What is the maximum wavelength of electromagnetic radiation that can eject photoelectrons from a metal with work function of 3.9 eV?
The maximum wavelength of electromagnetic radiation that can eject photoelectrons from a metal is
Ephot=hcλmax=AoutE_{phot}=\displaystyle \frac{hc} {\lambda_{max}}= A_{out}Ephot=λmaxhc=Aout
λmax=hcAout\displaystyle \lambda_{max} = \frac{hc}{A_{out}}λmax=Aouthc
λmax=6.626⋅10−34⋅3⋅1083.9⋅1.6⋅10−19=3.185⋅10−7=318.5⋅10−9 m=318.5 nm.\displaystyle \lambda_{max} = \frac{6.626 \cdot 10^{-34}\cdot 3 \cdot 10^8}{3.9 \cdot 1.6 \cdot 10^{-19}} = 3.185 \cdot 10^{-7} = 318.5 \cdot 10^{-9} \;m = 318.5 \; nm.λmax=3.9⋅1.6⋅10−196.626⋅10−34⋅3⋅108=3.185⋅10−7=318.5⋅10−9m=318.5nm.
Answer: 318.5 nm.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment