The work function for a photosensitive metal is 1.5 ev. the metal is illuminated with a light of wavelength 500 nm. find: (a) maximum kinetic energy of the emitted electrons (b) maximum speed of the emitted electron, and (c) stopping potential

A) We want to find the maximum kinetic energy of the photoelectrons.

E = 1243/λ - Φ

E = kinetic energy of photoelectrons, λ = light wavelength, Φ = work function

Given values:

λ = 500nm, Φ = 1.5eV

Plug in and solve for E:

E = 1243/500 - 1.5

E = 0.986eV

B) We want to find the maximum speed of the photoelectrons.

KE = 0.5mv²

KE = max kinetic energy, m = electron mass, v = max speed

Given values:

KE = 0.986eV = 1.58×10⁻¹⁹J, m = 9.11×10⁻³¹kg

Plug in and solve for v:

1.58×10⁻¹⁹ = 0.5(9.11×10⁻³¹)v²

v = 5.89×10⁵m/s

C) We want to find the stopping potential for the photoelectrons:

Vq = E

V = stopping potential, q = electron charge, E = max kinetic energy of photoelectrons

Given values:

q = e (elementary charge), E = 0.986eV

There isn't much work to do to find V. The energy E is already given in electron-volts which is defined as the amount of energy an electron gains when passed through a 1V potential difference. All you have to do is remove the "e" from "eV" and you'll have your answer.

V = 0.986V

ok

explanation:

connect the batteries in series and hope