Question

You are performing a photoelectric effect experiment. Using a photocathode made of cesium, you first illuminate it with a green laser beam \((\lambda=514.5 \mathrm{nm})\) of power \(100 \mathrm{~m} \mathrm{~W}\). Next, you double the power of the laser beam, to \(200 \mathrm{~mW}\). How will the energies per electron of the electrons emitted by the photo cathode compare for the two cases?

Short answer

Question: Compare the energies per electron of the electrons emitted by the photocathode in the photoelectric effect experiment with two different powers of a green laser beam (100 mW and 200 mW) using a cesium photocathode. The wavelength of the laser light is 514.5 nm. Answer: The energies per electron of the electrons emitted by the photocathode for both laser powers (100 mW and 200 mW) are the same. This is because the energy per electron is determined by the energy per photon, which remains constant as the wavelength does not change. However, the number of electrons emitted per second will be different, with twice the number of electrons emitted per second for the 200 mW laser power compared to the 100 mW laser power.

Step-by-step solution

Determine energy per photon

Using the Planck's equation, we can find the energy per photon in the incident light: \(E_{photon} = \frac{hc}{\lambda}\), where \(h\) is the Planck's constant \((6.626 \times 10^{-34} \ \text{Js})\), \(c\) is the speed of light \((3 \times 10^8 \ \text{m/s})\), and \(\lambda\) is the wavelength of the incident light (514.5 nm).

Convert wavelength to meters

Since we have the wavelength in nanometers, we need to convert it to meters: \(\lambda = 514.5 \ \text{nm} \times \frac{1 \ \text{m}}{10^9 \ \text{nm}} = 5.145 \times 10^{-7} \ \text{m}\)

Calculate the energy per photon

Now, we can find the energy per photon using the values of \(h\), \(c\), and \(\lambda\): \(E_{photon} = \frac{6.626 \times 10^{-34} \ \text{Js} \times 3 \times 10^8 \ \text{m/s}}{5.145 \times 10^{-7} \ \text{m}} = 3.868 \times 10^{-19} \ \text{J}\)

Calculate the number of photons per second

Next, we need to find the number of photons per second for both powers: \(n_1 = \frac{100 \ \text{mW}}{E_{photon}}\) for the first case (100 mW laser power) \(n_2 = \frac{200 \ \text{mW}}{E_{photon}}\) for the second case (200 mW laser power)

Convert power to Joules per second

In order to use Joules consistently, we need to convert the given power in milliwatts to Joules per second: \(100 \ \text{mW} = 100 \times 10^{-3} \ \text{J/s}\) \(200 \ \text{mW} = 200 \times 10^{-3} \ \text{J/s}\)

Find the number of photons per second for each case

Now, we can calculate the number of photons per second for both powers: \(n_1 = \frac{100 \times 10^{-3} \ \text{J/s}}{3.868 \times 10^{-19} \ \text{J}} = 2.584 \times 10^{17} \ \text{photons/s}\) \(n_2 = \frac{200 \times 10^{-3} \ \text{J/s}}{3.868 \times 10^{-19} \ \text{J}} = 5.168 \times 10^{17} \ \text{photons/s}\)

Compare the energies per electron

The energy per electron in both cases is determined by the energy per photon, which remains constant as the wavelength does not change. Thus, the energies per electron are the same for both cases. However, the number of electrons emitted per second will be different, with twice the number of electrons emitted per second for the 200 mW laser power compared to the 100 mW laser power.