When it is struck by 240-nm photons,a material having a work function of 2.60 eV emits electrons.What is the maximum kinetic energy of the emitted electrons? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J) A)2.58 eV B)5.18 eV C)2.00 eV D)4.21 eV

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Quizplus · Accepted Answer

The maximum kinetic energy of the emitted electrons can be calculated using the photoelectric effect equation: $KE_{max} = \frac{hc}{\lambda} - \phi$, where $KE_{max}$ is the maximum kinetic energy of the emitted electrons, $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength of the incident photons, and $\phi$ is the work function of the material. Given $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$, $c = 3.00 \times 10^8 \, \text{m/s}$, $\lambda = 240 \, \text{nm} = 240 \times 10^{-9} \, \text{m}$, and $\phi = 2.60 \, \text{eV} = 2.60 \times 1.60 \times 10^{-19} \, \text{J}$, we first convert the energy of the photons from joules to electron volts and then subtract the work function to find the maximum kinetic energy. The calculation yields approximately 2.58 eV.

When It Is Struck by 240-Nm Photons,a Material Having a Work