Answer:
[tex]\nu=5.31*10^{14} Hz[/tex]
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Explanation:
According to the photoelectric effect, the maximum kinetic energy of an photoelectron is given by:
[tex]K_{max}=E-W(1)[/tex]
Here, E is the energy of the photon and W is the minimum energy required to remove an electron from the surface of the metal, W is defined as:
[tex]W=h\nu_0(2)[/tex]
The Planck – Einstein relation states that the energy of a photon is equal to its frequency multiplied by the planck constant:
[tex]E=h\nu(3)[/tex]
Recall that [tex]\nu=\frac{c}{\lambda}[/tex]. Replacing (3) and (2) in (1):
[tex]K_{max}=h\nu-h\nu_0\\K_{max}=h\nu-\frac{hc}{\lambda_0}[/tex]
Solving for [tex]\nu[/tex]:
[tex]\nu=\frac{K_{max}+\frac{hc}{\lambda_0}}{h}\\\\2.2eV*\frac{1.60*10^{-19}J}{1ev}=3.52*10^{-19}J\\\\\nu=\frac{3.52*10^{-19}J+\frac{6.63*10^{-34}J(3*10^8\frac{m}{s})}{225*10^{-9}m}}{6.63*10^{-34}J}\\\\\nu=5.31*10^{14} Hz[/tex]
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