Modern Physics for Engineers

Problem Set 6

Problem Set - Due in class, Wednesday, Nov. 3, 1999.

1. SM&M 8-5

transit time = .01 sec
deflection = 1/2 a t^2 = 1/2 F/m t^2 = .001 m
--< F = 2 .001 m 1.77e-25 kg / (.01 sec)^2 = 3.55e-24 N
dB/dx = F/mu_e = 3.55e-24 N/9.27e-24 J/T = .38 T/m

2. SM&M 8-21

A)	1s^2 2s^2 2p^4

B)    n    l   m_l   m_s
      1    0    0    +1/2
      1    0    0    -1/2
      2    0    0    +1/2
      2    0    0    -1/2
      2    1    1    +1/2
      2    1    1    -1/2
      2    1    0    +1/2
      2    1    -1   +1/2

Note: The fourth electron can go anywhere in the 2p shell with opposite spin to the electron already in that level. The first three have to go in different orbitals.

3. SM&M 10-8

Part A -- First calculate the center of mass with respect to the Cl atom:

	r_cm = (1 * .128 + 35 * 0)/(1 + 35) = .004 nm

Then calculate the moment of inertia:

	I = 1 * (.128 - .004)^2 + 35 * .004^2
	  = .016 u nm^2 = 2.66e-47 kg m^2

The the rotational levels are:

	E = hbar^2 l (l+1)/ (2 I) = 2.1e-22 Joules l (l+1)
	  = 1.3e-3 eV l (l+1)

Part B) When calculating the classical frequency of vibration, you must use the the reduced mass, mu=(m_1 m_2)/(m_1+m_2) = 35/36 amu ~ 1 amu.

	k = 2 U/x^2 = 2 x .15 eV/(.01 nm)^2 = 3e3 eV/nm^2
	k = mu omega^2 = mu (2 Pi f)^2
-->	f = Sqrt[k c^2/(mu c^2)]/(2 Pi) 
  = Sqrt[3e3 eV/(1e-9 m)^2 (3e8 m/sec)^2/(35/36 931e6 eV)]/(2 Pi)
	  = 8.7e13 Hz = 87 THz

Part C) the two lowest states are:

	E1 = h f/2 = 4.14e-15 eV sec * 8.7e13 /sec /2 = .18 eV
	E2 = 3 h f/2 = 3 E1 = .54 eV

Part D)
Vibrational transition = E2-E1 = .36 eV
Rotational = E(l=1) --> E(l=0) = (2 - 0) 1.3e-3 eV = 2.6e-3 eV

4. SM&M 10-13 Mathematica Solution This solution shows how to calculate the change in energy introduced By this potential. [Incidentially, I get + 7 (hbar w)^2/(64 U0) instead of the book's - (hbar w)^2/(16 U0).]
   Part A)
   

   	As r --> Infinity, U --> U_0 because Exp[- beta (r-R_0)] --> 0. 	dU/dr = 0 = 2 U_0 (1 - Exp[- beta(r-R_0)] beta Exp[-beta (r-R_0)]
   --> 	Exp[-beta ( r-R_0)] = 1 --> r = R_0 at equilibrium.
   

   Part B)
   Near equilibrium, K is the second derivative of the potential evaluated at the equilibrium distance. Thus

   	K = d^2 U/dr^2| r=R_0 = 2 beta^2 U_0 Exp[-beta (r - R_0)] 
   (1- Exp[-beta (r-R0)]|r = R_0
   	    = 2 beta^2 U_0
   

   Part C)

    
   E_dis = U0 - E_vib
         = U0 - 1/2 hbar w + (hbar w)^2/(16 U0)
   

   Part D)
   The reduced mass m for hydrogen is 1.67e-27 kg /2 = 8.4e-28 kg. Given K = 573 N/m --> w = 8.3e14 Hz and hbar w = 8.8e-20 joules = .54 eV.< Thus:

   	4.52 = U0 -.27 + (.54)^2/(16 U0)  (quadratic equation!)
   	     ~= U0 - .27
   	U0 ~= 4.79eV
   

5. SM&M 10-17 -- Hint: Use Psi[x] = Sin[k x] for x<L/2 and Psi[x] = +/- Sin[k (L-x)] for x>L/2. This form comes from the fact that Psi must be continuous at L/2.