A) \[8\times {{10}^{5}}\,V/m;\text{ }0.8\times {{10}^{-}}^{14}J\]
B) \[4\times {{10}^{5}}\,V/m;\,\,1.6\times {{10}^{-}}^{14}J\]
C) \[8\times {{10}^{5}}\,V/m;\,\,3.2\times {{10}^{-}}^{14}J\]
D) \[4\times {{10}^{5}}\,V/m;\,\,0.8\times {{10}^{-}}^{14}J\]
Correct Answer: C
Solution :
The electric field is given by \[E=\frac{V}{d}\,=\,\frac{200\times {{10}^{3}}}{25\times {{10}^{-2}}}=8\times {{10}^{5}}\,V/m\] The energy imparted to the electrons as they move from cathode to the anode is \[=\,\,eE.d\] \[=\text{ }1.6\times {{10}^{-19}}\times 8\times {{10}^{5}}\times 25\times {{10}^{-}}^{2}\] \[=\,\,3.2\times {{10}^{-}}^{14}\,J\]
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