Violet light with a wavelength of $430 \mathrm{~nm}$ falls on a sodium plate $(W=2.75 \mathrm{eV})$ and a potassium plate $(W=$ $2.30 \mathrm{eV})$. Which of of the following statements is true?
A) Both plates emit electrons, but those emitted by the sodium plate have the larger kinetic energy.
B) Both plates emit electrons, but those emitted by the potassium plate have the larger kinetic energy.
C) Only the sodium plate emits electrons.
D) Only the potassium plate emits electrons.
E) Neither plate emits electrons.

Question

Quizplus · Accepted Answer

To determine which plates emit electrons and the kinetic energy of the emitted electrons, we use the photoelectric effect equation: $E_{photon} = W + K.E.$, where $E_{photon}$ is the energy of the incoming photon, $W$ is the work function of the material, and $K.E.$ is the kinetic energy of the emitted electron. The energy of the incoming photon can be calculated using $E_{photon} = \frac{hc}{\lambda}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{m}^2 \text{kg} / \text{s}$), $c$ is the speed of light ($3.00 \times 10^8 \, \text{m/s}$), and $\lambda$ is the wavelength of the light ($430 \, \text{nm}$ or $430 \times 10^{-9} \, \text{m}$). First, calculate the energy of the violet light photon:$$E_{photon} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{430 \times 10^{-9}} = \frac{1.9878 \times 10^{-19}}{430} \approx 4.62 \times 10^{-19} \, \text{J}$$Converting this energy to electron volts (eV) (1 eV = $1.602 \times 10^{-19} \, \text{J}$):$$E_{photon} \approx \frac{4.62 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.88 \, \text{eV}$$Since the work function of sodium is $2.75 \, \text{eV}$ and for potassium is $2.30 \, \text{eV}$, the energy of the incoming photon is sufficient to overcome the work function of both metals. Therefore, both plates emit electrons.The kinetic energy of the emitted electrons is the difference between the photon energy and the work function:- For sodium: $K.E._{Na} = 2.88 \, \text{eV} - 2.75 \, \text{eV} = 0.13 \, \text{eV}$- For potassium: $K.E._{K} = 2.88 \, \text{eV} - 2.30 \, \text{eV} = 0.58 \, \text{eV}$Thus, both plates emit electrons, but those emitted by the potassium plate have the larger kinetic energy.

Violet Light with a Wavelength Of 430 nm430 \mathrm{~nm}430 nm Falls on a Sodium Plate

Violet Light with a Wavelength Of $430 \mathrm{~nm}$ Falls on a Sodium Plate