Question

Molybdenum metal must absorb radiation with a minimum frequency of \(1.09 \times 10^{15} \mathrm{~s}^{-1}\) before it can eject an electron from its surface via the photoelectric effect. (a) What is the minimum energy needed to eject an electron? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(120 \mathrm{~nm}\), what is the maximum possible kinetic energy of the emitted electrons?

Short answer

(a) The minimum energy needed to eject an electron from the molybdenum surface is \(7.23 \times 10^{-19} J\). (b) The wavelength of radiation providing a photon of this energy is \(275 nm\). (c) The maximum possible kinetic energy of the emitted electrons when molybdenum is irradiated with light of wavelength 120 nm is \(9.38 \times 10^{-19} J\).

Step-by-step solution

a) Finding the minimum energy needed to eject an electron

To find the minimum energy needed to eject an electron, we will use Planck's equation \(E = h \nu\). We are given the frequency as \(1.09 \times 10^{15} s^{-1}\) and Planck's constant \(h = 6.63 \times 10^{-34} J s\). So, we can find the energy: \(E = h \nu = (6.63 \times 10^{-34} J s)(1.09 \times 10^{15} s^{-1})\) \(E = 7.23 \times 10^{-19} J\) The minimum energy needed to eject an electron is \(7.23 \times 10^{-19} J\).

b) Finding the wavelength of radiation providing a photon of this energy

Now we can use the equation that relates the wavelength, frequency and speed of light: \(c = \lambda \nu\). We need to find the wavelength \(\lambda\) of the radiation that provides the minimum energy needed to eject an electron. Speed of light \(c = 3.00 \times 10^8 m/s\), and the frequency is given as \(1.09 \times 10^{15} s^{-1}\). So we can find the wavelength by isolating \(\lambda\): \(\lambda = \frac{c}{\nu} = \frac{3.00 \times 10^8 m/s}{1.09 \times 10^{15} s^{-1}}\) \(\lambda = 2.75 \times 10^{-7} m\) or \(275 nm\) The wavelength of radiation providing a photon of this energy is \(275 nm\).

c) Finding the maximum possible kinetic energy of the emitted electrons

To find the maximum possible kinetic energy for the emitted electrons, we need to use the Planck's equation and the photoelectric effect equation. First, we need to find the energy of the photon when molybdenum is irradiated with light of wavelength 120 nm. Given wavelength \( \lambda = 120 * 10^{-9} m\), we can find the frequency using \(c = \lambda \nu\): \(\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 m/s}{120 \times 10^{-9} m}\) \(\nu = 2.50 \times 10^{15} s^{-1}\) Now, we can find the energy of the photon using Planck's equation: \(E = h \nu = (6.63 \times 10^{-34} J s)(2.50 \times 10^{15} s^{-1})\) \(E = 1.66 \times 10^{-18} J\) Now we can find the maximum possible kinetic energy of the emitted electrons using the photoelectric effect equation: \(K_\text{max} = E - W\), where \(E = 1.66 \times 10^{-18} J\) and the work function \(W = 7.23 \times 10^{-19} J\). \(K_\text{max} = 1.66 \times 10^{-18} J - 7.23 \times 10^{-19} J = 9.38 \times 10^{-19} J\) The maximum possible kinetic energy of the emitted electrons is \(9.38 \times 10^{-19} J\).