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  • For a thermionic emitter (metallic) if  represents the current density and <img alt="\mathrm{T}"

For a thermionic emitter (metallic) if \mathrm{J} represents the current density and \mathrm{T} is its absolute temperature then the correct curve between \mathrm{\log _e \frac{J}{T^2} \text { and } \frac{1}{T}} is

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)

best_answer

According to Richardson-Dushman equation
\mathrm{J=A T^2 e^{-b / T}}
Taking log of this equation
\mathrm{\log _e \frac{J}{T^2}=\log _e A-\frac{b}{T}}

\mathrm{i. e.,} graph between \mathrm{log_{e}\frac{J}{T^{2}}\ and\ \frac{1}{T}} will be a straight line having negative slope and positive intercept \mathrm{(log_{e}A)} on \mathrm{log_{e}\frac{J}{T^{2}}} axis
 

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shivangi.shekhar

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