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