MCAT Physics Question 121: Answer and Explanation

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Question: 121

1. If the work function of a metal is 6.622 × 10^–20 J and a ray of electromagnetic radiation with a frequency of 1.0 × 10¹⁴ Hz is incident on the metal, what will be the speed of the electrons ejected from the metal? (Note: h = 6.626 × 10^-34 J·s and m_e^- = 9.1 × 10^-31 kg)

Correct Answer: C

Explanation:

To determine the speed of the electrons ejected, we must first calculate their kinetic energy:

Using the formula for the kinetic energy, we can now calculate the speed of the ejected electrons:

Notice the wide range in the exponents for the answer choices. While the math in this question may seem complex, this allows us to round significantly.

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