Question

Light with a wavelength of 5.0 · 10-7 m strikes a surface that requires 2.0 ev to eject an electron

Answer 1

Answer on Question 60320, Physics, Quantum Mechanics

Question:

Light with a wavelength of $5.0 \cdot 10^{-7} \, \text{m}$ strikes a surface that requires $2.0 \, \text{eV}$ to eject an electron. Calculate the maximum kinetic energy (in electron volts) of the emitted photoelectron.

Solution:

Using the mathematical description of the photoelectric effect, we can write the maximum kinetic energy $E_{Kmax}$ of an emitted photoelectron as follows:

$$E_{Kmax} = h f - \varphi = h \frac{c}{\lambda} - \varphi,$$

here, $h = 4.135 \cdot 10^{-15} \, \text{eV} \cdot \text{s}$ is the Planck constant, $f$ is the frequency of the incident photon, $c$ is the speed of light, $\lambda$ is the wavelength of the light and $\varphi$ is the work function for the metal (energy required to eject an electron from the surface).

Then, we get:

$$\begin{array}{l} E_{Kmax} = h \frac{c}{\lambda} - \varphi = 4.135 \cdot 10^{-15} \, \text{eV} \cdot \text{s} \cdot \frac{3 \cdot 10^{8} \, \frac{\text{m}}{\text{s}}}{5.0 \cdot 10^{-7} \, \text{m}} - 2.0 \, \text{eV} = \\ = 2.5 \, \text{eV} - 2.0 \, \text{eV} = 0.5 \, \text{eV}. \end{array}$$

Answer:

$$E_{Kmax} = 0.5 \, \text{eV}.$$

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