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E = hfλ = h/pW₀e⁻
Class 12 Physics · Chapter 11 · Worksheet
Dual Nature of Radiation and Matter
Twenty-eight multiple choice questions with an answer key at the end. No question from this chapter appeared in any paper I could extract or recover, so every question here is built to the AAI pattern and to the difficulty the domain section has shown elsewhere.
30questions
35 minsuggested time
0verified PYQs
MCQformat throughout
Score0 correct · 0 of 30 attempted
Before you begin
Formulas you will need
Key formulas
Photon energy: E = hf = hc ∕ λ · h = 6.626 × 10−34 J s · 1 eV = 1.6 × 10−19 J
Einstein's equation: KEmax = hf − W₀ · Work function W₀ = hf₀ · Stopping potential: eV₀ = KEmax
de Broglie wavelength: λ = h ∕ p = h ∕ mv · For an electron accelerated through V volts: λ ≈ 1.227 ∕ √V nm
Photon momentum p = h ∕ λ = E ∕ c · photons have zero rest mass
How to use this worksheet. Frequency decides the energy of each photoelectron; intensity decides how many are emitted. Keeping those two apart answers most of this chapter. And when frequency is doubled, the kinetic energy more than doubles, because the work function is subtracted only once.
Section A · Level 1
Photoelectric effect and de Broglie waves
Ten questions. None should take more than twenty seconds.
Q1
The photoelectric effect demonstrates the:
Light delivers its energy in discrete packets, which is why the effect depends on frequency rather than intensity. That is particle behaviour.
Q2
The minimum energy required to remove an electron from a metal surface is called the:
The work function W₀ is the energy barrier the electron must overcome to escape.
Q3
Einstein's photoelectric equation is:
The photon supplies hf, the work function is spent escaping, and whatever remains appears as kinetic energy.
Q4
Photoemission occurs only if the frequency of the incident light is:
Below the threshold no single photon carries enough energy, and no amount of extra intensity helps.
Q5
Increasing the intensity of incident light while keeping the frequency fixed increases the:
More intensity means more photons, so more electrons are ejected, but each still carries the same maximum energy.
Q6
The maximum kinetic energy of the photoelectrons depends on the:
Only the frequency sets the energy per photon, and hence the maximum kinetic energy.
Q7
The stopping potential is the voltage needed to:
It is the reverse voltage that just stops even the most energetic electron, so eV₀ = KEmax.
Q8
The de Broglie wavelength of a particle of momentum p is:
λ = h ∕ p. The larger the momentum, the shorter the wavelength, which is why everyday objects show no wave behaviour.
Q9
The rest mass of a photon is:
Photons always travel at c and have no rest mass, though they do carry energy and momentum.
Q10
The momentum of a photon of wavelength λ is:
p = h ∕ λ, which is also E ∕ c.
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Section B · Level 2
Photon energy, stopping potential and wavelength
Ten questions. Work in electron volts wherever the question allows it.
Q11
The energy of a photon of wavelength 500 nm is closest to:
E = hc ∕ λ = (6.626 × 10−34 × 3 × 108) ∕ (500 × 10−9) = 3.98 × 10−19 J, which is about 2.5 eV.
Q12
A metal has a work function of 2 eV. Light of energy 3 eV falls on it. The maximum kinetic energy of the emitted electrons is:
KEmax = 3 − 2 = 1 eV. The work function is subtracted, never added.
Q13
For the same metal and light, the stopping potential is:
Since eV₀ = KEmax, a maximum energy of 1 eV corresponds to a stopping potential of 1 V.
Q14
If the frequency of incident light is doubled while the work function stays the same, the maximum kinetic energy:
KEmax = hf − W₀. Doubling f doubles hf but leaves W₀ unchanged, so the difference more than doubles. This is a favourite trap.
Q15
The threshold wavelength of a metal is the wavelength:
Longer wavelength means lower frequency, so above the threshold wavelength the photons are too weak to cause emission.
Q16
A particle has momentum 6.626 × 10−27 kg m/s. Its de Broglie wavelength is:
λ = h ∕ p = (6.626 × 10−34) ∕ (6.626 × 10−27) = 10−7 m.
Q17
If the speed of a particle is doubled, its de Broglie wavelength:
Wavelength is inversely proportional to momentum, so doubling the speed halves it.
Q18
Which of the following has the longest de Broglie wavelength at the same speed?
At equal speed, the lightest particle has the smallest momentum and therefore the longest wavelength. The electron is by far the lightest of the four.
Q19
The photoelectric current depends on:
Current measures how many electrons are emitted per second, which is set by the number of photons arriving, that is the intensity.
Q20
The value of Planck's constant is closest to:
Planck's constant is 6.626 × 10−34 J s. Option 4 is the electronic charge in coulombs.
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Section C · Level 3
Why wave theory fails, and graphical relationships
Eight questions. Ask what each graph's slope and intercept represent.
Q21
The photoelectric effect cannot be explained by the wave theory of light because wave theory predicts that emission would:
On the wave picture, energy accumulates gradually, so any frequency should eventually work. Experiment shows a sharp threshold instead, which only the photon picture explains.
Q22
The emission of photoelectrons is:
Emission begins within about 10−9 s of illumination, far too fast for energy to have accumulated gradually as wave theory would require.
Q23
A graph of maximum kinetic energy against frequency for a metal is a straight line whose slope is:
Rearranging KE = hf − W₀ shows the slope is h and the intercept on the energy axis is −W₀. Remarkably, the slope is the same for every metal.
Q24
On that same graph, the intercept on the frequency axis gives the:
Where the kinetic energy falls to zero, the frequency is exactly the threshold value f₀.
Q25
The de Broglie wavelength of an electron accelerated through a potential difference V is proportional to:
The electron gains energy eV, so its momentum goes as √V and the wavelength as 1 ∕ √V. Higher accelerating voltage gives a shorter wavelength, which is why electron microscopes use high voltages.
Q26
Electron diffraction experiments provide evidence for the:
Diffraction is a wave phenomenon, so observing it with electrons confirms de Broglie's hypothesis.
Q27
Why is the wave nature of a moving cricket ball never observed?
λ = h ∕ p, and for an everyday object p is enormous compared with h, giving a wavelength far too small to detect by any means.
Q28
Two metals have work functions of 2 eV and 4 eV. Light of energy 3 eV falls on both. Emission occurs from:
The photon energy must exceed the work function. 3 eV clears the 2 eV barrier but not the 4 eV one, so only the first metal emits.
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Answer key
Answers
Check only after you have attempted every question. Tap any question above to see the full worked solution.
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