Problem 02. X-rays for human applications can be created via electronic transitions by running current to shoot electrons at a screen with a large $Z$ material coating (See X-ray vacuum tubes, cathode ray tubes for more details). (a) A common material for these coatings is Tungsten, what is the energy of the $n=1$ and $n=2$ orbitals? (b) What is the wavelength of a photon emitted when a free electron $n=\infty, 1 / \infty=0$ falls into the $n=1$ and $n=2$ orbitals? (c) Are these x-rays? (x-rays have wavelength of roughly 0.03 and 3 nanometers)

Step 1 ：Calculate the energy of the n=1 and n=2 orbitals for Tungsten using the formula: \(E_n = -\frac{Z^2 * e^4 * m_e}{2 * h^2 * n^2}\) with Z = 74, e = 1.602 * 10^{-19} C, $m_e$ = 9.109 * 10^{-31} kg, and h = 6.626 * 10^{-34} Js. This gives \(E_1 = -3.74 * 10^{-36} J\) and \(E_2 = -9.35 * 10^{-37} J\).

Step 2 ：Calculate the wavelength of the emitted photons for the n=1 and n=2 orbitals using the formula: \(\lambda = \frac{h * c}{E}\) with h = 6.626 * 10^{-34} Js, c = 3 * 10^8 m/s, and the calculated energies. This gives \(\lambda_1 = 53.13 nm\) and \(\lambda_2 = 212.51 nm\).

Step 3 ：\boxed{\text{Final Answer:}} (a) \boxed{E_1 = -3.74 * 10^{-36} J, E_2 = -9.35 * 10^{-37} J} (b) \boxed{\lambda_1 = 53.13 nm, \lambda_2 = 212.51 nm} (c) \boxed{\text{Not x-rays}}