What is the potential difference needed to accelerate a He+ ion (charge +e, mass 4 u) from rest to a speed of 1.6×10^6 m/s?

• A. -5.344×10^4 V
• B. +5.344×10^4 V
• C. -1.0×10^3 V
• D. 0 V

Answer: (A)

The question tests how electric potential difference relates to kinetic energy for a charged particle. The work done by the electric field on a charge moving through a potential difference ΔV equals qΔV, and that work becomes the particle’s kinetic energy when starting from rest: (1/2) m v^2 = q ΔV. Since the ion is positively charged, reaching the given speed by moving through a potential difference indeed requires the end potential to be lower than the start, so ΔV is negative.

Compute the mass and energy: m = 4 u ≈ 4 × 1.6605×10^-27 kg ≈ 6.642×10^-27 kg. Speed v = 1.6×10^6 m/s, so v^2 = 2.56×10^12. Kinetic energy (non-relativistic) is (1/2) m v^2 ≈ 0.5 × 6.642×10^-27 × 2.56×10^12 ≈ 8.5×10^-15 J.

With q = e ≈ 1.602×10^-19 C, the required voltage magnitude is ΔV = KE / q ≈ (8.5×10^-15) / (1.602×10^-19) ≈ 5.3×10^4 V.

Because the potential difference must be negative to accelerate a positive ion in this setup, the answer is about -5.34×10^4 V.