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Sagot :
Radiation with a wavelength of 263.9 nm can eject a 2.00 -eV electron from a calcium electrode. Calcium has a piece function of 2.seventy one eV.
The expelled photoelectrons are E = 2.0 eV, so
⇒ [tex]2.0 eV = E_{in} -2.71 eV[/tex]
⇒ [tex]E_{in}=2.0 eV +2.71eV[/tex]
⇒ [tex]E_{in} = 4.71 eV[/tex]
in Joules, it is
⇒ [tex]E_{in} =4.71*1.6*10^{-19} J[/tex]
⇒ [tex]E_{in} =7.536*10^{-19} J[/tex]
Since [tex]E_{in}=h\nu=\frac{hc}{\lambda}[/tex], the wavelength must be
[tex]\Rightarrow \lambda = \frac{hc}{E_{in} }[/tex]
[tex]\Rightarrow \lambda=\frac{(6.64*10^{-34}Js)(3*10^{8}m/s) }{7.536*10^{-19} J}[/tex]
[tex]\Rightarrow \lambda=263.9\times10^{-9} \; m[/tex]
[tex]\Rightarrow \lambda=263.9 \;n m[/tex]
inside the photoelectric impact, the electricity of the photoelectrons chosen from metallic with work function ∅ thanks to misguided radiation of the entirety
If radiation moves a calcium electrode with the work feature ∅ = 2.seventy one eV and the power of radiation equals
Know more about the wavelength of radiation:
https://brainly.com/question/9108146
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