Photoelectric Effect Examples in Physics

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Photoelectric Effect.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Physics.

Concept Recap

The photoelectric effect is the emission of electrons from a material when light of high enough frequency shines on it.

Light can hit a surface like tiny packets of energy and knock electrons out.

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How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Photoelectric Effect asks whether the system is nuclear, quantum, or relativistic before using an everyday model.

Common stuck point: Students often know a formula related to photoelectric effect but skip the recognition step: Does the situation involve particles, nuclei, photons, or relativistic speeds where everyday mechanics is not enough? That leads to a correct-looking substitution attached to the wrong physical model.

Sense of Study hint: Ask: Does the situation involve particles, nuclei, photons, or relativistic speeds where everyday mechanics is not enough?

Worked Examples

Example 1

medium
A photon of frequency 1.0×1015 Hz1.0\times10^{15}\text{ Hz} strikes sodium (ϕ=3.65×1019 J\phi = 3.65\times10^{-19}\text{ J}). Find KEmaxKE_{\max}. (h=6.6×1034h = 6.6\times10^{-34})

Answer

KEmax2.95×1019 JKE_{\max} \approx 2.95\times10^{-19}\text{ J}

First step

1
Ephoton=hf=(6.6×1034)(1.0×1015)=6.6×1019 JE_\text{photon} = hf = (6.6\times10^{-34})(1.0\times10^{15}) = 6.6\times10^{-19}\text{ J}.

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Example 2

medium
A photon of wavelength 300 nm300\text{ nm} strikes a metal with ϕ=3.0×1019 J\phi = 3.0\times10^{-19}\text{ J}. Find KEmaxKE_{\max}. (h=6.6×1034h = 6.6\times10^{-34}, c=3×108c = 3\times10^8)

Example 3

medium
A photon of E=5.0 eVE = 5.0\text{ eV} hits a metal with ϕ=2.0 eV\phi = 2.0\text{ eV}. Find KEmaxKE_{\max} in joules (1 eV=1.6×1019 J1\text{ eV} = 1.6\times10^{-19}\text{ J}).

Example 4

hard
A photon of frequency 1.5×1015 Hz1.5\times10^{15}\text{ Hz} ejects an electron from a metal of ϕ=4.0×1019 J\phi = 4.0\times10^{-19}\text{ J}. Find the electron's maximum speed (h=6.6×1034h = 6.6\times10^{-34}, me=9.1×1031 kgm_e = 9.1\times10^{-31}\text{ kg}).

Example 5

hard
Light of frequency 2f02 f_0 hits a metal with threshold f0f_0. Express KEmaxKE_{\max} in terms of ϕ\phi.

Example 6

challenge
In a Millikan-style experiment, stopping voltages are V1=0.5 VV_1 = 0.5\text{ V} at f1=5×1014 Hzf_1 = 5\times10^{14}\text{ Hz} and V2=1.5 VV_2 = 1.5\text{ V} at f2=7.5×1014 Hzf_2 = 7.5\times10^{14}\text{ Hz}. Find Planck's constant. (e=1.6×1019 Ce = 1.6\times10^{-19}\text{ C})

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Find the energy of a photon of frequency f=5×1014 Hzf = 5\times10^{14} \text{ Hz}. (h=6.6×1034h = 6.6\times10^{-34})

Example 2

easy
In the photoelectric effect, electrons are emitted only when light has a high enough what?

Example 3

easy
Write the photoelectric equation relating photon energy, work function, and maximum kinetic energy.

Example 4

easy
If a metal's work function is ϕ=3×1019 J\phi = 3\times10^{-19} \text{ J} and a photon delivers 5×1019 J5\times10^{-19} \text{ J}, find the maximum kinetic energy of the electron.

Example 5

easy
Does increasing the brightness (intensity) of light below the threshold frequency cause electron emission?

Example 6

easy
Above the threshold, does increasing intensity increase the number or the energy of emitted electrons?

Example 7

easy
Find the threshold frequency for a metal with work function ϕ=3.3×1019 J\phi = 3.3\times10^{-19} \text{ J}. (h=6.6×1034h = 6.6\times10^{-34})

Example 8

easy
In the photoelectric effect, light behaves more like a wave or like particles (photons)?

Example 9

medium
A photon of frequency 8×1014 Hz8\times10^{14} \text{ Hz} hits a metal with work function 3.3×1019 J3.3\times10^{-19} \text{ J}. Find KEmaxKE_{\max}. (h=6.6×1034h = 6.6\times10^{-34})

Example 10

medium
A metal has threshold frequency 4×1014 Hz4\times10^{14} \text{ Hz}. Find its work function. (h=6.6×1034h = 6.6\times10^{-34})

Example 11

medium
The stopping potential for ejected electrons is 1.5 V1.5 \text{ V}. Find their maximum kinetic energy (e=1.6×1019 Ce = 1.6\times10^{-19} \text{ C}).

Example 12

medium
A photon has wavelength λ=4×107 m\lambda = 4\times10^{-7} \text{ m}. Find its energy. (h=6.6×1034h = 6.6\times10^{-34}, c=3×108c = 3\times10^8)

Example 13

medium
Light of energy 6×1019 J6\times10^{-19} \text{ J} ejects electrons with KEmax=2×1019 JKE_{\max} = 2\times10^{-19} \text{ J}. Find the work function.

Example 14

medium
Two metals have work functions ϕ1=2×1019\phi_1 = 2\times10^{-19} and ϕ2=5×1019 J\phi_2 = 5\times10^{-19} \text{ J}. For the same photon energy, which emits faster electrons?

Example 15

medium
A photon of frequency 6imes1014extHz6 imes10^{14} ext{ Hz} hits a metal with work function 2.6imes1019extJ2.6 imes10^{-19} ext{ J}. Find KEmaxKE_{\max}. (h=6.6imes1034h = 6.6 imes10^{-34})

Example 16

medium
A metal emits electrons with KEmax=1.4imes1019extJKE_{\max} = 1.4 imes10^{-19} ext{ J} under a photon of 4imes1019extJ4 imes10^{-19} ext{ J}. Find the work function.

Example 17

medium
The stopping potential is 2extV2 ext{ V}. Find the maximum kinetic energy of the electrons (e=1.6imes1019extCe = 1.6 imes10^{-19} ext{ C}).

Example 18

challenge
A photon of 7×1019 J7\times10^{-19} \text{ J} hits a metal with ϕ=4×1019 J\phi = 4\times10^{-19} \text{ J}. Find the electron's maximum speed (me=9.1×1031 kgm_e = 9.1\times10^{-31} \text{ kg}).

Example 19

challenge
Light of wavelength 3×107 m3\times10^{-7} \text{ m} strikes a metal with work function 4×1019 J4\times10^{-19} \text{ J}. Find KEmaxKE_{\max}. (h=6.6×1034h = 6.6\times10^{-34}, c=3×108c = 3\times10^8)

Example 20

challenge
Photon energy EE just exceeds work function ϕ\phi so KEmax=0.1ϕKE_{\max} = 0.1\phi. Express EE in terms of ϕ\phi.

Example 21

easy
Find the energy of a photon with frequency f=6×1014 Hzf = 6\times10^{14}\text{ Hz}. (h=6.6×1034h = 6.6\times10^{-34})

Example 22

easy
The work function of a metal is 4.5×1019 J4.5\times10^{-19}\text{ J}. Find its threshold frequency. (h=6.6×1034h = 6.6\times10^{-34})

Example 23

easy
A metal has ϕ=3.0×1019 J\phi = 3.0\times10^{-19}\text{ J}. A photon arrives with E=5.0×1019 JE = 5.0\times10^{-19}\text{ J}. Find KEmaxKE_{\max}.

Example 24

easy
Express the work function ϕ=4.8×1019 J\phi = 4.8\times10^{-19}\text{ J} in electronvolts. (1 eV=1.6×1019 J1\text{ eV} = 1.6\times10^{-19}\text{ J})

Example 25

easy
A photon has wavelength λ=500 nm\lambda = 500\text{ nm}. Find its energy. (h=6.6×1034h = 6.6\times10^{-34}, c=3×108c = 3\times10^8)

Example 26

medium
The stopping voltage for ejected electrons is Vs=0.8 VV_s = 0.8\text{ V}. Find their maximum kinetic energy (e=1.6×1019 Ce = 1.6\times10^{-19}\text{ C}).

Example 27

medium
A metal emits electrons with KEmax=1.5×1019 JKE_{\max} = 1.5\times10^{-19}\text{ J} under light of f=7×1014 Hzf = 7\times10^{14}\text{ Hz}. Find ϕ\phi. (h=6.6×1034h = 6.6\times10^{-34})

Example 28

medium
A metal has threshold wavelength λ0=500 nm\lambda_0 = 500\text{ nm}. Find its work function. (h=6.6×1034h = 6.6\times10^{-34}, c=3×108c = 3\times10^8)

Example 29

medium
Find the threshold wavelength for a metal with ϕ=6.0×1019 J\phi = 6.0\times10^{-19}\text{ J}. (h=6.6×1034h = 6.6\times10^{-34}, c=3×108c = 3\times10^8)

Example 30

medium
A photon ejects an electron with KEmax=4.0×1019 JKE_{\max} = 4.0\times10^{-19}\text{ J}. Find the stopping voltage (e=1.6×1019 Ce = 1.6\times10^{-19}\text{ C}).

Example 31

medium
A graph of KEmaxKE_{\max} versus ff has slope hh and yy-intercept ϕ-\phi. If the slope is 6.6×1034 J s6.6\times10^{-34}\text{ J s} and intercept is 3.3×1019 J-3.3\times10^{-19}\text{ J}, find the threshold frequency.

Example 32

medium
Two metals have ϕA=2.0 eV\phi_A = 2.0\text{ eV} and ϕB=4.0 eV\phi_B = 4.0\text{ eV}. Identical photons of E=5.0 eVE = 5.0\text{ eV} hit each. Which emits faster electrons?

Example 33

medium
A laser delivers 1×10181\times10^{18} photons per second of energy 4×1019 J4\times10^{-19}\text{ J} each. Find the laser's power.

Example 34

hard
A photon of λ=250 nm\lambda = 250\text{ nm} hits a metal with threshold wavelength λ0=350 nm\lambda_0 = 350\text{ nm}. Find KEmaxKE_{\max}. (h=6.6×1034h = 6.6\times10^{-34}, c=3×108c = 3\times10^8)

Example 35

hard
A photon ejects an electron whose stopping voltage is 1.2 V1.2\text{ V}. The metal's work function is 3.0×1019 J3.0\times10^{-19}\text{ J}. Find the photon frequency (h=6.6×1034h = 6.6\times10^{-34}, e=1.6×1019 Ce = 1.6\times10^{-19}\text{ C}).

Example 36

hard
A metal of work function ϕ=2.8×1019 J\phi = 2.8\times10^{-19}\text{ J} is illuminated with light of λ=400 nm\lambda = 400\text{ nm}. Will electrons be emitted? (h=6.6×1034h = 6.6\times10^{-34}, c=3×108c = 3\times10^8)

Example 37

hard
A photoelectric cell is illuminated by red light (f=4×1014 Hzf = 4\times10^{14}\text{ Hz}) and ejected electrons have KEmax=1.0×1019 JKE_{\max} = 1.0\times10^{-19}\text{ J}. The light is replaced with blue light of f=7×1014 Hzf = 7\times10^{14}\text{ Hz}. Find the new KEmaxKE_{\max}. (h=6.6×1034h = 6.6\times10^{-34})

Example 38

challenge
Light of λ=200 nm\lambda = 200\text{ nm} shines on a metal with ϕ=3.0 eV\phi = 3.0\text{ eV}. Find KEmaxKE_{\max} in eV. (h=6.6×1034h = 6.6\times10^{-34}, c=3×108c = 3\times10^8, 1 eV=1.6×1019 J1\text{ eV} = 1.6\times10^{-19}\text{ J})

Background Knowledge

These ideas may be useful before you work through the harder examples.

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