Physical Chemistry II â Exam 1 SOLUTIONS

More documents

Recommendations

Info

1.) continued2These frequency values may now be substituted, along with values of the kinetic energy, intothe equation for the slope,h = ΔKEΔν=1.362 ×10 −21 J – 2.960 ×10 −19 J5.552 ×10 14 s –1 – 9.993 ×10 14 s –1h = 6.634 ×10 −34 J s .€
2.) (17 points) Consider two operators,ˆ A = d /dx andˆ B = x. For the functionf (x) = x e −ax , where a is a constant, evaluate the following quantities.3a.)A ˆ B ˆ f ( x)€€A ˆ B ˆ f ( x) = d dx x ( x ) e−ax= d (dx x 2 e −ax)= 2x e −ax − a x 2 e −axA ˆ B ˆ f ( x) = ( 2 − ax) x e −ax .€b.) B ˆ A ˆ f ( x)( )( )B ˆ A ˆ f ( x) = x d dx x e−axB ˆ A ˆ f x= x e −ax − a x e −ax( ) = ( 1− ax) x e −ax .€c.) Do the operatorsˆ A andˆ B commute?(a yes or no answer is sufficient)NO. The operators€ €ˆ A andˆ B do not commute in this case.€€
Page 1: Chemistry 362Spring 2013Dr. Jean M.
Page 5 and 6: 3.) continued5Substituting this ene
Page 7 and 8: 4.) continued7b.) Verify whether or
Page 9 and 10: 6.) (16 points)9a.) For the particl

Physical Chemistry II â Exam 1 SOLUTIONS

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?

Physical Chemistry II â Exam 1 SOLUTIONS