The photoelectric work function for a metal surface is \(2.3 eV\). If the light of wavelength 6800Å is incident on the surface of metal, find threshold frequency and incident frequency. Will there be an emission of photoelectrons or not? [Velocity of light \(c=3 \times 10^8 m / s\), Planck's constant, \(h=6.63 \times 10^{-34} Js\) ]
(Feb. 2016)
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Given:
\(\phi_0=2.3 eV =2.3 \cdot 1.6 \cdot 10^{-19} J =3.68 \cdot 10^{-19} J\)
$\lambda =6800 Å=6800*10^{-10}m,$
$c=3*10^{8 m/s},$
$h=6.63*10^{-34} Js$
We know that the incident frequency is given as
\(v=\frac{c}{\lambda}\)
\(\therefore v=\frac{3 \cdot 10^8}{6800 \cdot 10^{-10}}=4.41 \cdot 10^{14} Hz\)
Now, if the incident frequency is greater than the threshold frequency, then photoelectrons will be emitted from the metal surface. The threshold frequency is given from work function as
\(v_0=\frac{\phi_0}{h}\)
\(v_0=\frac{3.68 \cdot 10^{-19}}{6.63 \cdot 10^{-34}}=5.55 \cdot 10^{14} Hz\)
Since, photoelectrons $v<v_0$ will not be emitted.
\(\phi_0=2.3 eV =2.3 \cdot 1.6 \cdot 10^{-19} J =3.68 \cdot 10^{-19} J\)
$\lambda =6800 Å=6800*10^{-10}m,$
$c=3*10^{8 m/s},$
$h=6.63*10^{-34} Js$
We know that the incident frequency is given as
\(v=\frac{c}{\lambda}\)
\(\therefore v=\frac{3 \cdot 10^8}{6800 \cdot 10^{-10}}=4.41 \cdot 10^{14} Hz\)
Now, if the incident frequency is greater than the threshold frequency, then photoelectrons will be emitted from the metal surface. The threshold frequency is given from work function as
\(v_0=\frac{\phi_0}{h}\)
\(v_0=\frac{3.68 \cdot 10^{-19}}{6.63 \cdot 10^{-34}}=5.55 \cdot 10^{14} Hz\)
Since, photoelectrons $v<v_0$ will not be emitted.
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