Question

Follow the steps outlined in this problem to estimate the time lag (predicted classically but not observed experimentally) in the photoelectric effect. Let the intensity of the incident radiation be $0.01 \mathrm{W} / \mathrm{m}^{2} .\( (a) If the area of the atom is \)(0.1 \mathrm{nm})^{2},$ find the energy per second falling on the atom. (b) If the work function is 2.0 eV, how long would it take (classically) for enough energy to fall on this area to liberate one photoelectron? (c) Explain briefly, using the photon model, why this time lag is not observed.

Short answer

Why is this time lag not observed using the photon model? Answer: Classically, it would take 32 seconds to liberate one photoelectron. However, this time lag is not observed in the photon model because the energy transfer from a photon to an electron is instantaneous and all-or-nothing, unlike the gradual energy transfer predicted by classical physics.

Step-by-step solution

1. Calculate the energy falling on the atom per second.

To find the energy falling on the atom per second, we will first calculate the energy received per square meter and then multiply it with the area of the atom. We have given the intensity of the radiation as \(0.01 \mathrm{W/m^2}\). The area of the atom is \((0.1 \mathrm{nm})^2\). First, we need to convert the area of the atom from nanometers to meters:\((0.1 \times 10^{-9} \mathrm{m})^2 = 10^{-18} \mathrm{m^2}\). Now we can find the energy falling on the atom per second: \(0.01 \mathrm{\frac{W}{m^2}} \times 10^{-18} \mathrm{m^2} = 10^{-20} \mathrm{W}\).

2. Calculate the time to liberate one photoelectron.

We are given the work function as 2.0 eV. First, we need to convert this energy to Joules using the conversion factor \(1 \mathrm{eV} = 1.6 \times 10^{-19} \mathrm{J}\): \(2.0 \mathrm{eV} \times 1.6 \times 10^{-19} \mathrm{\frac{J}{eV}} = 3.2 \times 10^{-19} \mathrm{J}\). Next, we want to find out how long it would take for this amount of energy to fall on the atom. We can do this by dividing the required energy by the energy per second falling on the atom: \(\frac{3.2 \times 10^{-19} \mathrm{J}}{10^{-20} \mathrm{W}} = 3.2 \times 10^1 \mathrm{s}\) So, classically, it would take 32 seconds to liberate one photoelectron.

3. Explain why this time lag is not observed using the photon model.

According to the photon model, a photoelectron is emitted instantaneously when a single photon having energy equal to or greater than the work function of the material interacts with an electron. This is because, in the quantum model, the energy of light is quantized and packaged into photons. When a photon interacts with an electron, it transfers its entire energy to the electron instantaneously. If the energy of the photon is greater than the work function, the electron is immediately ejected from the atom as a photoelectron. Therefore, there is no time lag observed because the energy transfer happens instantaneously in an all-or-nothing manner, unlike the gradual energy transfer predicted by the classical model.