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JEE Main & Advanced Physics Semiconducting Devices Question Bank Critical Thinking

  • question_answer
    A metallic surface with work function of 2 eV, on heating to a temperature of 800 K gives an emission current of 1 mA. If another metallic surface having the same surface area, same emission constant but work function 4 eV is heated to a temperature of 1600 K, then the emission current will be

    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 \[{{i}_{2}}=4{{i}_{1}}=4\,mA.\]


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