Question Detail
Question
Find the potential difference required to give a helium nucleus (charge q = 3.2 x 10⁻¹⁹ C) a kinetic energy of 4.8 × 10³ eV, where 1 eV = 1.6 × 10⁻¹⁹ J.
Correct Answer
Option D is the correct answer.
Detailed Explanation
The relationship between kinetic energy (KE), charge (q), and potential difference (V) is KE = qV.Given:KE = 4.8 × 10³ eVq = 3.2 × 10⁻¹⁹ C1 eV = 1.6 × 10⁻¹⁹ JFirst, convert KE from eV to Joules:KE_J = 4.8 × 10³ eV × (1.6 × 10⁻¹⁹ J/eV) = 7.68 × 10⁻¹⁶ JNow, calculate the potential difference V:V = KE_J / qV = (7.68 × 10⁻¹⁶ J) / (3.2 × 10⁻¹⁹ C)V = (7.68 / 3.2) × 10³ VV = 2.4 × 10³ V.
Hint
The kinetic energy gained by a charge moving through a potential difference is given by KE = qV. Remember to convert eV to Joules.