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Kmax = (6.63x1034 J.s) (7.09x10¹4 s¹)-(2.17x10 J) h. y = 2.25×108 m/s 3.00×108 m/s 1.

Kmax = (6.63x1034 J.s) (7.09x10¹4 s¹)-(2.17x10 J) h. y = 2.25×108 m/s 3.00×108 m/s 1.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Here is a transcription of the provided text, formatted for an educational website: --- ### Physics Calculations **Problem g:** Maximum Kinetic Energy Calculation Given: \[ K_{\text{max}} = (6.63 \times 10^{-34} \, \text{J.s}) (7.09 \times 10^{14} \, \text{s}^{-1}) - (2.17 \times 10^{-19} \, \text{J}) \] **Problem h:** Lorentz Factor Calculation Given: \[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{2.25 \times 10^{8} \, \text{m/s}}{3.00 \times 10^{8} \, \text{m/s}}\right)^2}} \] **Diagrams and Equations Explained:** For Problem g, the formula used to compute the maximum kinetic energy (\( K_{\text{max}} \)) is based on the photoelectric effect equation. Here, \(K_{\text{max}}\) is determined by subtracting the work function (in Joules) from the product of Planck’s constant (\(6.63 \times 10^{-34} \, \text{J.s}\)) and the frequency (\(7.09 \times 10^{14} \, \text{s}^{-1}\)). For Problem h, the equation calculates the Lorentz factor (\(\gamma\)) necessary for understanding relativistic effects as objects approach the speed of light. The formula includes the velocity of the object (\(2.25 \times 10^{8} \, \text{m/s}\)) divided by the speed of light (\(3.00 \times 10^{8} \, \text{m/s}\)), and ensures the calculation is done using the square root of one minus the squared velocity term. These equations are essential for studying modern physics and understanding fundamental principles like the photoelectric effect and special relativity. ---
Transcribed Image Text:Here is a transcription of the provided text, formatted for an educational website: --- ### Physics Calculations **Problem g:** Maximum Kinetic Energy Calculation Given: \[ K_{\text{max}} = (6.63 \times 10^{-34} \, \text{J.s}) (7.09 \times 10^{14} \, \text{s}^{-1}) - (2.17 \times 10^{-19} \, \text{J}) \] **Problem h:** Lorentz Factor Calculation Given: \[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{2.25 \times 10^{8} \, \text{m/s}}{3.00 \times 10^{8} \, \text{m/s}}\right)^2}} \] **Diagrams and Equations Explained:** For Problem g, the formula used to compute the maximum kinetic energy (\( K_{\text{max}} \)) is based on the photoelectric effect equation. Here, \(K_{\text{max}}\) is determined by subtracting the work function (in Joules) from the product of Planck’s constant (\(6.63 \times 10^{-34} \, \text{J.s}\)) and the frequency (\(7.09 \times 10^{14} \, \text{s}^{-1}\)). For Problem h, the equation calculates the Lorentz factor (\(\gamma\)) necessary for understanding relativistic effects as objects approach the speed of light. The formula includes the velocity of the object (\(2.25 \times 10^{8} \, \text{m/s}\)) divided by the speed of light (\(3.00 \times 10^{8} \, \text{m/s}\)), and ensures the calculation is done using the square root of one minus the squared velocity term. These equations are essential for studying modern physics and understanding fundamental principles like the photoelectric effect and special relativity. ---
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