A metal has a work function of 4.50 eV. Find the maximum kinetic energy of the photoelectrons if light of wavelength 250 nm shines on the metal. (1 eV = 1.60 × 10−19 J, c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s) A) 0.00 eV.
B) 0.37 eV.
C) 0.47 eV.
D) 0.53 eV.

0.46 eV

Explanation:

The maximum kinetic energy of the photoelectrons can be found using the formula

K.E = h(c)/λ - h(c)/λ(T), where

K.E = maximum kinetic energy needed

h(c) = Planck's constant, 6.626*10^-34

λ = Wavelength of the light, 250 nm

To start with, we convert our Planck's constant to nm and we have

h(c) = 1240 eV nm

Again, from the question, we already have h(c)/λ(T) given as 4.5 eV so all we need to do is substitute for it.

Plugging in all our values, we have

K.E = 1240/250 - 4.5

K.E = 4.96 - 4.5

K.E = 0.46 eV

answer: good job it's all correct !

explanation:

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.