Q: Which equation correctly expresses the maximum kinetic energy of photoelectrons in the photoelectric effect?

$$KE_{\text{max}} = hf - \phi$$

Q: A metal has a work function $$\phi = 4.31\ \text{eV}. $$What is the cutoff wavelength $$\lambda_c$$ for this metal?

$$\lambda_c = \frac{hc}{\phi} = \frac{(6.63 \times 10$$^{-34}$$\ \text{J} \cdot \text{s})(3.00 \times 10^8$\ \text{m/s})}{4.31\ \text{eV} \times 1.60 \times 10$$^{-19}$$\ \text{J/eV}} = 2.88 \times 10$$^{-7}$$\ \text{m} = 288\ \text{nm}$$

Q: If the energy of incident photons is $$5.50\ \text{eV}$$ and the work function of the surface is $$4.31\ \text{eV}$$, what is the maximum kinetic energy of the ejected electrons?

$$KE_{\text{max}} = 5.50\ \text{eV} - 4.31\ \text{eV} = 1.19\ \text{eV}$$

Q: A surface has a work function $$\phi = 6.35\ \text{eV}. $$What is the cutoff frequency $$f_c$$ for this surface?

$$f_c = \frac{\phi}{h} = \frac{6.35\ \text{eV} \times 1.60 \times 10$$^{-19}$$\ \text{J/eV}}{6.63 \times 10$$^{-34}$$\ \text{J} \cdot \text{s}} = 1.53 \times 10$$^{15}$$\ \text{Hz}$$

Q: What is the stopping potential $$V_s if $$the maximum kinetic energy of photoelectrons is $$2.15\ \text{eV}$$?

$$V_s = \frac{KE_{\text{max}}}{e} = \frac{2.15\ \text{eV}}{1\ \text{eV}} = 2.15\ \text{V}$$

Q: If the maximum kinetic energy of photoelectrons is $$2.50\ \text{eV}$$, what is their maximum speed?

$$v_{\text{max}} = \sqrt{\frac{2 \times 2.50\ \text{eV} \times 1.60 \times 10$$^{-19}$$\ \text{J/eV}}{9.11 \times 10$$^{-31}$$\ \text{kg}}} = 9.37 \times 10^5$\ \text{m/s}$$

Q: A photon with wavelength $$400\ \text{nm}$$ strikes a surface. What is the energy of the photon?

$$E_{\text{photon}} = \frac{hc}{\lambda} = \frac{(6.63 \times 10$$^{-34}$$\ \text{J} \cdot \text{s})(3.00 \times 10^8$\ \text{m/s})}{400 \times 10$$^{-9}$$\ \text{m}} \times \frac{1\ \text{eV}}{1.60 \times 10$$^{-19}$$\ \text{J}} = 3.11\ \text{eV}$$

Q: If the work function of silver is $$4.73\ \text{eV}$$ and the maximum kinetic energy of photoelectrons is $$2.50\ \text{eV}$$, what is the wavelength of the incident light?

$$\lambda = \frac{hc}{KE_{\text{max}} + \phi_{\text{silver}}} = \frac{(6.63 \times 10$$^{-34}$$\ \text{J} \cdot \text{s})(3.00 \times 10^8$\ \text{m/s})}{(2.50 + 4.73)\ \text{eV} \times 1.60 \times 10$$^{-19}$$\ \text{J/eV}} = 1.72 \times 10$$^{-7}$$\ \text{m} = 172\ \text{nm}$$

Q: Given a stopping potential $$\Delta V_s = 0.216\ \text{V}$$, what is the maximum kinetic energy of the photoelectrons?

$$KE_{\text{max}} = e \Delta V_s = 0.216\ \text{eV}$$

Q: If the cutoff frequency for a surface is $$1.53 \times 10$$^{15}$$\ \text{Hz}$$, what is the cutoff wavelength?

$$\lambda_c = \frac{c}{f_c} = \frac{3.00 \times 10^8$\ \text{m/s}}{1.53 \times 10$$^{15}$$\ \text{Hz}} = 1.96 \times 10$$^{-7}$$\ \text{m} = 196\ \text{nm}$$

Question 1

Which equation correctly expresses the maximum kinetic energy of photoelectrons in the photoelectric effect?

Accepted Answer

$$KE_{	ext{max}} = hf - \phi$$

Question 2

A metal has a work function $$\phi = 4.31\ 	ext{eV}. $$What is the cutoff wavelength $$\lambda_c$$ for this metal?

Accepted Answer

$$\lambda_c = \frac{hc}{\phi} = \frac{(6.63 	imes 10$$^{-34}$$\ 	ext{J} \cdot 	ext{s})(3.00 	imes 10^8$\ 	ext{m/s})}{4.31\ 	ext{eV} 	imes 1.60 	imes 10$$^{-19}$$\ 	ext{J/eV}} = 2.88 	imes 10$$^{-7}$$\ 	ext{m} = 288\ 	ext{nm}$$

Question 3

If the energy of incident photons is $$5.50\ 	ext{eV}$$ and the work function of the surface is $$4.31\ 	ext{eV}$$, what is the maximum kinetic energy of the ejected electrons?

Accepted Answer

$$KE_{	ext{max}} = 5.50\ 	ext{eV} - 4.31\ 	ext{eV} = 1.19\ 	ext{eV}$$

Question 4

A surface has a work function $$\phi = 6.35\ 	ext{eV}. $$What is the cutoff frequency $$f_c$$ for this surface?

Accepted Answer

$$f_c = \frac{\phi}{h} = \frac{6.35\ 	ext{eV} 	imes 1.60 	imes 10$$^{-19}$$\ 	ext{J/eV}}{6.63 	imes 10$$^{-34}$$\ 	ext{J} \cdot 	ext{s}} = 1.53 	imes 10$$^{15}$$\ 	ext{Hz}$$

Question 5

What is the stopping potential $$V_s if $$the maximum kinetic energy of photoelectrons is $$2.15\ 	ext{eV}$$?

Accepted Answer

$$V_s = \frac{KE_{	ext{max}}}{e} = \frac{2.15\ 	ext{eV}}{1\ 	ext{eV}} = 2.15\ 	ext{V}$$

Question 6

If the maximum kinetic energy of photoelectrons is $$2.50\ 	ext{eV}$$, what is their maximum speed?

Accepted Answer

$$v_{	ext{max}} = \sqrt{\frac{2 	imes 2.50\ 	ext{eV} 	imes 1.60 	imes 10$$^{-19}$$\ 	ext{J/eV}}{9.11 	imes 10$$^{-31}$$\ 	ext{kg}}} = 9.37 	imes 10^5$\ 	ext{m/s}$$

Question 7

A photon with wavelength $$400\ 	ext{nm}$$ strikes a surface. What is the energy of the photon?

Accepted Answer

$$E_{	ext{photon}} = \frac{hc}{\lambda} = \frac{(6.63 	imes 10$$^{-34}$$\ 	ext{J} \cdot 	ext{s})(3.00 	imes 10^8$\ 	ext{m/s})}{400 	imes 10$$^{-9}$$\ 	ext{m}} 	imes \frac{1\ 	ext{eV}}{1.60 	imes 10$$^{-19}$$\ 	ext{J}} = 3.11\ 	ext{eV}$$

Question 8

If the work function of silver is $$4.73\ 	ext{eV}$$ and the maximum kinetic energy of photoelectrons is $$2.50\ 	ext{eV}$$, what is the wavelength of the incident light?

Accepted Answer

$$\lambda = \frac{hc}{KE_{	ext{max}} + \phi_{	ext{silver}}} = \frac{(6.63 	imes 10$$^{-34}$$\ 	ext{J} \cdot 	ext{s})(3.00 	imes 10^8$\ 	ext{m/s})}{(2.50 + 4.73)\ 	ext{eV} 	imes 1.60 	imes 10$$^{-19}$$\ 	ext{J/eV}} = 1.72 	imes 10$$^{-7}$$\ 	ext{m} = 172\ 	ext{nm}$$

Question 9

Given a stopping potential $$\Delta V_s = 0.216\ 	ext{V}$$, what is the maximum kinetic energy of the photoelectrons?

Accepted Answer

$$KE_{	ext{max}} = e \Delta V_s = 0.216\ 	ext{eV}$$

Question 10

If the cutoff frequency for a surface is $$1.53 	imes 10$$^{15}$$\ 	ext{Hz}$$, what is the cutoff wavelength?

Accepted Answer

$$\lambda_c = \frac{c}{f_c} = \frac{3.00 	imes 10^8$\ 	ext{m/s}}{1.53 	imes 10$$^{15}$$\ 	ext{Hz}} = 1.96 	imes 10$$^{-7}$$\ 	ext{m} = 196\ 	ext{nm}$$

Question 11

A graph of $$KE_{	ext{max}} vs. $$frequency $$f$$ for the photoelectric effect shows a straight line above a certain threshold frequency. What does the slope of this line represent?

Accepted Answer

Planck's constant $$h$$

Question 12

If the wavelength of incident light is $$546.1\ 	ext{nm}$$ and the maximum kinetic energy of photoelectrons is $$0.376\ 	ext{eV}$$, what is the work function $$\phi$$?

Accepted Answer

$$\phi = \frac{hc}{\lambda} - KE_{	ext{max}} = 1.90\ 	ext{eV}$$

Question 13

If the stopping potential is $$0.216\ 	ext{V}$$ and the work function is $$1.90\ 	ext{eV}$$, what is the wavelength of the incident light?

Accepted Answer

$$\lambda = \frac{hc}{KE_{	ext{max}} + \phi} = \frac{(6.63 	imes 10$$^{-34}$$\ 	ext{J} \cdot 	ext{s})(3.00 	imes 10^8$\ 	ext{m/s})}{(0.216 + 1.90)\ 	ext{eV} 	imes 1.60 	imes 10$$^{-19}$$\ 	ext{J/eV}} = 587.5 	imes 10$$^{-9}$$\ 	ext{m}$$

Question 14

Which of the following best describes the physical meaning of the work function $$\phi$$?

Accepted Answer

The minimum energy required to remove an electron from the surface of a material

Question 15

If the cutoff wavelength for a material is $$555\ 	ext{nm}$$, what is the work function $$\phi$$?

Accepted Answer

$$\phi = \frac{hc}{\lambda_c} = \frac{(6.63 	imes 10$$^{-34}$$\ 	ext{J} \cdot 	ext{s})(3.00 	imes 10^8$\ 	ext{m/s})}{555 	imes 10$$^{-9}$$\ 	ext{m}} 	imes \frac{1\ 	ext{eV}}{1.60 	imes 10$$^{-19}$$\ 	ext{J}} = 2.24\ 	ext{eV}$$

Question 16

A photon with energy $$E_{	ext{photon}} = 3.11\ 	ext{eV}$$ strikes a surface with work function $$\phi = 2.24\ 	ext{eV}. $$What is the maximum kinetic energy of the ejected electron?

Accepted Answer

$$KE_{	ext{max}} = 3.11\ 	ext{eV} - 2.24\ 	ext{eV} = 0.87\ 	ext{eV}$$

Question 17

If the maximum kinetic energy of photoelectrons is $$4.00 	imes 10$$^{-19}$$\ 	ext{J}$$, what is the corresponding speed of the electrons?

Accepted Answer

$$v_{	ext{max}} = \sqrt{\frac{2 	imes 4.00 	imes 10$$^{-19}$$\ 	ext{J}}{9.11 	imes 10$$^{-31}$$\ 	ext{kg}}} = 9.37 	imes 10^5$\ 	ext{m/s}$$

Photoelectric Effect: Cutoff Wavelength, Frequency, and Kinetic Energy

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