If the temperature of a tungsten filament is raised from 2000 K to 2010 K, by what factor does the emission current change? Work function of tungsten is 4.5 eV.

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Solution

Given:
Work function of tungsten = 4.5 eV
Initial temperature of tungsten filament, T = 2000 K
Final temperature of tungsten filament, T' = 2010 K
Let the emission current at (T = 2000 K) be i.
Let the emission current at (T' = 2010 K) be i'.
The emission currents are given by
$i = A S T^{2} e^{- ϕ / k T}$
$i' = A S T'^{2} e^{- e / k T'}$
Dividing i by i', we get:
$\frac{i}{i'} = \frac{T^{2}}{T'^{2}} \frac{e^{- ϕ / k T'}}{e^{- ϕ / k T'}} \frac{i}{i'} = {(\frac{T}{T'})}^{2} e^{- ϕ / k T + ϕ / k T'} = {(\frac{T}{T'})}^{2} e^{\frac{ϕ}{k} (\frac{1}{T'} - \frac{1}{T})} \frac{i}{i'} = {(\frac{2000}{2010})}^{2} e^{\frac{4.5 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23}} (\frac{1}{2010} - \frac{1}{2000}) .} = \frac{40000}{(201)^{2}} e^{\frac{4.5 \times 1.6}{1.38} (- 0.0248)} = \frac{0.8786}{(201)^{2}} \times 40000 = 0.8699 \frac{i}{i'} = \frac{1}{0.8699} = 1.1495 = 1.14$

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