3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A clean metal surface is illuminated with monochromatic light of wavelength λ and the stopping potential is found to be V. When the wavelength is changed to 2λ/3, the stopping potential becomes 3V. The threshold wavelength of the metal is ______. Exactly one option is correct.
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Answer: B
Einstein: eV0=hc/λ−ϕ.
hc/λ−ϕ=eV and 3hc/(2λ)−ϕ=3eV.
Subtracting: hc/(2λ)=2eV, so eV=hc/(4λ) and ϕ=hc/λ−hc/(4λ)=3hc/(4λ).
Threshold wavelength λ0=hc/ϕ=4λ/3 (option B).
Check: at 2λ/3, 3hc/(2λ)−3hc/(4λ)=3hc/(4λ)=3eV.
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
A photocell is illuminated by monochromatic light of frequency ν (above the threshold frequency of the cathode) and fixed intensity (power per unit area) I. The saturation photocurrent is i0 and the stopping potential is V0. The frequency is now changed to 2ν while the intensity is kept equal to I and the illuminated area is unchanged. Assume that every incident photon has the same probability of ejecting an electron, independent of frequency, and that all emitted electrons are collected at saturation. The new saturation current and stopping potential are ______. Exactly one option is correct.
Ai0 and exactly 2V0
Bi0/2 and exactly 2V0
Ci0/2 and greater than 2V0
D2i0 and greater than 2V0
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Answer: C
Saturation current ∝ number of photons per second =IA/(hν). At fixed I, doubling ν halves the photon rate, so the saturation current becomes i0/2.
Stopping potential: eV0=hν−ϕ; new eV0′=2hν−ϕ=2(eV0+ϕ)−ϕ=2eV0+ϕ. Since ϕ>0, V0′>2V0.
Hence: i0/2 and a stopping potential greater than 2V0 (option C).
Trap: "doubling frequency doubles stopping potential" ignores the work function; "intensity fixed so current fixed" confuses energy flux with photon flux.
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
Monochromatic light falls on a metal of work function 2.0 eV. A narrow slit system selects only those photoelectrons whose velocities are perpendicular to a uniform magnetic field of magnitude B=2.0×10−4 T, in which they move on circles. The largest radius of these circles is 2.0 cm. Take e=1.6×10−19 C, me=9.1×10−31 kg, hc=1240 eVnm, and treat the electrons nonrelativistically. The wavelength of the light is ______. Exactly one option is correct.
A364 nm
B620 nm
C459 nm
D258 nm
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Answer: A
Largest radius belongs to the fastest electrons: r=mevmax/(eB), so vmax=eBr/me=9.1×10−311.6×10−19×2.0×10−4×0.020=7.03×105 m/s.
Kmax=21mevmax2=(eBr)2/(2me)=2.25×10−19 J=1.41 eV.
Photon energy =2.0+1.41=3.41 eV, so λ=1240/3.41=364 nm (option A).
Trap: 620 nm is just the threshold wavelength; using K=mev2 or 41mev2 gives the other options.
Difficulty: Very hard