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PHY 102 Test Compilation PHY 102 14 Objective Question
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Question

Find the potential difference required to give a helium nucleus (q = 3.2 x 10-19 C) whose kinetic energy is 4.8 × 103eV where 1eV = 1.6 x 10-19J.
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 relationship between kinetic energy (KE), charge (q), and potential difference (V) is KE = qV.Given: Charge q = 3.2 × 10-19 C. Kinetic Energy KE = 4.8 × 103 eV.First, convert the kinetic energy from electronvolts (eV) to Joules (J) using 1 eV = 1.6 × 10-19 J:KEJ = (4.8 × 103) × (1.6 × 10-19) J = 7.68 × 10-16 J.Now, calculate the potential difference V:V = KEJ / q = (7.68 × 10-16 J) / (3.2 × 10-19 C) = 2.4 × 103 V.

Hint

Ensure consistent units; convert kinetic energy from eV to Joules before using KE = qV.

Question Info

Course: PHY 102
Topic: Test Compilation PHY 102 14
Type: Objective
Difficulty: Not specified
Question 1022 of 1819
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