MCQ
When a proton is accelerated with 1 volt potential difference, then its kinetic energy is
- A$\frac{1}{1840} \mathrm{eV}$
- B1840 eV
- ✓1 eV
- D$1840~c^2~\mathrm{eV}$
✓
Answer
Correct option: C.
1 eV
(c) 1 eV
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