A) 1 mA
B) 2 mA
C) 4 mA
D) None of these
Correct Answer: C
Solution :
The emission current \[i=A{{T}^{2}}S{{e}^{-\varphi /kT}}\] For the two surfaces A1 = A2, S1 = S2, T1 = 800 K, \[{{T}_{2}}=1600\,K,\,{{\varphi }_{1}}/{{T}_{1}}={{\varphi }_{2}}/{{T}_{2}}\] Therefore, \[\frac{{{i}_{2}}}{{{i}_{1}}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{2}}\]= (2)2 = 4 Þ \[{{i}_{2}}=4{{i}_{1}}=4\,mA.\]You need to login to perform this action.
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