The photoelectric effect is studied using a tungsten target. The work function of tungsten is 4.5 eV. The incident photons have energy \(4.8 \mathrm{eV} .\) (a) What is the threshold frequency? (b) What is the stopping potential? (c) Explain why, in classical physics, no threshold frequency is expected.

Short Answer

Expert verified
Answer: The threshold frequency is approximately \(1.08 * 10^{15} \text{Hz}\), the stopping potential is \(0.3 \text{V}\), and the discrepancy is due to classical physics assuming a continuous energy transfer dependent on the amplitude of the incident light, rather than the quantized relationship between energy and frequency observed in the photoelectric effect.

Step by step solution

01

Convert the energies to Joules.

First, it is important to convert the given energies from electron volts (eV) to joules (J) since we will be working with Planck's constant, which is in units of joule-seconds. To convert eV to J, we can use the conversion factor of 1 eV = 1.6*10^(-19) J. Work function, \(\phi = 4.5 \text{eV} = 4.5 * 1.6*10^(-19) \text{J}\) Incident photon energy, \(E_{photon} = 4.8 \text{eV} = 4.8 * 1.6*10^(-19) \text{J}\)
02

Calculate the threshold frequency (a).

To find the threshold frequency, \(\nu_t\), we'll use the work function equation: \(\phi = h\nu_t\) We know that Planck's constant, \(h = 6.626*10^{-34} \text{J s}\). Now, isolating \(\nu_t\) to find its value: \(\nu_t = \frac{\phi}{h}\) Substitute the values of \(\phi\) and \(h\): \(\nu_t = \frac{4.5 * 1.6*10^{-19} \text{J}}{6.626*10^{-34} \text{J s}} = 1.08 * 10^{15} \text{Hz}\) The threshold frequency is approximately \(1.08 * 10^{15} \text{Hz}\).
03

Calculate the stopping potential (b).

To find the stopping potential, \(V_0\), we will use the following equation: \(eV_0 = E_{photon} - \phi\) We know the electron charge, \(e = 1.6*10^{-19} \text{C}\). Solve for \(V_0\): \(V_0 = \frac{E_{photon} - \phi}{e}\) Substitute the values of \(E_{photon}\), \(\phi\), and \(e\): \(V_0 = \frac{(4.8 - 4.5) * 1.6*10^{-19} \text{J}}{1.6*10^{-19} \text{C}} = 0.3 \text{V}\) The stopping potential is \(0.3 \text{V}\).
04

Explain why classical physics does not expect a threshold frequency (c).

In classical physics, it was assumed that the energy of an incident electromagnetic wave (light) was proportional to its amplitude, not its frequency. Therefore, a continuous energy transfer was expected between the incident wave and the electrons, independent of the frequency of the light. This means that electrons should be emitted regardless of the frequency, as long as the intensity of light is high enough. However, the photoelectric effect and subsequent experiments demonstrated that there is a threshold frequency below which no photoelectrons are emitted, even with very high-intensity light. This implies that there is a quantized relationship between the energy of incident photons and their frequency, as described by Planck's theory of light quanta and Einstein's interpretation of the photoelectric effect. In conclusion, classical physics does not predict a threshold frequency because it assumes a continuous energy transfer dependent on the amplitude of the incident light, rather than the quantized relationship between energy and frequency observed in the photoelectric effect.

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