Determining Planck's Constant From Photoelectric Data
In a photoelectric experiment the stopping potential rises from 0.50 V to 2.15 V as the frequency of incident light is increased from 6.0×1014 Hz to 1.0×1015 Hz. What value of Planck's constant do these data yield?
Select the correct option:
Solution
6.6×10−34 J s
Since V0=ehν−eϕ0 is linear in frequency, the slope of V0 versus ν equals eh, and the unknown work function cancels when we take differences between two data points. The change in stopping potential is ΔV0=2.15−0.50=1.65 V, while the change in frequency is Δν=(1.0×1015)−(6.0×1014)=4.0×1014 Hz. The slope is therefore ΔνΔV0=4.0×10141.65=4.125×10−15 V s. Multiplying by the electronic charge gives h=ΔνΔV0×e=4.125×10−15×1.6×10−19=6.6×10−34 J s. The value 3.3×10−34 would result from halving the slope. The value 1.6×10−19 is the electronic charge, not Planck's constant. The value 9.1×10−31 is the electron mass, an unrelated constant. This difference method, which eliminates the work function entirely, is exactly how Millikan extracted h from photoelectric measurements, as recounted in NCERT. The recovered value matches the accepted Planck constant, confirming the data and the linear photoelectric model are mutually consistent. A valuable feature of this difference method is that it is completely insensitive to the particular metal used, because the work function affects only the intercept and is differenced away. This robustness is why the photoelectric determination of Planck's constant is considered one of the cleanest in physics, requiring only accurate stopping-potential and frequency measurements rather than any knowledge of the cathode material.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- physics
- Chapter
- dual nature of radiation and matter
- Topic
- determining planck's constant from photoelectric data
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
6.6×10−34 J s
Since V0=ehν−eϕ0 is linear in frequency, the slope of V0 versus ν equals eh, and the unknown work function cancels when we take differences between two data points. The change in stopping potential is ΔV0=2.15−0.50=1.65 V, while the change in frequency is Δν=(1.0×1015)−(6.0×1014)=4.0×1014 Hz. The slope is therefore ΔνΔV0=4.0×10141.65=4.125×10−15 V s. Multiplying by the electronic charge gives h=ΔνΔV0×e=4.125×10−15×1.6×10−19=6.6×10−34 J s. The value 3.3×10−34 would result from halving the slope. The value 1.6×10−19 is the electronic charge, not Planck's constant. The value 9.1×10−31 is the electron mass, an unrelated constant. This difference method, which eliminates the work function entirely, is exactly how Millikan extracted h from photoelectric measurements, as recounted in NCERT. The recovered value matches the accepted Planck constant, confirming the data and the linear photoelectric model are mutually consistent. A valuable feature of this difference method is that it is completely insensitive to the particular metal used, because the work function affects only the intercept and is differenced away. This robustness is why the photoelectric determination of Planck's constant is considered one of the cleanest in physics, requiring only accurate stopping-potential and frequency measurements rather than any knowledge of the cathode material.
This hard difficulty physics question is from the chapter dual nature of radiation and matter, covering the topic of determining planck's constant from photoelectric data. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse dual nature of radiation and matter questions on RankGuru.