Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV.What is the maximum kinetic energy of the photoelectrons emitted from the metal? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s = 4.141 × 10^-15 ev ∙ s,1 eV = 1.60 × 10^-19 J) A)4.52 eV B)3.11 eV C)1.41 eV D)2.82 eV E)1.70 eV

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Quizplus · Accepted Answer

The maximum kinetic energy of the photoelectrons can be calculated using the photoelectric equation: $KE_{max} = \frac{hc}{\lambda} - \phi$, where $KE_{max}$ is the maximum kinetic energy, $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength of the incident light, and $\phi$ is the work function of the metal. Given $h = 4.141 \times 10^{-15} \, \text{eV} \cdot \text{s}$, $c = 3.00 \times 10^{8} \, \text{m/s}$, $\lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}$, and $\phi = 1.70 \, \text{eV}$, we find $KE_{max} = \frac{(4.141 \times 10^{-15})(3.00 \times 10^{8})}{400 \times 10^{-9}} - 1.70 \, \text{eV} = 3.11 - 1.70 = 1.41 \, \text{eV}$.

Light of Wavelength 400 Nm Falls on a Metal Surface