A) 1540 eV
B) 13.6 eV
C) 1 eV
D) zero
Correct Answer: C
Solution :
When a proton is accelerated through IV, its kinetic energy \[K=qV\] For a proton, \[q=e=1.6\times {{10}^{-19}}\,C\] and V = 1 volt Thus, \[K=(1.6\times {{10}^{-19}}\times 1)\,J=1\,eV\] (as 1 eV = \[1.6\times {{10}^{-19}}\,J\])
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