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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 × 103 eV. (Given 1eV = 1.6 × 10-19 J).
Options
A 4.8 &times; 10<sup>3</sup> V
B 1.2 &times; 10<sup>3</sup> V
C 3.6 &times; 10<sup>3</sup> V
D 2.4 &times; 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. First, convert the kinetic energy from electron-volts to Joules:KE = 4.8 × 103 eV × (1.6 × 10-19 J/eV) = 7.68 × 10-16 J.Now, calculate the potential difference:V = KE / q = (7.68 × 10-16 J) / (3.2 × 10-19 C)V = (7.68 / 3.2) × 103 VV = 2.4 × 103 V.

Hint

Remember the conversion factor between eV and Joules, and the formula relating kinetic energy, charge, and potential difference.

Question Info

Course: PHY 102
Topic: Test Compilation PHY 102 14
Type: Objective
Difficulty: Not specified
Question 1011 of 1819
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