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QPages 352-354: Q9: HC Verma: Concepts of Physics - Volume 2 Electric Current through Gases Pages 352-354: Q9 Solution: The constant A in the Richardson-Dushman equation for tungsten…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently substitute the work function value directly in electron-volts (eV) into the exponential term $e^{-\phi/kT}$ without converting it to Joules, leading to a massive numerical error. Always convert $\phi$ to Joules by multiplying by $1.6 \times 10^{-19}$ to match the units of Boltzmann's constant $k$.

The constant A in the Richardson-Dushman equation for tungsten is 60×104 A m−2 K−260 \times 10^4 \text{ A m}^{-2} \text{ K}^{-2}. The work function of tungsten is 4.5 eV. A tungsten cathode with a surface area 2.0×10−5 m22.0 \times 10^{-5} \text{ m}^2 is heated by a 24 W electric heater. In steady state, the heat radiated by the heater and the cathode equals the energy input by the heater and the temperature becomes constant. Assuming that the cathode radiates like a blackbody, calculate the saturation current due to thermions. Take Stefan's Constant = 6×10−8 W m−2 K−46 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}. Assume that the thermions take only a small fraction of the heat supplied.

Step-by-Step Verified Solution

**Answer:** Saturation current i=1 mAi = 1 \text{ mA}. **Step-by-Step Explanation:** Given: A=60×104 A m−2 K−2A = 60 \times 10^4 \text{ A m}^{-2} \text{ K}^{-2}, ϕ=4.5\phi = 4.5 eV, S=2.0×10−5 m2S = 2.0 \times 10^{-5} \text{ m}^2, H=24H = 24 W, σ=6×10−8\sigma = 6 \times 10^{-8}. Stefan's law: P=σST4⇒24=6×10−8×2.0×10−5×T4P = \sigma S T^4 \Rightarrow 24 = 6 \times 10^{-8} \times 2.0 \times 10^{-5} \times T^4. T4=2×1013⇒T=2114.7T^4 = 2 \times 10^{13} \Rightarrow T = 2114.7 K. i=AST2e−ϕ/kT=60×104×2×10−5×(2114.7)2×e−4.5×1.6×10−19/(1.38×10−23×2114.7)i = AST^2 e^{-\phi/kT} = 60 \times 10^4 \times 2 \times 10^{-5} \times (2114.7)^2 \times e^{-4.5 \times 1.6 \times 10^{-19}/(1.38 \times 10^{-23} \times 2114.7)} i=1.03×10−3 A≈1 mAi = 1.03 \times 10^{-3} \text{ A} \approx 1 \text{ mA} .

All Questions in this Chapter

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