Early photoelectric light detectors were based on cesium. Cesium metal has a work function of 3.43 *10^-¹⁹ J. What is the minimum frequency of electromagnetic radiation required to eject photoelectrons from a cesium surface? A) 1.75 *10¹⁴/s B) 3.15 * 10¹⁴/s C) 5.17 * 10¹⁴/s D) 1.55 * 10¹³/s E) 7.24 * 10¹⁵/s

Question

Quizplus · Accepted Answer

The minimum frequency of electromagnetic radiation required to eject photoelectrons from a metal surface is given by the equation: $f_{min}=\frac{W}{h}$ where $W$ is the work function of the metal and $h$ is Planck's constant. Substituting the given values, we get: $f_{min}=\frac{3.43 imes10^{-19} ext{ J}}{6.626 imes10^{-34} ext{ J}\cdot ext{s}}\approx5.17 imes10^{14} ext{ Hz}$ Converting to F1F1F1S1F1F1F10 10^-¹⁹/s, we get: $f_{min}\approx5.17 imes10^{14} ext{ Hz}=5.17 ext{ PHz}$. Therefore, the best choice is C.

Early Photoelectric Light Detectors Were Based on Cesium