Einstein's Photoelectric Equation
Einstein's Photoelectric Equation revolutionized our understanding of the photoelectric effect by providing a clear relationship between the energy of incident photons and the kinetic energy of emitted electrons. Building on the work of Heinrich Hertz, Einstein proposed that the energy of a photon is directly proportional to its frequency, described by the equation E=hν, where E is the energy, h is Planck's constant, and ν\nuν is the frequency. This equation also introduced the concept that any excess energy beyond the work function of a material translates into the kinetic energy of the emitted electrons. In real-life applications, Einstein's equation underpins technologies such as photovoltaic cells in solar panels, which convert light into electrical energy, and advanced photoelectric sensors used in various electronic devices and safety systems. In this article, we will discuss the concept of Einstein's Photoelectric Equation with solved examples for better understanding.
- Einstein's Photoelectric Equation
- Solved Examples Based on Einstein's Photoelectric Equation
- Example 1: The threshold frequency of a metal is f0 . When the light of frequency 2f0 is incident on the metal plate, the maximum velocity of photoelectrons is v1. When the frequency of incident radiation is increased to 5f0, the maximum velocity of photoelectrons emitted is v2. The ratio of v1 to v2 is:
- Summary
Einstein's Photoelectric Equation
Einstein's photoelectric equation is a fundamental principle that describes the relationship between the energy of incident photons and the kinetic energy of the electrons ejected from a material during the photoelectric effect. This equation was a pivotal development in quantum physics and contributed to the understanding of light as being composed of discrete packets of energy called photons. Einstein's Photoelectric Equation provides a quantitative description of the photoelectric effect, where electrons are emitted from a material's surface when exposed to light.
As we studied E=hv is the equation of energy of each photon. Now we have also studied that the threshold frequency is the frequency below which the electrons won’t come out of the metallic surface. From the above equation, we see that Energy is a function of frequency. Now one question will come to mind when the electron gets ejected then where does all the electron go? Does that electron have some energy to go anywhere? If yes then which type of energy?
All these types of questions were well answered by the great scientist Albert Einstein, According to the experiment performed by Albert Einstein, there are some conclusions that those electrons have kinetic energy only. Also, the energy absorbed by the photons is partly used to overcome the force of the metallic surface. Since there is no electric field present outside the metallic surface so there only energy present is pure kinetic energy.
So, we have K.E. of the photo-electrons = (Energy obtained from the Photon) – (The energy used to escape the metallic surface)
Here, The energy used to escape the metallic surface is the work function (ϕ) which we have discussed already. So Einstein’s Photoelectric equation can also be written as
K.E. =hν−Φ
We can understand the work function more clearly like this
As we know an electron needs some minimum energy to be extracted from a metallic surface. So from the above equation, if ν = threshold frequency (ν0) then the electrons get just enough quantum energy to come out of the metal. This means that the kinetic energy of such an electron is zero. So we can write that -
K.E. =hν−Φ
This is the relation between the threshold frequency and the work function. We can also change this equation in terms of the threshold wavelength.
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Solved Examples Based on Einstein's Photoelectric Equation
Example 1: The threshold frequency of a metal is f0 . When the light of frequency 2f0 is incident on the metal plate, the maximum velocity of photoelectrons is v1. When the frequency of incident radiation is increased to 5f0, the maximum velocity of photoelectrons emitted is v2. The ratio of v1 to v2 is:
1) v1v2=18
2) v1v2=14
3) v1v2=116
4) v1v2=12
Solution:
Using photoelectric equation hf−hf0=eV0
As per question
h(2f0)−h(f0)=eV1 h(2f0−f0)=eV1hf0=eV1…(1)h(5f0)−hf0=eV2 h(5f0−f0)=eV24hf0=eV2….(2)
Equation
21⇒4hf0hf=eV2eV1V2 V1=4
As we know
KEmax=eV=12mvmax2vmax∝V∴v2v1=V2 V1=4=2v1v2=12
Hence, the answer is the option (4).
Example 2: A photon with a wavelength of 400 nm strikes a metal surface, producing an electron with a kinetic energy of 2.5 eV. What is the work function of the metal surface?
1) 0.5 eV
2) 2.5 eV
3) 3.3 eV
4) 4.1 eV
Solution:
The energy of a photon is given by
E=hc/λ.
Therefore, the energy of a photon with a wavelength of 400 nm is given by
E=(6.626×10−34Js)(3.00×108 m/s)/(400×10−9 m)=4.97×10−19 JE=3eV
The kinetic energy of the emitted electron is given by
KE=hf−W
where W is the work function of the metal surface.
Solving for W , we get
W=hf−KE=(3−2.5)eV=0.5eV
Hence, the answer is the option (1).
Example 3: What is the work function of a metal if it emits electrons with a kinetic energy of 2.5 eV when exposed to light with a wavelength of 500 nm? (Given: h=6.626×10−34 Js, s=3.0×108 m/s,1eV=1.6×10−19 J)
1) 3.94 eV
2) 4.47 eV
3) 2.13 eV
4) 1.48 eV
Solution:
The energy of the incident light can be calculated using the equation
E=hcλ.E=(6.626×10−34 Js)×(3.0×108 m/s)(500×10−9 m)E=3.98eV
The work function of the metal can be calculated using the equation
ϕ=E−KE,
where KE is the kinetic energy of the emitted electrons.
ϕ=3.98eV−2.5eV=1.48eV
Hence, the answer is the option (4).
Example 4: What is the kinetic energy of an electron emitted from a metal surface when the frequency of the incident radiation is 1.2×1015 Hz and the work function of the metal is 1.8 eV?
1) 5 eV
2) 3.2 eV
3) 120 eV
4) 9 eV
Solution:
The kinetic energy of an electron emitted from a metal surface by incident radiation is given by the equation
KE=hν−ϕ
where h is Planck's constant, ν is the frequency of the incident radiation, and ϕ is the work function of the metal.
Substituting the given values, we get
KE=(6.626×10−34Js)(1.2×1015 Hz)−(1.8eV)=3.2eV
Hence, the answer is the option (2).
Example 5: Given below are two statements; one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): In Vernier Calliper if a positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading.
Reasons (R): The zero error in Vernier Calliper might have happened due to a manufacturing defect or due to rough handling.
In the light of the above statements, choose the correct answer from the options given below:
1) Both (A) and (R) are correct and (R) is the correct explanation of (A)
2) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
3) (A) is true but (R) is false
4) (A) is false but (R) is true
Solution:
Actual reading = observed reading – zero error
⇒ observed reading > actual reading
Hence, the answer is the option (2).
Summary
Einstein's Photoelectric Equation describes how photons cause the emission of electrons from a material's surface. It asserts that the energy of a photon is used to overcome the material's work function ϕ, with any excess energy converting into the kinetic energy of the emitted electrons (Ek=hν−ϕ). This equation explains the relationship between photon frequency, work function, and the kinetic energy of emitted electrons. It has practical applications in technologies like solar panels and photoelectric sensors, which harness the photoelectric effect to convert light into electrical energy.