The work function for a metal surface is \(2.2 eV\). If light of wavelength 5000Å is incident on the surface of the metal, find the threshold frequency and incident frequency. Will there be an emission of photoelectrons or not? \(\left(c=3 \times 10^8 m / s , 1 eV =1.6 \times 10^{-19} J , h =6.63 \times 10^{-34} J . s\right.\). \()\)
(March 2018)
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Given work function \(=\Phi_0=2.2 eV\)
\(=2.2 \times 1.6 \times 10^{-19} J\)
\(=3.52 \times 10^{-19}\)
\(\lambda=500Å=5000 \times 10^{-10} m , c =3 \times 10^8 m / s , h =6.63 \times 10^{-34} Js\)
Incident frequency \(v =\frac{c}{\lambda}=\frac{3 \times 10^8}{5000 \times 10^{-10}}\)
\(=\frac{3 \times 10^8}{5 \times 10^{-7}}=\frac{3}{5} \times 10^{-15} Hz\)
\(=\frac{30}{5} \times 10^{14} Hz =6 \times 10^{14} Hz\)
Threshold frequency \(v _0=\frac{\Phi_0}{h}=\frac{3.52 \times 10^{-19} J }{6.63 \times 10^{-34} Js }\)
\(=0.53 \times 10^{15} Hz =5.3 \times 10^{14} Hz\)
Emission of photoelectron takes place only if incident frequency is greater than the threshold frequency. But, in this case $n>n_0$therfore, photoelectric emission takes place.
\(=2.2 \times 1.6 \times 10^{-19} J\)
\(=3.52 \times 10^{-19}\)
\(\lambda=500Å=5000 \times 10^{-10} m , c =3 \times 10^8 m / s , h =6.63 \times 10^{-34} Js\)
Incident frequency \(v =\frac{c}{\lambda}=\frac{3 \times 10^8}{5000 \times 10^{-10}}\)
\(=\frac{3 \times 10^8}{5 \times 10^{-7}}=\frac{3}{5} \times 10^{-15} Hz\)
\(=\frac{30}{5} \times 10^{14} Hz =6 \times 10^{14} Hz\)
Threshold frequency \(v _0=\frac{\Phi_0}{h}=\frac{3.52 \times 10^{-19} J }{6.63 \times 10^{-34} Js }\)
\(=0.53 \times 10^{15} Hz =5.3 \times 10^{14} Hz\)
Emission of photoelectron takes place only if incident frequency is greater than the threshold frequency. But, in this case $n>n_0$therfore, photoelectric emission takes place.
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