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:
  • A
    920 keV
  • B
    $\frac{1}{1840} keV$
  • C
    1 keV
  • D
    1840 keV

Answer

(c) 1 keV
Explanation: K.E. gained $= qV = e \times 1 kV =1 keV$

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