Q.The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Photoelectric Effect
The Photoelectric Effect: When Light Knocks Electrons Loose
Imagine you're throwing tennis balls at a wall covered in loose pebbles. If you throw hard enough, a pebble might get knocked off. That's the basic picture — but the photoelectric effect is the quantum version of this, and it completely shattered classical physics.
The Intuition
Light is made of tiny packets of energy called photons. Each photon carries a specific amount of energy, determined by its colour (frequency). When a photon hits a metal surface, it can transfer its energy to an electron inside the metal. If that energy is enough, the electron breaks free and flies out.
Think of electrons in a metal like people in a room with a high window. To escape, they need enough energy to reach the window sill. A photon is like a boost — but only if it gives enough energy in one shot. No amount of weak boosts (dim light) will work if each individual boost is too small.
The Precise Statement
Ephoton=hf=ϕ+Kmax
Where:
- Ephoton=hf is the energy of a photon (Planck's constant h=6.63×10−34 J⋅s, f is frequency)
- ϕ is the work function — the minimum energy needed to remove an electron from that metal
- Kmax is the maximum kinetic energy of the ejected electron
What Classical Physics Got Wrong
Before Einstein (1905), physicists thought light was a continuous wave. They expected:
- Brighter light → more energy per electron → faster electrons
- Any colour would eventually eject electrons if you waited long enough
But experiments showed the opposite:
| Observation | Classical Prediction | Actual Result |
|---|---|---|
| Effect of intensity | Brighter light → faster electrons | Brighter light → more electrons, same speed |
| Threshold frequency | None — any light works eventually | Below a certain frequency, no electrons no matter how bright |
| Time delay | Electrons need time to absorb energy | Electrons appear instantly (within 10−9 s) |
The Key Insight
Einstein said: light behaves like a stream of particles (photons), each with energy hf. One photon interacts with one electron. If hf<ϕ, the electron cannot escape — period. If hf>ϕ, the excess energy becomes kinetic energy:
Kmax=hf−ϕ
This is why:
- Increasing intensity (more photons) ejects more electrons, but each electron still gets the same energy per photon — so their speed doesn't change.
- Below threshold frequency, even a trillion photons per second can't help — each one is too weak individually.
The photoelectric effect proved that light is quantized — it comes in discrete packets. This was the birth of quantum mechanics. Einstein won the 1921 Nobel Prize for this, not for relativity.
A Worked Example
Problem: A metal has work function ϕ=2.0 eV. Light of frequency f=6.0×1014 Hz shines on it. Find the maximum kinetic energy of ejected electrons. (h=4.14×10−15 eV⋅s)
Step 1: Photon energy
E=hf=(4.14×10−15)(6.0×1014)=2.48 eV …
Why this formula?
Photoelectric Effect: Why the Key Formulas Hold
The photoelectric effect is a cornerstone of quantum physics. It showed that light behaves as particles (photons) , not just waves. Let's build the reasoning step-by-step.
1. The Core Idea: Energy Conservation
When a photon hits a metal surface, it transfers all its energy to a single electron inside the metal.
- The photon's energy is E=hf, where h is Planck's constant and f is the frequency of light.
- The electron needs a minimum energy to escape the metal — this is called the work function, ϕ.
Why only one electron?
Einstein proposed that light is quantized into discrete packets (photons). A single photon cannot split its energy among multiple electrons — it interacts with one electron at a time.
2. The Photoelectric Equation
If the photon's energy is greater than the work function, the excess energy becomes the electron's kinetic energy after escape:
hf=ϕ+Kmax
Where:
- hf = energy of incident photon
- ϕ = work function (minimum energy to remove electron)
- Kmax = maximum kinetic energy of ejected electron
Why "maximum" kinetic energy?
- Electrons inside the metal have different binding energies.
- Some electrons are near the surface (loosely bound) → get maximum K.
- Others are deeper → lose energy in collisions before escaping → lower K.
3. The Stopping Potential Connection
We measure Kmax using a stopping potential Vs:
Kmax=eVs
Where e is the electron charge. This is because:
- An electric field opposing the electron's motion does work eVs to stop it.
- At the stopping potential, the electron's kinetic energy is exactly balanced by the electric potential energy.
Combining:
hf=ϕ+eVs
This is the Einstein photoelectric equation in its most testable form.
4. Why the Threshold Frequency Exists
From the equation:
hf=ϕ+eVs
If f is too low, hf<ϕ. Then:
- The photon cannot supply enough energy to overcome the work function.
- No electron is ejected, regardless of light intensity.
The threshold frequency f0 is when Kmax=0:
hf0=ϕ⇒f0=hϕ
Why intensity doesn't matter for ejection?
- Intensity = number of photons per second.
- Each photon still has energy hf. If hf<ϕ, even a billion photons won't eject an electron — each photon is individually too weak.
5. Why Kinetic Energy Depends on Frequency, Not Intensity
From Kmax=hf−ϕ:
- Frequency f directly determines Kmax.
- Intensity only affects the number of electrons ejected (more photons → more electrons), not their individual energy.
This was the key experimental contradiction with classical wave theory:
- Classical: Higher intensity = bigger wave amplitude = more energy to electrons.
- Reality: Higher frequency = more energy per electron; intensity only changes current.
6. Summary of Key Relationships …
Concept: Maximum Kinetic Energy — photoelectric emission occurs only if the incident photon energy exceeds the work function.
Step 1: Find the photon energy.
Wavelength λ=330 nm=330×10−9 m.
Photon energy E=λhc. Using hc=1240 eV⋅nm:
E=330 nm1240 eV⋅nm≈3.76 eV
Step 2: Compare with work function ϕ=4.2 eV. …
The key is to compare the incident photon energy with the metal's work function. For a 330 nm wavelength, the photon energy is about 3.76 eV, which is less than the work function of 4.2 eV. Therefore, no photoelectric emission will occur.
Why this comparison works
Photoelectric emission happens only when an incident photon has enough energy to overcome the binding energy holding an electron in the metal. That minimum required energy is the work function (ϕ). If the photon's energy (E) is less than ϕ, the electron simply cannot be freed — no matter how many photons hit the surface.
So the entire problem reduces to one question: Is the photon energy from a 330 nm wave greater than or equal to 4.2 eV?
Step-by-step solution
- Find the photon energy in joules first.
The energy of a single photon is given by E=λhc, where:
- h=6.63×10−34 J⋅s (Planck's constant)
- c=3.00×108 m/s (speed of light)
- λ=330 nm=330×10−9 m
E=330×10−9(6.63×10−34)(3.00×108)
Compute step by step:
E=3.30×10−71.989×10−25=6.027×10−19 J
- Convert this energy into electronvolts. Since 1 eV=1.602×10−19 J, we divide:
E=1.602×10−196.027×10−19≈3.76 eV
A faster route: use the handy constant hc=1240 eV⋅nm. Then E=λ (nm)1240 gives the energy directly in eV. Here: E=3301240≈3.76 eV. This shortcut saves time in exams — just remember the constant is 1240 eV⋅nm.
- Compare with the work function. …
Method: Photoelectric Effect Threshold Condition
We check whether the incident photon has enough energy to overcome the metal's work function. If the photon energy is greater than or equal to the work function, emission occurs.
Step 1: Convert work function to joules (or keep in eV — we'll compare in eV).
Work function ϕ=4.2 eV.
Step 2: Find the energy of the incident photon.
Photon energy E=λhc, where h=6.63×10−34 J⋅s, c=3×108 m/s, and λ=330 nm=330×10−9 m.
First compute in joules:
E=330×10−9(6.63×10−34)(3×108)=3.30×10−71.989×10−25=6.027×10−19 J
Convert to eV (1 eV = 1.6×10−19 J):
E=1.6×10−196.027×10−19=3.77 eV
Step 3: Compare photon energy with work function.
Photon energy =3.77 eV
Work function =4.2 eV …
The most common mistake here is rushing to compare the work function directly with the photon energy without checking units. The work function is given in eV, but the wavelength is in nm — you must convert everything to a consistent unit system before comparing.
Mistake 1: Forgetting to convert wavelength to energy in eV
Students often calculate the photon energy in joules and then compare it directly to 4.2 eV without converting. That gives a meaningless comparison.
How to avoid: Always compute the photon energy in the same unit as the work function. Use the formula
E=λhc
with h=6.63×10−34 J⋅s, c=3×108 m/s, and λ=330 nm=330×10−9 m.
First get E in joules:
E=330×10−9(6.63×10−34)(3×108)=6.03×10−19 J
Now convert to eV using 1 eV=1.6×10−19 J:
E=1.6×10−196.03×10−19=3.77 eV
A shortcut many students try is using hc=1240 eV⋅nm directly. That works, but only if you remember the constant correctly. The exact value is 1240 eV⋅nm, so E=3301240≈3.76 eV. This is faster and less error-prone — but only if you trust the constant.
Mistake 2: Comparing the wrong quantities
Some students compare the photon energy to the threshold frequency or to the maximum kinetic energy instead of the work function. The condition for emission is simple: photoelectric emission occurs only if the incident photon energy E is greater than or equal to the work function ϕ.
How to avoid: State the condition clearly before plugging numbers. Write:
For emission: E≥ϕ
Here E=3.77 eV and ϕ=4.2 eV. Since 3.77<4.2, emission does not occur.
Mistake 3: Confusing wavelength and frequency
A student might compute the frequency from λ and then compare it to the threshold frequency. That's fine in principle, but it adds an extra step where unit errors can creep in. The threshold frequency is f0=ϕ/h, and you'd need to check if f>f0. It's safer to work directly with energy.
How to avoid: Stick to one method. The energy comparison is the most direct: compute E in eV, compare to ϕ in eV.
Mistake 4: Misinterpreting "work function" …
Showing the 12 most recent of 21 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.The work function of a material is 6.6 eV. Then, the threshold wavelength of the metal is approximately (Take h=6.6×10−34 J.s) (A) 108 nm (B) 188 nm (C) 208 nm (D) 228 nm (E) 250 nm
›Reveal solutionSolution
Threshold wavelength satisfies W=hc/λ0. With W=6.6 eV this gives about 188 nm.
Convert the work function to joules:
W=6.6 eV=6.6×1.6×10−19=1.056×10−18 J.
Threshold wavelength: …
- KEAM 2026Set eng-2026-04184 marksMCQQ.Two light rays of wavelength λ and 4λ incident on the surface of a photo sensitive material emit electrons with max kinetic energy E and 6E respectively. The work function of the material is (h = Planck's constant, c = velocity of light in free space) (A) λhc (B) 2λhc (C) 5λ2hc (D) 5λ3hc (E) 3λhc
›Reveal solutionSolution
Writing Einstein's equation for both wavelengths and eliminating E gives the work function W=5λ2hc.
Einstein's photoelectric equation, KE=λhc−W:
For wavelength λ:
E=λhc−W(1)
For wavelength λ/4 (photon energy 4hc/λ):
6E=λ4hc−W(2)
Subtract (1) from (2): …
- KEAM 2026Set eng-2026-04194 marksMCQQ.If the threshold wavelengths of two metals are in the ratio 1:3, then the work functions of these metals are in the ratio (A) 1:3 (B) 2:1 (C) 3:1 (D) 1:2 (E) 3:2
›Reveal solutionSolution
Work function is inversely proportional to threshold wavelength, so wavelengths 1:3 give work functions 3:1.
The threshold (work function) relation is
ϕ=λ0hc⇒ϕ∝λ01.
Given λ0,1:λ0,2=1:3, …
- KEAM 2026Set eng-2026-04204 marksMCQQ.Stopping potentials for the metals A and B are 0.4 V and 1.6 V, respectively. When illuminated by same light, the difference in their work functions is: (A) 2.0 eV (B) 1.2 eV (C) 6.4 eV (D) 0.4 eV (E) 2.0 V
›Reveal solutionSolution
With identical incident light, the work-function difference equals e times the stopping-potential difference: 1.2 eV.
Photoelectric equation. eV0=hν−ϕ, so ϕ=hν−eV0 (same hν for both metals).
Difference. …
- KEAM 2026Set eng-2026-04214 marksMCQQ.If a radiation of energy 5.2eV falls on the photosensitive surfaces of Mo and Ni, they emit photoelectrons with maximum kinetic energy of 0.5eV and 1eV, respectively. Then the work function of (A) Mo is 2.6eV (B) Ni is 6.2eV (C) Mo is 6.2eV (D) Ni is 4.2eV (E) Mo is 4.2eV
›Reveal solutionSolution
Photoelectric equation ϕ=E−KEmax: Ni =4.2eV.
Using KEmax=E−ϕ:
ϕMo=5.2−0.5=4.7eV,ϕNi=5.2−1=4.2eV. …
- KEAM 2026Set eng-2026-04224 marksMCQQ.If the energy of the incident radiation on a metal is increased by 10 %, the kinetic energy of the emitted photoelectrons increases from 0.5eV to 0.75 eV, then the work function of the metal is (A) 2 eV (B) 1.5 eV (C) 1 eV (D) 2.5 eV (E) 1.8 eV
›Reveal solutionSolution
Einstein's equation E=W+KE. Increasing E by 10% raises KE from 0.5 to 0.75 eV; solving 1.1(W+0.5)=W+0.75 gives W=2 eV.
Initially E=W+0.5. When the incident energy is increased by 10%:
1.1E=W+0.75.
Substituting E=W+0.5: …
- KEAM 2026Set pha-2026-0419F4 marksMCQQ.The work function of a material that has the threshold frequency of 5×1014 Hz is (h=6.626×10−34 Js) (A) 3.09 eV (B) 5.35 eV (C) 4.14 eV (D) 2.07 eV (E) 1.03 eV
›Reveal solutionSolution
Work function ϕ0=hν0=3.313×10−19 J≈2.07 eV.
The work function equals Planck's constant times the threshold frequency:
ϕ0=hν0=(6.626×10−34J s)(5×1014Hz)=3.313×10−19J. …
- KEAM 2026Set pha-2026-0420F4 marksMCQQ.If the stopping potential in a photoelectric experiment is measured to be 1.82 V, the maximum speed of the emitted electrons, in ms−1, is (mass of the electron = 9.1×10−31 kg) (A) 8.0×105 (B) 2.3×105 (C) 3.0×105 (D) 7.3×106 (E) 5.3×1011
›Reveal solutionSolution
eV0=21mv2; v=2eV0/m=2(1.6×10−19)(1.82)/9.1×10−31≈8.0×105 m/s.
The stopping potential equals the maximum kinetic energy of the photoelectrons:
eV0=21mvmax2⇒vmax=m2eV0.
Substituting e=1.6×10−19C, V0=1.82V, m=9.1×10−31kg: …
- KEAM 2025Set eng-2025-04234 marksMCQQ.Pick out the INCORRECT statement from the following : In photoelectric phenomenon, (A) the value of stopping potential is the same for radiations of all frequencies (B) the stopping potential is more negative for the incident radiation of higher frequency (C) the value of saturation current depends on the intensity of incident radiation (D) the value of saturation current is independent of frequency of incident radiation (E) the emission of electrons is instantaneous
›Reveal solutionSolution
The incorrect statement is that the stopping potential is the same for all frequencies.
Concept and Intuition
Einstein's photoelectric equation gives eV_0 = h*nu - phi, so the stopping potential increases with frequency. Saturation current depends on intensity, not frequency, and emission is instantaneous. Statement (A) contradicts the frequency dependence of stopping potential and is therefore the false one.
Step-by-Step Solution
- eV_0 = h*nu - phi, so V_0 depends on frequency.
- Hence (A) 'same for all frequencies' is wrong. …
- KEAM 2025Set eng-2025-04254 marksMCQQ.If light waves of wavelengths λ and λ/3 are incident on the surface of a material, photoelectrons are emitted with maximum kinetic energy E and 4E respectively, then the work function of the material is (A) 2λhc (B) 3λhc (C) λhc (D) 2λ3hc (E) λ2hc
›Reveal solutionSolution
Apply Einstein's photoelectric equation to both wavelengths and eliminate E: the work function comes out as hc/3λ.
Einstein's equation KE=λhc−ϕ:
E=λhc−ϕ(1)
4E=λ/3hc−ϕ=λ3hc−ϕ(2)
Subtracting (1) from (2): …
- KEAM 2025Set eng-2025-04264 marksMCQQ.Which one of the following statements is INCORRECT? In photoelectric effect (A) Threshold frequency is different for different metals (B) The same metal gives same response to light of different wavelengths (C) The emission of photoelectrons is an instantaneous process (D) Above the threshold frequency the number of photoelectrons emitted per sec is directly proportional to the intensity of incident radiation (E) The maximum K.E. of the photoelectrons is independent of the intensity of incident radiation
›Reveal solutionSolution
A metal's photoelectric response depends strongly on the wavelength/frequency of light, so the claim that it responds identically to different wavelengths is false.
In the photoelectric effect, emission and the maximum kinetic energy of photoelectrons depend on the frequency (wavelength) of the incident light: KEmax=hν−ϕ0.
Statements (A), (C), (D) and (E) are all correct standard features of the photoelectric effect. …
- KEAM 2025Set eng-2025-04274 marksMCQQ.The plot of maximum kinetic energy of photo-electrons to the energy of the incident photon above its threshold frequency on a photo-sensitive material of work function φ is (A) an oblique straight line with a positive slope. (B) an oblique straight line with a negative slope. (C) an oblique straight line passing through the origin. (D) an exponential curve. (E) a polynomial curve of order 2.
›Reveal solutionSolution
Einstein's equation Kmax=hν−φ makes maximum KE a straight line in photon energy hν with positive slope +1.
Einstein's photoelectric equation is:
Kmax=hν−φ,
where hν is the incident photon energy and φ the work function. Plotting Kmax (y-axis) against the photon energy E=hν (x-axis) gives a straight line of slope +1 (positive) and intercept −φ on the KE axis. …
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