Question Detail
Question
The threshold wavelength of photoelectric emission of a metal is 4000 Angstroms. What is the minimum energy required to eject a photoelectron?
Correct Answer
Option B is the correct answer.
Detailed Explanation
E=hc/lambda0=(6.626e-34)(3e8)/(4e-7)=4.97e-19J=3.10eV.
Hint
E=hc/lambda0=(6.626e-34)(3e8)/(4e-7)=4.97e-19J=3.10eV.