A metal surface is illuminated with a wavelength of 550nm, and the suppression voltage is measured to be 0.19V. What is the fugitive work of this metal
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According to the photoelectric effect theory, the energy $E$ of a photon is equal to the frequency of the incident light $f$ multiplied by Planck's constant $h$, which can also be expressed in terms of the wavelength of the light $\lambda$: $E = h f = \frac{hc}{\lambda}$. When the metal surface is illuminated, only the evolved work of the metal surface can emit electrons, so the photon energy $E with a wavelength of 550 nm should be greater than or equal to the evolved work of the metal surface $W$, that is, $E \geq W$. 因此,可以可用出功$W \leq \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} J \cdot s \times 3.00 \times 10^8 m/s}{550 \times 10^{-9} m} = 3.61 eV$. Therefore, the metal has a maximum work of 3.61 eV.
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