QUIZ 27/28 (TEST 4)  AND (TEST 3)

Chapter 27
CQ's: 12, 13, 14, 15, 16, 19, 22, 25, 26, 27
Problems -  Planck: 5, 6, 7; Einstein:  10, 11, 17, 18, 19, 20, 24, 32; Matter waves: 37, 38, 39, 40, 48, 49, 52, 54, 55
Chapter 28 Problems: 1, 2
Turn-in Chapter 27 -  6 (easy -see examples 27-1, 27-2), 10, 18, 20, 32 (easy-see example 27-9), 40, 48, 54
Turn-in Chapter 28 - Problem 2--- Scroll down to the very bottom for hint. 
6. See examples 27-1, 27-2.  Blackbodies are intriguing. The absorb all but reflect no radiation, and EMIT radiation at the same rate it's  absorbed if the body's  temperature is the same as the surroundings. Emission does not equal reflection. Reflection of incident radiation occurs on the body's surface. But emission  is radiation from  molecular vibrations inside the body.

Yo, see equation 14-6. This represents the NET rate of radiation leaving and entering the black body. Note  the negative term represents the radiation leaving the body. When the body (at T1) has  the same temperature as its surroundings (at T2), the net rate is zero.  In equation 14-5 for the emission rate T = T1, whether or not T1 = T2.  

Equation 14-5's  temperature T appears in  examples  27-1 and 27-2. 
7. Read section 27-3, especially page 758. f is given.  See equation 27-3. The difference in energy between adjacent energy levels is hf.
10.  Equation 27-4.
18.  Equation 27-5b. Set KE = 0 since at this wavelength and frequency the electron cannot escape because the photon is not providing enough energy. Solve for f and compute wavelength = c/f. Convert the work function from  eV to Joules before starting the problem.
20. Before starting, the "stopping potential" is just a fancy way of saying "kinetic energy" of the emitted electron freed by photon energy.  

SEE EXAMPLE 27-3 AND 27-4. 

(a) KE = hf - Wo.  Set KE and solve for Wo. The frequency is given.
(b)  KE = hf - Wo  . Solve for KE at the new frequency now that you have the work function from part (a).  
32. Easy-see example 27-9; the photon , which is pure energy and has no rest mass, decays into 2 particles with mass. In other words , some of the photon energy is converted into mass: hf = KE + 2moc2, where mo = electron rest mass 
(The electron and the positron have opposite charges but the *same* mass.)  Solve for KE the kinetic energy of the two particles.
40. Note that KE= = (1/2)*m*v2 = (1/2)*(m*v)2/m  = (1/2)*p2/m , where p = h/lambda, where lambda = wavelength.  
The two particles, proton and electron,  are assumed in this case to have the same wavelength, thus they have the same momentum p and thus the ratio of  kinetic energies is simply the inverse mass ratio:   (1/2)*p2/mp / (1/2)*p2/me  = me/mp.
48. For absorption, see example 27-14. For absorption h*f +  Eni = Enf    or h*f = Enf  - Eni , where nf > ni. In  example 27-14, nf = 3 and ni = 2.   For emission, see example 27-13:  Eni = Enf + h*f or h*f = Eni  - Enf   , where ni > nf    .  Thus,  we can categorize the following

(a) absorption
(b) emission
(c) absorption

Note the formula for hydrogen's energy = -13.6 eV/n2, where n = ni or nf  .  

(a) See example 27-14. But now, Eni = -13.6 eV/12 = -13.6 eV and Enf =  -13.6 eV/32  = -13.6 eV/9  = -1.51.  This part is exactly like example 27-14, so you can almost copy it if you wanted the wavelength. But they only want the difference in energy. Compute  h*f = Enf  - Eni . 
(b)  Compute h*f = Eni  - Enf     . 
(c) More hints soon.   Compute h*f = Enf  - Eni     .   
54. See example 27-15. The only difference between this and that problem is Z; Z = 2 in example 27-15 and Z = 1 for this problem.  Note you are  essentially computing h*f = Enf  - Eni    where nf = infinity and ni = 1; see example 27-15 part (a).  After that computation get the wavelength as shown in example 27-15 part (b). 
Chapter 28
2. This problem mimics the single slit diffraction experiments of Chapter 24, but now the waves are electrons'. See section 24.5 and also Quiz 24, problem 20. Just consider the width of the central maximum like you did in #20, Ch. 28. 
Use :

 0.010 m = Y1 - Y-1 = 2*L*lambda / a .  Find  L given a = 3.0 mm. Convert to meters (m). You can get lambda from equation 27-8. Remember p = m*v, where m and v are given. 

To be entirely accurate, assume the "slit"  is a hole. See page 712, section 25- 7. In that case, write,
 0.010 m =  2*1.22*L*lambda / D.  Find  L given D = 3.0 mm. Convert to meters (m). You can get lambda from equation 27-8. Remember p = m*v, where m and v are given.