MCQ
A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1 kV, its kinetic energy will be:
- A920 keV
- B$\frac{1}{1840} keV$
- C1 keV
- D1840 keV
✓
Answer
(c) 1 keV
Explanation: K.E. gained $= qV = e \times 1 kV =1 keV$
Explanation: K.E. gained $= qV = e \times 1 kV =1 keV$
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