Q.Light of frequency 7.21×1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0×105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Maximum Kinetic Energy
Maximum Kinetic Energy – From Intuition to Precision
Kinetic energy is the energy of motion: the faster something moves, the more kinetic energy it has. In many physical situations there is a maximum possible kinetic energy a particle can reach — set either by energy conservation or by an external energy constraint.
The Precise Statement
Kmax=21mvmax2
Where:
- Kmax = maximum kinetic energy (in joules)
- m = mass of the object (in kg)
- vmax = maximum speed reached (in m/s)
This formula alone doesn't tell you why there's a maximum — the physics lies in energy conservation or in an external constraint that limits the speed.
Where Does the Maximum Come From?
1. Energy conservation (no friction)
In a closed system, total mechanical energy E=K+U is constant, so
Kmax=E−Umin
The maximum kinetic energy occurs when the potential energy U is at its minimum — for example, a falling object is fastest (and U smallest) just before it lands.
2. External constraints (e.g., the photoelectric effect)
In modern physics, electrons in a metal absorb light energy. Each photon delivers a fixed energy hf. The electron must spend part of that energy escaping the metal (the work function ϕ); the rest becomes kinetic energy:
Kmax=hf−ϕ
Here the maximum is set entirely by the photon energy — no matter how intense the light, no single electron can gain more kinetic energy than this.
A Common Mistake
Students often think "maximum kinetic energy" means the fastest speed possible in the universe. It doesn't. The "maximum" is relative to the given system — the highest value under the stated conditions (height, spring compression, photon energy, and so on), not a universal speed limit. …
Why this formula?
Maximum Kinetic Energy — Why the Formula Holds
The idea of "maximum kinetic energy" appears in two very different contexts in your syllabus: photoelectric effect (modern physics) and simple harmonic motion (oscillations). I'll cover both, because the why is different in each case.
1. In the Photoelectric Effect
The formula you must know:
Kmax=hν−ϕ
where h is Planck's constant, ν is the frequency of incident light, and ϕ is the work function of the metal.
Why this formula? It comes from Einstein's photon model and energy conservation.
A single photon carries energy E=hν. When it strikes a metal surface, it can transfer all of its energy to one electron. That electron must first overcome the binding force holding it in the metal — the minimum energy needed for this is the work function ϕ. Any leftover energy becomes the electron's kinetic energy after it escapes.
So:
Photon energy = Energy to escape + Kinetic energy of ejected electron
hν=ϕ+K
If the electron just barely escapes (with zero kinetic energy), the photon frequency is the threshold frequency ν0, where hν0=ϕ.
For a higher frequency, the maximum kinetic energy an ejected electron can have is when it absorbs the photon's full energy and loses nothing to collisions inside the metal. That gives:
Kmax=hν−ϕ
Kmax does not depend on light intensity. Intensity only increases the number of electrons, not their maximum energy. This was the key puzzle that classical physics couldn't explain.
2. In Simple Harmonic Motion (SHM)
For a particle executing SHM, the maximum kinetic energy is:
Kmax=21mω2A2
where m is mass, ω is angular frequency, and A is amplitude.
Why this formula? It follows directly from the velocity equation.
In SHM, displacement is x=Asin(ωt+ϕ). Differentiating gives velocity:
v=dtdx=Aωcos(ωt+ϕ)
The velocity is maximum when cos(ωt+ϕ)=±1, i.e., at the equilibrium position (x=0):
vmax=Aω …
Concept: Maximum Kinetic Energy — the photoelectric equation relates the incident photon energy to the work function and the maximum kinetic energy of the ejected electron.
The photoelectric equation is:
hf=ϕ+21mvmax2
where ϕ=hf0 is the work function and f0 is the threshold frequency.
Step 1: Write the equation in terms of f0:
hf=hf0+21mvmax2
Step 2: Solve for f0:
f0=f−2hmvmax2
Step 3: Substitute values. Use m=9.1×10−31 kg, h=6.63×10−34 J⋅s:
f0=7.21×1014−2×6.63×10−34(9.1×10−31)(6.0×105)2 …
The threshold frequency is found by equating the maximum kinetic energy of ejected electrons to the difference between the incident photon energy and the work function. Using Kmax=hf−hf0, we get f0=f−hKmax. The result is f0=4.74×1014 Hz.
The core idea here is the photoelectric effect: when light hits a metal, each photon gives its energy hf to an electron. The electron uses some of that energy to escape the metal (the work function ϕ=hf0), and the rest becomes kinetic energy. The maximum kinetic energy occurs for electrons that escape without losing energy to collisions inside the metal.
So the equation is:
Kmax=hf−hf0
where f0 is the threshold frequency — the minimum frequency needed to eject any electron at all.
We know f=7.21×1014 Hz and the maximum speed vmax=6.0×105 m/s. We need f0.
- Find the maximum kinetic energy. The kinetic energy is Kmax=21mevmax2, where me=9.11×10−31 kg (electron mass).
Kmax=21(9.11×10−31)(6.0×105)2
First square the speed: (6.0×105)2=3.6×1011.
Then multiply: 9.11×10−31×3.6×1011=3.2796×10−19.
Half of that: Kmax=1.6398×10−19 J.
You can also work in electronvolts if you prefer, but joules are fine here since Planck's constant is in J·s. Just be consistent.
- Write the photoelectric equation.
hf=hf0+Kmax
So
hf0=hf−Kmax
and
f0=f−hKmax
- Plug in the numbers. Planck's constant h=6.626×10−34 J⋅s. First compute hf:
hf=(6.626×10−34)(7.21×1014)=4.777×10−19 J
(Check: 6.626×7.21≈47.77, and 10−34×1014=10−20, so 4.777×10−19 — correct.) …
Method: Photoelectric Equation Approach
This problem uses Einstein's photoelectric equation, which connects the incident photon energy, the work function (or threshold frequency), and the maximum kinetic energy of ejected electrons.
Step 1 – Write the photoelectric equation
The maximum kinetic energy of ejected electrons is given by:
Kmax=hf−hf0
where h is Planck's constant, f is the incident frequency, and f0 is the threshold frequency.
Step 2 – Express Kmax in terms of the given speed
The maximum kinetic energy is also:
Kmax=21mvmax2
where m is the electron mass (9.1×10−31 kg) and vmax=6.0×105 m/s.
Step 3 – Equate and solve for f0
From the two expressions:
21mvmax2=hf−hf0
Rearranging for f0:
f0=f−2hmvmax2
Step 4 – Substitute values
Take h=6.63×10−34 J⋅s, m=9.1×10−31 kg, f=7.21×1014 Hz, and vmax=6.0×105 m/s.
First compute the kinetic energy term:
2mvmax2=2(9.1×10−31)(6.0×105)2 …
Students often lose marks on this question not because the photoelectric equation is hard, but because they rush or mis-handle units and constants. Here are the most common mistakes and how to avoid each.
Mistake 1: Forgetting to convert electron volts to joules (or vice versa)
The photoelectric equation Kmax=hf−ϕ uses h=6.63×10−34 J⋅s. If you try to work in eV without converting consistently, you'll get a wrong numerical answer. Many students compute hf in joules, then subtract a work function in eV — that's mixing units.
How to avoid: Stick entirely to SI units (joules, kg, m/s) throughout the calculation. Only convert to eV at the very end if the question asks for it. Here, the threshold frequency is asked in Hz, so stay in joules.
Mistake 2: Using the wrong expression for kinetic energy
The maximum kinetic energy of ejected electrons is Kmax=21mvmax2, where m is the electron mass (9.11×10−31 kg). Some students mistakenly use mv (momentum) or forget the 21 factor.
How to avoid: Write the kinetic energy formula explicitly before plugging numbers. Double-check that you've squared the speed and multiplied by half.
Mistake 3: Confusing threshold frequency with threshold wavelength
The threshold frequency f0 is related to the work function by ϕ=hf0. Some students try to use λ0=c/f0 prematurely, or they solve for wavelength when the question asks for frequency.
How to avoid: Read the question carefully — it asks for threshold frequency. Solve directly from f0=hϕ after finding ϕ. Don't introduce wavelength unless needed.
Mistake 4: Arithmetic or exponent errors with large/small numbers
The numbers here are typical: h≈6.63×10−34, me≈9.11×10−31, speeds around 105–106 m/s, frequencies around 1014 Hz. A single exponent slip (e.g., writing 10−20 instead of 10−19) changes the answer completely.
How to avoid: Work step by step, writing each intermediate result in scientific notation. Use your calculator carefully — enter the full expression at once if possible, or check the exponent after each multiplication.
Mistake 5: Forgetting that Kmax is the maximum kinetic energy …
Showing the 12 most recent of 19 on this concept.
- CBSE 2026Set A1 markMCQQ.The maximum kinetic energy of the photoelectrons varies (A) linearly with the frequency and the intensity of the incident radiation (B) linearly with the frequency and is independent of the intensity of the incident radiation (C) inversely with the frequency and is independent of the intensity of the incident radiation (D) inversely with the intensity and is independent of the frequency of the incident radiation
›Reveal solutionSolution
Einstein's photoelectric equation makes Kₘₐₓ a linear function of frequency, independent of intensity.
Einstein's photoelectric equation is:
Kmax=hν−ϕ0
where ν is the frequency of incident light and ϕ0 the work function. This is a straight line in ν (slope h), so Kmax varies linearly with frequency. Intensity only sets the number of photons (hence the photocurrent), …
- CBSE 2026Set A1 markMCQQ.The maximum velocity of an electron emitted from a metal surface becomes two times when the frequency v of the incident light is doubled. The work function of the metal is (A) 2hv/3 (B) hv/3 (C) zero (D) hv/2
›Reveal solutionSolution
Using KE ∝ v_max², doubling v_max means KE becomes 4×; solving Einstein's equation at frequency v and 2v gives φ = 2hv/3.
Einstein's photoelectric equation is
21mvmax2=hν−ϕ.
At frequency ν, maximum speed vmax:
21mvmax2=hν−ϕ.(1)
When the frequency is doubled to 2ν, the maximum speed doubles to 2vmax, so the kinetic energy becomes four times: …
- CBSE 2026Set ANNUAL1 markMCQQ.The specific charge of electron is :(a) 1.9 × 10^31 C/kg(b) 1.6 × 10^19 C/kg(c) 1.76 × 10^11 C/kg(d) 1.76 × 10^-11 C/kg
›Reveal solutionSolution
e/m = (1.6 × 10⁻¹⁹)/(9.1 × 10⁻³¹) ≈ 1.76 × 10¹¹ C/kg.
The specific charge of the electron is the ratio of its charge to its mass, e/m. Using e = 1.6 × 10⁻¹⁹ C and m = 9.1 × 10⁻³¹ kg:
…
- CBSE 2026Set ANNUAL1 markQ.Write True or False: By suitably heating, sufficient thermal energy can be imparted to the free electrons to enable them to come out of the metal.
›Reveal solutionSolution
True — this describes thermionic emission.
Free electrons in a metal are normally held inside by the surface potential barrier (the work function). If the metal is heated to a high temperature, the free electrons gain thermal energy; some acquire enough energy to overcome the work function and are emitted from the metal surface. T …
- CBSE 2025Set ANNUAL1 markMCQQ.Assertion (A): When ultraviolet light is incident on two photosensitive metals having different work functions then maximum kinetic energy of the photo electrons is greater for metal having low work function. Reason (R): Kinetic energy =21mv2(a) Both A and R are correct and R is the correct explanation of A.(b) Both A and R are correct but R is not correct explanation of A.(c) A is correct but R is incorrect.(d) A and R both are incorrect.
›Reveal solutionSolution
A is true by Einstein's photoelectric equation; R is a true but generic definition of KE that, by itself, does not explain why a lower work function gives a higher maximum photoelectron KE.
Assertion (A) is true. By Einstein's photoelectric equation, the maximum kinetic energy of an emitted photoelectron is
KEmax=hν−ϕ
where ϕ is the work function of the metal. For the same incident frequency ν, a metal with a lower work function ϕ gives a larger KEmax, so A is correct.
…
- CBSE 2025Set IMPROVEMENT1 markMCQQ.Assertion (A): Photoelectric current is zero when the stopping potential (V₀) is sufficient to repel even the most energetic (K_max) photoelectrons. Reason (R): According to Einstein's photoelectric equation, K_max = hν − ϕ₀, where hν is the energy of each quantum of radiation incident on the metal surface and ϕ₀ is the work function of the metal. Select the correct option.(a) Both A and R are correct and R is the correct explanation of A.(b) Both A and R are correct but R is not the correct explanation of A.(c) A is correct but R is incorrect.(d) Both A and R are incorrect.
›Reveal solutionSolution
Both statements are individually true, but Einstein's equation for K_max does not itself explain why the photoelectric current becomes zero at the stopping potential.
Assertion (A) is true by the very definition of the stopping potential V0: it is the minimum retarding potential that is just able to stop even the fastest (most energetic, Kmax) photoelectrons, so at V=V0 no photoelectron — however energetic — can reach the collector, and the photoelectric current drops to zero.
Reason (R) states Einstein's photoelectric equation, Kmax=hν−ϕ0, which is also true — it correctly gives the maximum kinetic energy of an emitted photoelectron in terms of the incident photon energy hν and the work function ϕ0.
…
- CBSE 2025Set ANNUAL1 markMCQQ.The energy of a photon is 18 eV and the work function of the material is 8 eV. The value of stopping potential is(a) zero(b) 8 V(c) 10 V(d) 26 V
›Reveal solutionSolution
The stopping potential (in volts) equals the maximum kinetic energy of the photoelectrons (in eV), which is the photon energy minus the work function.
Einstein's photoelectric equation:
Kmax=hν−ϕ0
where hν is the photon energy and ϕ0 is the work function. Given hν=18 eV and ϕ0=8 eV:
…
- CBSE 2025Set ANNUAL1 markMCQQ.The maximum kinetic energy of the photoelectrons depends only on :(a) incident angle.(b) frequency.(c) pressure.(d) potential.
›Reveal solutionSolution
By Einstein's photoelectric equation, KEmax=hν−ϕ0, so for a given metal (fixed work function ϕ0) the maximum KE depends only on the frequency ν of incident light, not on its intensity.
Einstein's photoelectric equation is
KEmax=hν−ϕ0
where h is Planck's constant, ν is the frequency of incident radiation, and ϕ0 is the work function of the metal (a fixed property of the material). Increasing the intensity of light increases the number of photoelectrons emitted (photocurrent) but not the maximum kinetic energy o …
- CBSE 2025Set ANNUAL1 markMCQQ.The maximum kinetic energy of photoelectrons emitted from a metal surface when photons of energy 5.6 eV fall on it, is 4eV. The stopping potential in volts is –(a) 1.6 V(b) 3.2V(c) 4V(d) 5.6V
›Reveal solutionSolution
Stopping potential V0=KEmax/e, and KEmax is already given directly as 4 eV.
The stopping potential is defined by eV0=KEmax. Here KEmax is given directly as 4eV (not needing to be computed from the photon energy and work function, both of which are alread …
- CBSE 2025Set ANNUAL1 markMCQQ.In photoelectric emission, a radiation whose frequency is 4 times threshold frequency of a certain metal is incident on the metal. Then, the maximum possible velocity of the emitted electron will be :(a) 2mhν0(b) mhν0(c) 2mhν0(d) m6hν0
›Reveal solutionSolution
With incident frequency 4ν0, Einstein's photoelectric equation gives KEmax=3hν0, so vmax=6hν0/m.
Working
Einstein's photoelectric equation: KEmax=hν−hν0.
Given ν=4ν0:
KEmax=h(4u0)−hu0=3hu0
Since KEmax=21mvmax2: …
- CBSE 2024Set ANNUAL1 markMCQQ.The maximum kinetic energy of a photo electron emitted from a metal is 1.8 eV. The value of stopping potential (cut-off voltage) will be -(a) 3.6 V(b) 2.0 V(c) 1.8 V(d) 0.9 V
›Reveal solutionSolution
The stopping potential V₀ is defined by eV₀ = KE_max, so measuring KE_max in electron-volts directly gives V₀ in volts.
The stopping potential V0 is the retarding potential just sufficient to stop even the fastest photoelectrons, defined by:
eV0=KEmax
Here KEmax=1.8 eV, i.e. KEmax=1.8e (in joules, using the electron-volt definition). Substituting:
…
- CBSE 2024Set ANNUAL1 markMCQQ.Photons of energies 1eV and 2eV are successively incident on a metallic surface of work function 0.5 eV. The ratio of kinetic energy of most energetic photoelectrons in the two cases will be -(a) 1:2(b) 1:3(c) 1:1(d) 1:4
›Reveal solutionSolution
Einstein's photoelectric equation gives KEmax=Ephoton−ϕ; subtract the same work function from each photon energy and take the ratio.
Work function ϕ=0.5 eV.
KE1=1−0.5=0.5 eV …
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