Practice Photoelectric Effect in Physics

Use these practice problems to test your method after reviewing the concept explanation and worked examples.

Worksheet PDF Free Answer-key PDF Family

Quick 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.

Showing a random 20 of 50 problems.

Example 1

medium
Doubling the intensity of monochromatic light above threshold doubles the ___ but does not change KEmax⁡KE_{\max}.

Example 2

medium
A photon of frequency 8×1014 Hz8\times10^{14} \text{ Hz} hits a metal with work function 3.3×10−19 J3.3\times10^{-19} \text{ J}. Find KEmax⁡KE_{\max}. (h=6.6×10−34h = 6.6\times10^{-34})

Example 3

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 4

medium
The stopping potential for ejected electrons is 1.5 V1.5 \text{ V}. Find their maximum kinetic energy (e=1.6×10−19 Ce = 1.6\times10^{-19} \text{ C}).

Example 5

easy
Find the energy of a photon with frequency f=6×1014 Hzf = 6\times10^{14}\text{ Hz}. (h=6.6×10−34h = 6.6\times10^{-34})

Example 6

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

Example 7

medium
Two metals have work functions ϕ1=2×10−19\phi_1 = 2\times10^{-19} and ϕ2=5×10−19 J\phi_2 = 5\times10^{-19} \text{ J}. For the same photon energy, which emits faster electrons?

Example 8

medium
A graph of KEmax⁡KE_{\max} versus ff has slope hh and yy-intercept −ϕ-\phi. If the slope is 6.6×10−34 J s6.6\times10^{-34}\text{ J s} and intercept is −3.3×10−19 J-3.3\times10^{-19}\text{ J}, find the threshold frequency.

Example 9

easy
Express the work function ϕ=4.8×10−19 J\phi = 4.8\times10^{-19}\text{ J} in electronvolts. (1 eV=1.6×10−19 J1\text{ eV} = 1.6\times10^{-19}\text{ J})

Example 10

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 11

medium
Light of energy 6×10−19 J6\times10^{-19} \text{ J} ejects electrons with KEmax⁡=2×10−19 JKE_{\max} = 2\times10^{-19} \text{ J}. Find the work function.

Example 12

easy
Above threshold, increasing light intensity changes the ___ of emitted electrons.

Example 13

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

Example 14

medium
A photon of frequency 6imes1014extHz6 imes10^{14} ext{ Hz} hits a metal with work function 2.6imes10−19extJ2.6 imes10^{-19} ext{ J}. Find KEmax⁡KE_{\max}. (h=6.6imes10−34h = 6.6 imes10^{-34})

Example 15

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 KEmax⁡KE_{\max} in eV. (h=6.6×10−34h = 6.6\times10^{-34}, c=3×108c = 3\times10^8, 1 eV=1.6×10−19 J1\text{ eV} = 1.6\times10^{-19}\text{ J})

Example 16

easy
If a metal's work function is ϕ=3×10−19 J\phi = 3\times10^{-19} \text{ J} and a photon delivers 5×10−19 J5\times10^{-19} \text{ J}, find the maximum kinetic energy of the electron.

Example 17

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

Example 18

medium
A metal emits electrons with KEmax⁡=1.5×10−19 JKE_{\max} = 1.5\times10^{-19}\text{ J} under light of f=7×1014 Hzf = 7\times10^{14}\text{ Hz}. Find ϕ\phi. (h=6.6×10−34h = 6.6\times10^{-34})

Example 19

medium
A photon has wavelength λ=4×10−7 m\lambda = 4\times10^{-7} \text{ m}. Find its energy. (h=6.6×10−34h = 6.6\times10^{-34}, c=3×108c = 3\times10^8)

Example 20

medium
The stopping voltage for ejected electrons is Vs=0.8 VV_s = 0.8\text{ V}. Find their maximum kinetic energy (e=1.6×10−19 Ce = 1.6\times10^{-19}\text{ C}).
← Back to Photoelectric Effect Worked Examples Formula Explained