Question Detail

PHY 102 Test Compilation PHY 102 14 Objective Question
Take CBT

Question

Find the potential difference required to give a helium nucleus (q = 3.2 x 10-19 C) whose kinetic energy is 4.8 × 103 eV.
Options
A 4.8 x 10<sup>3</sup> V
B 1.2 x 10<sup>3</sup> V
C 3.6 x 10<sup>3</sup> V
D 2.4 x 10<sup>3</sup> V
Correct Answer
Correct Answer

Option D is the correct answer.

Detailed Explanation

The kinetic energy gained by a charge moving through a potential difference is given by KE = qV, where KE is kinetic energy, q is the charge, and V is the potential difference. First, convert the kinetic energy from electronvolts (eV) to Joules (J) using 1 eV = 1.6 × 10-19 J. KE = (4.8 × 103 eV) × (1.6 × 10-19 J/eV) = 7.68 × 10-16 J. Then, calculate the potential difference: V = KE / q = (7.68 × 10-16 J) / (3.2 × 10-19 C) = 2.4 × 103 V.

Hint

Remember the relationship between kinetic energy, charge, and potential difference (KE = qV), and the conversion factor between eV and Joules.

Question Info

Course: PHY 102
Topic: Test Compilation PHY 102 14
Type: Objective
Difficulty: Not specified
Question 1018 of 1819
Previous Course Details Next