A)
B)
C)
D)
Correct Answer: A
Solution :
According to Richardson-Dushman equation \[J=A{{T}^{2}}{{e}^{-b/T}}\] Taking log of this equation \[{{\log }_{e}}\frac{J}{{{T}^{2}}}={{\log }_{e}}A-\frac{b}{T}\] i.e. graph between \[{{\log }_{e}}\frac{J}{{{T}^{2}}}\] and \[\frac{1}{T}\] will be a straight line having negative slope and positive intercept (logeA) on \[{{\log }_{e}}\frac{J}{{{T}^{2}}}\text{axis}\text{.}\]You need to login to perform this action.
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