 4.