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?

To calculate the wavelength of the photon torpedo, we will use the work function given and the Planck's equation. The Planck's equation states that the energy of a photon is related to its wavelength by: \(E = \dfrac{hc}{\lambda}\), where \(E\) is the energy of the photon, \(h\) is the Planck's constant (\(6.63 \times 10^{-34} \; \mathrm{Js}\)), \(c\) is the speed of light (\(3.00 \times 10^{8} \; \mathrm{m/s}\)), and \(\lambda\) is the wavelength. We are given the work function of the enemy vessel's hull, which is the energy needed to eject an electron: \(W = 7.52 \times 10^{-19} \mathrm{~J}\). We can set the work function equal to the energy of the photon and then solve for the wavelength: \(W = \dfrac{hc}{\lambda}\). Rearranging the equation to find \(\lambda\), we get: \(\lambda = \dfrac{hc}{W}\). Now, we can plug in the values: \[ \lambda = \dfrac{(6.63 \times 10^{-34} \; \mathrm{Js})(3.00 \times 10^{8} \; \mathrm{m/s})}{7.52 \times 10^{-19} \mathrm{~J}} = \dfrac{1.989 \times 10^{-25} \; \mathrm{Jm}}{7.52 \times 10^{-19} \mathrm{~J}} \] Finally, simplify the expression to obtain the wavelength: \[ \lambda \approx 2.65\times 10^{-7} \; \mathrm{m} = 265\; \mathrm{nm} \]

We now have a photon torpedo with a wavelength of 259 nm. To determine if it can eject an electron, we need to calculate its energy and compare it to the work function. Using Planck's equation, we can calculate the energy of the photon torpedo: \[ E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34} \; \mathrm{Js})(3.00 \times 10^{8} \; \mathrm{m/s})}{259 \times 10^{-9} \; \mathrm{m}} \] Then, simplify the expression to obtain the energy: \[ E \approx 7.67 \times 10^{-19} \; \mathrm{J} \] Comparing this energy to the work function (7.52 x 10^{-19} J), we observe that the energy of the photon torpedo (7.67 x 10^{-19} J) is greater than the work function. Therefore, this photon torpedo can eject an electron and cause damage to the enemy vessel.

We are given the electron configuration of the element: [Ar] 4s^1 3d^10. To determine the element, we need to find the atomic number based on the electron configuration. The last electron that was added to reach this configuration is in the 4s orbital. In the periodic table, the noble gas [Ar] represents Argon with an atomic number of 18. Now, count the remaining electrons: 1 electron in the 4s orbital and 10 electrons in the 3d orbital. Total electrons for this element are 18 (core electrons) + 1 (4s) + 10 (3d) = 29. The element with an atomic number of 29 is Copper (Cu). Therefore, the hull of the enemy vessel is made of Copper.