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.

Answer 1

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:-

\[\frac{i}{i'} = \frac{T^2}{T '^2}\frac{e^{- \phi/kT'}}{e^{- \phi/kT'}}\]

\[\frac{i}{i'} =  \left( \frac{T}{T'} \right)^2  e^{- \phi/kT + \phi/kT'} \]

\[ =  \left( \frac{T}{T'} \right)^2  e^\frac{\phi}{k}\left( \frac{1}{T'} - \frac{1}{T} \right) \]

\[\frac{i}{i'} =  \left( \frac{2000}{2010} \right)^2  e^\frac{4 . 5 \times 1 . 6 \times {10}^{- 19}}{1 . 38 \times {10}^{- 23}}\left( \frac{1}{2010} - \frac{1}{2000} \right) . \]

\[ = \frac{40000}{(201 )^2} e^\frac{4 . 5 \times 1 . 6}{1 . 38}( - 0 . 0248) \]

\[ = \frac{0 . 8786}{(201 )^2} \times 40000 = 0 . 8699\]

\[\frac{i}{i'} = \frac{1}{0 . 8699} = 1 . 1495 = 1 . 14\]