Physics · Nuclear Physics
Work through this past-paper style MCQ, then read the full explanation. Practice more physics questions on mMCQ with adaptive practice and topic analytics.
The particle carrying a charge of (2e) falls through a potential difference of 3V. Energy required by the particle is:
- A
9.6 × 10^-19J
- B
1.6 ×10^-19 J
- C
3.2 ×10^-19 J
- D
6.9 × 10^-19 J
K.E = (1.6 ×10^-19) × (C×V) As K.E = (1.6×10^-19) (2×3) = 9.6 × 10^-19 J
A.9.6 × 10^-19 J: This is the correct answer, obtained by using the equation E = qV and substituting the values given in the question.
K.E = (1.6 ×10^-19) × (C×V) As K.E = (1.6×10^-19) (2×3) = 9.6 × 10^-19 J
A.9.6 × 10^-19 J: This is the correct answer, obtained by using the equation E = qV and substituting the values given in the question.
1.6 × 10^-19 J: This is the value of the charge of a single electron (e), not the energy required by the particle.
3.2 × 10^-19 J: This is twice the value of the charge of a single electron (2e), but does not take into account the potential difference through which the particle falls.
6.9 × 10^-19 J: This is a value between 3.2 × 10^-19 J and 9.6 × 10^-19 J, but is not the correct answer. The energy required by the particle is directly proportional to the potential difference, so a larger potential difference would result in a larger energy requirement.
Tagged under Physics · Nuclear Physics · 2009