Final answer:
(B) Both metals A and B will emit photo-electrons.
The photoelectric effect tells us that photoelectrons are emitted when the energy of incident light is greater than the metal's work function. Since a photon with a wavelength of 350 nm has energy higher than both the work functions of metals A (4.8 eV) and B (2.2 eV), both metals will emit photoelectrons.
Step-by-step explanation:
The question pertains to the photoelectric effect, which involves the emission of photoelectrons from a metal surface when exposed to light of a certain wavelength.
According to the photoelectric effect, electrons will be emitted if the energy of the photons (determined by the incident light's wavelength) is greater than the work function of the metal.
The energy of a photon (E) can be calculated using the formula E = hc/λ, where h is Planck's constant (6.626 x 10-34 J·s), c is the speed of light in vacuum (3.00 x 108 m/s), and λ is the wavelength of the incident light.
For metal A with a work function of 4.8 eV and metal B with a work function of 2.2 eV, and radiation of wavelength 350 nm (3.50 x 10-7 m), we first convert the wavelength to energy using the formula E = hc/λ and then compare it to each metal's work function.
The energy of the photon at 350 nm is:
E = (6.626 x 10-34 Js) (3.00 x 108 m/s) / (3.50 x 10-7 m)
= 5.66 eV
Since 5.66 eV is greater than both the work functions of metals A and B, we can conclude that:
(B) Both metals A and B will emit photo-electrons.