Final answer:
Photons that do not have a frequency high enough to meet or exceed the threshold frequency specific to the metal's work function will not cause the photoelectric effect. Frequencies less than those corresponding to gamma rays typically do not cause pair production, and energies too low for photoelectric effect similarly imply insufficient energy for ionization.
"the correct option is approximately option A"
Step-by-step explanation:
The work function of a metal is the minimum energy needed to eject an electron from the surface of that metal. The work function of the metal in question is stated to be 3.1×10⁻¹⁹ joules. To understand which frequency of photons will not cause various effects like the photoelectric effect, Compton scattering, pair production, and ionization of atoms, we need to consider the energy of photons and how it relates to these phenomena.
For the photoelectric effect to occur, photons must have energy equal to or greater than the metal's work function. The energy of a photon is given by the equation E = hf, where h is Planck's constant and f is the frequency of the photon. The threshold frequency (fc) for the photon to eject a photoelectron is given by fc = Work function / h. If the photon's frequency is lower than this threshold frequency, it will not have enough energy to eject an electron, hence the photoelectric effect will not occur.
For Compton scattering to occur, photons need to interact with electrons. While this can happen with photons of any energy, the effect is more noticeable with high-energy photons, such as X-rays. Lower energy photons might not provide noticeable scattering.
Pair production can only occur if the photon is within a strong electromagnetic field, like near the nucleus of an atom, and has enough energy to create an electron-positron pair. This requires a photon with at least 1.022 MeV, which is far greater than the energy associated with a work function in the range of a few electronvolts.
Lastly, ionization of atoms also requires that the incoming photon has enough energy to overcome the ionization potential of the atom in question. However, generally speaking, if a photon doesn't have enough energy for the photoelectric effect on a given metal, it is also unlikely to have sufficient energy for ionization, as ionization energies are typically higher than work functions.