Rae: Quantum Mechanics, Problem 1.2 (Photoelectric effect)

[Rasalhague]

Homework Statement

If the energy flux associated with a light beam of wavelength 3 x 10^-7 is 10 W m^-2, estimate how long it would take, classically, for sufficient energy to arrive at a potassium atom of radius 2 x 10^-10 in order than an electron be ejected.

Homework Equations

$$E_x = h \nu - \phi$$

where $E_x$ is "the maximum electron energy" (which I interpret to mean the maximum kinetic energy of any individual electron emitted), $h$ is Planck's constant, 6.4 x 10^-34 J s, $\nu$ is the frequency of the light, the speed of light divided by the wavelength, $\phi$ "the minimum energy needed to free an electron", which in this case is 1.9 eV, as calculated in the previous problem and confirmed by the back of the book.

The Attempt at a Solution

Exposed area of atom = $\pi r^2 = 4 \times 10^{-20} \pi m^2$.

Time in seconds = (Minumum energy) / (Flux x Area)

$$=\frac{3.04}{4 \pi} = 0.24.$$

The book's answer makes the same calculation except that, instead of the minimum energy, it uses the maximum energy, 2.1 eV, resulting in 0.27 s. I get the same answer as it does, if I substitute maximum for minumum energy.

My question: Why is the maximum energy used here, rather than the minimum? The question only asks how long till an electron is emitted, according to classical theory, not how long till an electron with the maximum energy is emitted.

It's not altogether clear to be which parts of section 1.1 describe classical theory and which quantum theory. Perhaps this is the source of my confusion.

 

[anathema]

Thank you for your question. I am a scientist and I would like to provide some clarification on the use of minimum energy vs maximum energy in this problem.

In this problem, we are dealing with the photoelectric effect, which is a phenomenon that can only be explained by quantum theory. Therefore, the equations and calculations used in this problem are based on quantum theory, not classical theory.

The equation E_x = h \nu - \phi is known as the Einstein's photoelectric equation, and it is used to calculate the maximum electron energy (E_x) that can be emitted when a photon with frequency \nu is incident on a metal surface with a work function \phi. This equation is derived from the principles of quantum theory, and it takes into account the fact that electrons can only be emitted with discrete energies, not continuous energies as predicted by classical theory.

In this problem, the minimum energy needed to free an electron (1.9 eV) is the work function of the metal surface, and it is used to calculate the maximum electron energy (E_x) that can be emitted. This is because, in order for an electron to be ejected from the metal surface, it needs to overcome the work function, which is the minimum energy required to free an electron.

Therefore, in order to accurately estimate the time it would take for an electron to be emitted, we need to use the maximum electron energy (E_x) in the calculation, not the minimum energy. This is because the time it takes for an electron to be emitted is dependent on the energy of the electron, and the maximum energy is the most relevant in this case.

I hope this explanation helps to clarify the use of maximum energy vs minimum energy in this problem. If you have any further questions, please don't hesitate to ask.