A particle carrying charge of 2e falls through a potential difference of 3.0 V. Calculate energy acquired by it.
Correct answer: A. 9.6 x 10^-19 J
- A. 9.6 x 10^-19 J
- B. 12.1 x 10^-19 J
- C. 14.5 x 10^-19 J
- D. 16.7 x 10^-19 J
- E. 18.5 x 10^-19 J
Explanation
To find the energy acquired by the particle, use the formula: Energy = qV, where q is the charge and V is the potential difference. The charge of the particle is 2e, where e is the elementary charge, approximately 1.6 × 10-19 C. The potential difference is given as 3.0 V. Thus, Energy = 2e × 3 V = 6eV. Converting eV to joules gives: Energy = 6 × 1.6 × 10-19 J = 9.6 × 10-19 J.Options B, C, D, and E are incorrect because they do not follow the correct calculation using the formula and given values.
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