HW14%20solutions

# HW14%20solutions - Homework 14 solutions Physics 204...

This preview shows pages 1–2. Sign up to view the full content.

Homework 14 solutions Physics 204, morning summer 2008 Chapter 29 Problems 8,21,38,43 Chapter 30 Problems 18,15 8. REASONING AND SOLUTION The work function of the material (using λ = 196 nm) is found from W 0 = hf = hc / λ = 1.01 × 10 –18 J The maximum kinetic energy of the ejected electron is (using λ = 141 nm) KE max = hf W 0 = hc / λ W 0 = 3.96 × 10 –19 J The speed of the electron is then v m = = × × = × 2 2 3 96 10 9 32 10 19 5 KE J 9.11 10 kg m / s max 31 c h c h . . 21. REASONING AND SOLUTION The de Broglie wavelength λ is given by Equation 29.8 as λ = h / p , where p is the magnitude of the momentum of the particle. The magnitude of the momentum is p = mv , where m is the mass and v is the speed of the particle. Using this expression in Equation 29.8, we find that λ λ = = = × × × = × h mv v h m or 6.63 10 J s 1.67 10 kg 0.282 10 m 1 41 10 m s –34 –27 –9 3 c hc h . / 38. REASONING The energy of a photon of frequency f is, according to Equation 29.2, E hf = , where h is Planck's constant. Since the frequency and wavelength are related by f c = / λ (see Equation 16.1), the energy of a photon can be written in terms of the wavelength as E hc = / λ . These expressions can be solved for both the wavelength and the frequency.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern