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Question
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.
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
\[i = AS T^2 e^{- \phi/kT}\]
\[i' = AST '^2 e^{- e/kT'}\]
Dividing i by i', we get:-
\[\frac{i}{i'} = \frac{T^2}{T '^2}\frac{e^{- \phi/kT'}}{e^{- \phi/kT'}}\]
\[\frac{i}{i'} = \left( \frac{T}{T'} \right)^2 e^{- \phi/kT + \phi/kT'} \]
\[ = \left( \frac{T}{T'} \right)^2 e^\frac{\phi}{k}\left( \frac{1}{T'} - \frac{1}{T} \right) \]
\[\frac{i}{i'} = \left( \frac{2000}{2010} \right)^2 e^\frac{4 . 5 \times 1 . 6 \times {10}^{- 19}}{1 . 38 \times {10}^{- 23}}\left( \frac{1}{2010} - \frac{1}{2000} \right) . \]
\[ = \frac{40000}{(201 )^2} e^\frac{4 . 5 \times 1 . 6}{1 . 38}( - 0 . 0248) \]
\[ = \frac{0 . 8786}{(201 )^2} \times 40000 = 0 . 8699\]
\[\frac{i}{i'} = \frac{1}{0 . 8699} = 1 . 1495 = 1 . 14\]
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