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Chapter 36: Problem 18

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 \(100 \mathrm{~mW}\). Next, you double the power of your laser beam, to \(200 \mathrm{~mW}\). How will the energies per electron of the electrons emitted by the cathode compare for the two cases?

Short Answer

Expert verified
Answer: Doubling the power of the laser beam does not change the energies per electron of the electrons emitted by the cesium photocathode. It only affects the number of emitted electrons.

Step by step solution

01

Find the energy of photons from the green laser beam emitting

To find the energy of the photons from the green laser beam, we can use the equation: \(E = \frac{hc}{\lambda}\) where \(E\) is the energy, \(h \approx 6.63 \times 10^{-34} \;\text{J} \cdot \text{s}\) is Planck's constant, \(c \approx 3.00 \times 10^8 \; \text{m} / \text{s}\) is the speed of light, and \(\lambda = 514.5 \times 10^{-9} \; \text{m}\) is the wavelength of the green laser beam. Plugging the values, we have: \(E = \frac{(6.63 \times 10^{-34} \;\text{J} \cdot \text{s})(3.00 \times 10^8 \; \text{m} / \text{s})}{514.5 \times 10^{-9} \; \text{m}}\) \(E \approx 3.87 \times 10^{-19} \; \text{J}\)
02

Calculate the number of photons emitted per second

We are given the power of the laser beam as 100 mW and 200 mW. We'll first calculate the number of photons emitted when the power is 100 mW. Power (P) = Energy (E) * Number of photons (n) / Time (t) n = P * t / E At 100 mW of power, we have: n = (100 \times 10^{-3} \; \text{J/s}) \times (\text{1 s}) / (3.87 \times 10^{-19} \; \text{J}) n \(\approx 2.58 \times 10^{17}\) photons/s
03

Calculate the number of emitted electrons in each case

As energy is conserved, one photon hits an electron in the photocathode, and thus we can assume the number of emitted electrons is equal to the number of photons for both the cases. Therefore, the number of emitted electrons when the power is 100 mW is: N1 = \(2.58 \times 10^{17}\) electrons/s Now, when the power is doubled to 200 mW: N2 = (200 \times 10^{-3} \; \text{J/s}) \times (\text{1 s}) / (3.87 \times 10^{-19} \; \text{J}) N2 \(\approx 5.16 \times 10^{17}\) electrons/s
04

Compare the energies per electron of the electrons emitted by the cathode

Since the energy of the photons is the same in both cases, the energy per emitted electron doesn't change as the individual energies of electrons remain same in both cases. Therefore, the energies per electron do not change when we double the power of the laser beam. Conclusion: Doubling the power of the laser beam does not change the energies per electron of the electrons emitted by the cathode; it only changes the number of emitted electrons.

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