A) 13.6 eV
B) 27.2 eV
C) 14.4 eV
D) 1.44eV
E) 28.8 eV
Correct Answer: C
Solution :
\[U=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{1}}}{r}\] \[\therefore \]\[U=\frac{9\times {{10}^{9}}\times (1.6\times {{10}^{-19}})(-1.6\times {{10}^{-19}})}{{{10}^{-10}}}J\] \[=-9\times {{10}^{9}}\times 1.6\times {{10}^{-19}}\times {{10}^{10}}eV\] \[=-14.4eV\]
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