# The Constant a in the Richardson−Dushman Equation for Tungsten is 60 × 104 a M−2k−2. the Work Function of Tungsten is 4.5 Ev. - Physics

Sum

The constant A in the Richardson−Dushman equation for tungsten is 60 × 104 A m−2K−2. The work function of tungsten is 4.5 eV. A tungsten cathode with a surface area 2.0 × 10−5 m2 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−1. Assume that the thermions take only a small fraction of the heat supplied.

#### Solution

Given:-
A = 60 × 104 A m−2 K−2
Work function of tungsten, ϕ = 4.5 eV
Stefan's Constant, σ = 6 × 10−8 W m−2 K−1
Surface area of tungsten cathode, S = 2.0 × 10−5 m2
Boltzmann's Constant, k = 1.38× 10−23 J/K
Heat supplied by the heater, H = 24 W

The cathode acts as a black body; thus, its emissivity is 1.
According Stefan's Law:-
The power (P) radiated by a blackbody with surface area (A) and temperature (T),

$P = \sigma A T^4$

$\Rightarrow T^4 = \frac{P}{\sigma A}$

$\Rightarrow T^4 = \frac{24}{(6 \times {10}^{- 8} ) \times (2 \times {10}^{- 5} )}$

$\Rightarrow T^4 = 2 \times {10}^{13} K = 20 \times {10}^{12} K$

$\Rightarrow T = 2 . 1147 \times {10}^3 = 2114 . 7 K$

According to the Richard-Dushmann equation, emission current,

$i = AS T^2 e^{- \phi/KT}$

$i = 6 \times {10}^5 \times 2 \times {10}^{- 5} \times (2114 . 7 )^2 \times e^\frac{- 4 . 5 \times 1 . 6 \times {10}^{- 19}}{1 . 38 \times {10}^{- 23} \times 2114 . 7}$

$i = 1 . 03456 \times {10}^{- 3} A \approx 1 . 0\text{ mA}$

Concept: Electron Emission
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 19 Electric Current through Gases
Q 9 | Page 353