Question

(a) Light of wavelength 300 nm is incident on a metal that has a work function of 1.4 eV. What is the maximum speed of the emitted electrons? (b) Repeat part (a) for light of wavelength 800 nm incident on a metal that has a work function of \(1.6 \mathrm{eV} .\) (c) How would your answers to parts (a) and (b) vary if the light intensity were doubled?

Short answer

Answer: The maximum speeds of the emitted electrons are 2.94 x 10^5 m/s for part (a) and 0 m/s for part (b). Doubling the light intensity does not change the maximum speeds of the electrons because it only increases the number of photons, not the energy of individual photons, which determines the speeds of the emitted electrons.

Step-by-step solution

Part (a)

1. Calculate the energy of the photons: \(E_{photon} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34} \ \text{Js})(3 \times 10^{8} \ \text{m/s})}{300 \times 10^{-9} \ \text{m}} = 6.626 \times 10^{-19} \ \text{J}\) 2. Compute the kinetic energy of the emitted electrons: \(K.E_{electron} = E_{photon} - W = 6.626 \times 10^{-19} \ \text{J} - 1.4 \ \text{eV} \times 1.6 \times 10^{-19} \ \text{J/eV} = 3.994 \times 10^{-19} \ \text{J}\) 3. Find the maximum speed of the emitted electrons: \(v = \sqrt{\dfrac{2K.E_{electron}}{m_{electron}}} = \sqrt{\dfrac{2(3.994 \times 10^{-19} \ \text{J})}{(9.109 \times 10^{-31} \ \text{kg})}} = 2.94 \times 10^5 \ \text{m/s}\) The maximum speed of the emitted electrons in part (a) is \(2.94 \times 10^5 \ \text{m/s}\).

Part (b)

1. Calculate the energy of the photons: \(E_{photon} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34} \ \text{Js})(3 \times 10^{8} \ \text{m/s})}{800 \times 10^{-9} \ \text{m}} = 2.481 \times 10^{-19} \ \text{J}\) 2. Compute the kinetic energy of the emitted electrons: \(K.E_{electron} = E_{photon} - W = 2.481 \times 10^{-19} \ \text{J} - 1.6 \ \text{eV} \times 1.6 \times 10^{-19} \ \text{J/eV} = 2.481 \times 10^{-19} \ \text{J} - 2.56 \times 10^{-19} \ \text{J}\) Since the photon energy is less than the work function of the metal, there will be no electrons emitted, and their speed will be zero. The maximum speed of the emitted electrons in part (b) is 0 m/s.

Part (c)

Doubling the light intensity means doubling the number of photons, but it does not change the energy of individual photons. The photoelectric effect depends on the energy of individual photons. Since the photons' energy remains the same in both cases, the electrons' maximum speed will also remain the same for parts (a) and (b). In conclusion, doubling the light intensity will not change the answers to parts (a) and (b).