Molybdenum metal must absorb radiation with an energy higher than $7.22 \times 10^{-19} \mathrm{~J}$ ( "energy threshold") before it can eject an electron from its surface via the photoelectric effect. (a) What is the frequency threshold for emission of electrons? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(240 \mathrm{nm}\), what is the maximum possible velocity of the emitted electrons?

Using the energy of a photon equation, we can solve for the frequency threshold: \(E_{threshold} = h\nu_{threshold}\) Solving for \(\nu_{threshold}\): \(\nu_{threshold} = \frac{E_{threshold}}{h}\) Given \(E_{threshold} = 7.22 \times 10^{-19} J\) and \(h = 6.63 \times 10^{-34} Js\), we can calculate the frequency threshold: \(\nu_{threshold} = \frac{7.22 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.09 \times 10^{15} \mathrm{Hz}\) The frequency threshold for emission of electrons is \(1.09 \times 10^{15} \mathrm{Hz}\).

Using the wavelength and frequency relationship, we can solve for the wavelength of radiation that provides the given photon energy: \(c = \lambda\nu\) Solving for \(\lambda\): \(\lambda = \frac{c}{\nu}\) Given \(c = 3 \times 10^8 \mathrm{m/s}\) and \(\nu_{threshold} = 1.09 \times 10^{15} \mathrm{Hz}\), we can calculate the wavelength of radiation: \(\lambda = \frac{3 \times 10^8}{1.09 \times 10^{15}} = 2.75 \times 10^{-7} \mathrm{m} = 275 \mathrm{nm}\) The wavelength of radiation that provides a photon of this energy is \(275 \mathrm{nm}\).

Given the wavelength of irradiated light, we can first calculate the energy of a photon at \(240 \mathrm{nm} \): \(E_{photon} = h\nu = \frac{hc}{\lambda}\) Using \(h = 6.63 \times 10^{-34} Js\), \(c = 3 \times 10^8 \mathrm{m/s}\) and \(\lambda = 240 \times 10^{-9} \mathrm{m}\), we can determine the energy of a photon: \(E_{photon} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{240 \times 10^{-9}} = 8.29 \times 10^{-19} J\) Now we can calculate the kinetic energy of emitted electrons: \(KE_{max} = E_{photon} - E_{threshold}\) Using \(E_{photon} = 8.29 \times 10^{-19} J\) and \(E_{threshold} = 7.22 \times 10^{-19} J\): \(KE_{max} = 8.29 \times 10^{-19} - 7.22 \times 10^{-19} = 1.07 \times 10^{-19} J\) Finally, we can find the maximum possible velocity of emitted electrons using the kinetic energy equation: \(KE_{max} = \frac{1}{2}mv^2\) Solving for \(v_{max}\): \(v_{max} = \sqrt{\frac{2KE_{max}}{m}}\) Assuming the mass of an electron is \(m = 9.11 \times 10^{-31} \mathrm{kg}\): \(v_{max} = \sqrt{\frac{2(1.07 \times 10^{-19})}{9.11 \times 10^{-31}}} = 4.38 \times 10^5 \mathrm{m/s}\) The maximum possible velocity of the emitted electrons is \(4.38 \times 10^5 \mathrm{m/s}\).