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

Option: 1

1 \mathrm{~mA}


Option: 2

2 \mathrm{~mA}


Option: 3

4 \mathrm{~mA}


Option: 4

None of these


Answers (1)

best_answer

The emission current \mathrm{i=A T^2 S e^{-\phi / k T} }
For the two surfaces\mathrm{ A_1=A_2, S_1=S_2, T_1=800 \mathrm{~K}, T_2=1600 \mathrm{~K}, \phi_1 / T_1=\phi_2 / T_2 }
Therefore, \mathrm{\frac{i_2}{i_1}=\left(\frac{T_2}{T_1}\right)^2=(2)^2=4 \Rightarrow i_2=4 i_1=4 \mathrm{~mA} }

Posted by

vishal kumar

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