As the weapons officer aboard the Starship Chemistry, it is your duty to configure a photon torpedo to remove an electron from the outer hull of an enemy vessel. You know that the work function (the binding energy of the electron) of the hull of the enemy ship is \(7.52 \times 10^{-19} \mathrm{J}\) a. What wavelength does your photon torpedo need to be to eject an electron? b. You find an extra photon torpedo with a wavelength of \(259 \mathrm{nm}\) and fire it at the enemy vessel. Does this photon torpedo do any damage to the ship (does it eject an electron)? c. If the hull of the enemy vessel is made of the element with an electron configuration of \([\mathrm{Ar}] 4 s^{1} 3 d^{10},\) what metal is this?

Given the work function of the enemy's hull, we can use the photoelectric effect equation to find the wavelength needed to eject an electron. The photoelectric effect equation is given by: \[E = hf\] where \(E\) is the energy of the photon, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), and \(f\) is the frequency of the light. Since \(f = \frac{c}{\lambda}\), where \(c\) is the speed of light in a vacuum (\(3 \times 10^8 \mathrm{m/s}\)), we can rewrite the equation as: \[E = h\left(\frac{c}{\lambda}\right)\] To find the wavelength (\(\lambda\)), we have: \[\lambda = \frac{hc}{E}\] Now we can plug in the given work function (\(E = 7.52 \times 10^{-19} \mathrm{J}\)): \[\lambda = \frac{(6.626 \times 10^{-34} \mathrm{Js})(3 \times 10^8 \mathrm{m/s})}{7.52 \times 10^{-19} \mathrm{J}}\]

Now, we calculate the wavelength: \[\lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{7.52 \times 10^{-19}} \approx 2.64 \times 10^{-7} \mathrm{m}\] For simplicity and convenience, we can convert the wavelength into nanometers (nm), by multiplying it by \(10^9\): \[\lambda = 2.64 \times 10^{-7} \times 10^9 = 264 \mathrm{nm}\] The required wavelength of the photon torpedo needed to eject an electron is approximately \(264 \mathrm{nm}\).

The found photon torpedo has a wavelength of \(259 \mathrm{nm}\), which is less than the required wavelength of \(264 \mathrm{nm}\). Since the photon energy is inversely proportional to the wavelength, the energy of the found photon torpedo is actually greater than required. Thus, the found photon torpedo can eject an electron and cause damage to the enemy vessel.

The electron configuration of the metal in the enemy's vessel is given as: \([\mathrm{Ar}] 4s^{1} 3d^{10}\) The electron configuration of Argon is \([Ne] 3s^2 3p^6\). Therefore, the complete electron configuration of the given metal is: \([Ne] 3s^2 3p^6 4s^1 3d^{10}\) After examining the periodic table, we find that the element with this electron configuration is Zinc (Zn), with atomic number 30. Thus, the enemy vessel's outer hull is made of Zinc.