The minimum energy needed to eject an electron from a potassium atom is 6.96 × 10−19...
The minimum energy needed to eject an electron from a potassium atom is 6.96 × 10−19 J. What is the maximum wavelength of light, in nanometers, that will show a photoelectric effect with potassium?
Homework Answers
Given:
E = 6.96*10^-19 J
use:
E = h*c/lambda
6.96*10^-19 J =(6.626*10^-34 J.s)*(3.0*10^8 m/s)/lambda
lambda = 2.856*10^-7 m
lambda = 286 nm
Answer: 286 nm
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