課程名稱︰物理化學下
課程性質︰農化系必修
課程教師︰張大釗
開課學院：理學院
開課系所︰化學系
考試日期（年月日）︰2013.6.6
考試時限（分鐘）：50分鐘
是否需發放獎勵金：是
（如未明確表示，則不予發放）
試題 :
1 35 -1
1. Vibrational frequency of H Cl : 2991 cm ,
(a) Calculate the force constant,
(b) using the same force constant to calculate the frequency of
2 35 1/2
D Cl. [v=1/2π(k/u) ] (10)
19 -1 35 -1
2. Vibrational frequency of H F:4138 cm , H Cl: 2991 cm , and
80 -1
H Br:2649 cm ; What is your observation? And give a reason for your
observation.(12)
3. Figure 1 show two potential curves, which one is harmonic potential and
which one is anharmonic potential? Explain why the later one is more
realistic.(12)
4. Why we normally see one vibrational peak for each normal mode, but many
rotational peaks in the infrared spectra?(10)
5. Following question 4 , the selection rule is given as △n = + 1 for
vibrational transition and △J = J -J = ±1 for rotational
final initial
transition, use a drawing to explain when △J = + 1 (R branch) and
△J = -1 (P branch) in the infrared spectra.(10)
6. The rotational energy on the quantum number, J is given by
2 2 2 -10
E=J(J+1)(h /8π ur ). If r = 1.27 X 10 m for HCl, calculate △E
for J = 1.
01 02
7. Calculate the transition dipole moment, μ and μ ,
x x
mn ∞ *
μ =∫ Ψ (x)μ Ψ (x) dx , where μ =ex is charge.(10)
x -∞ m e n e
2
1/4 -(1/2)αx
Ψ (x)=(α/π) e
0
2
3 1/4 -(1/2)αx
Ψ (x)=(4α /π) xe
1
2
1/4 2 -(1/2)αx
Ψ (x)=(α/4π) (2αx - 1)e
2
8. Figure shows infrared spectra of gaseous CO and CH4. How many vivratioanl
modes for each of them? Give the reason why they are different trom the
spectra.(18)
9.List three major steps (equations) to obtain the Einstein correlation:
2 3 3
A /B = 16π (h/2π)v /c , where A is spontaneous emission and B
12 12 12 21
is spontaneous emission and B is the stimulated emission. (B = B )
21 12 21
(18)
10.Stimulated emission and population inversion are the two key concepts for
the development of laser. Based on the equation in question 9, under what
condition one has better chance to construct a laser.(10)
11.Beer-Lambert law is given as I(x)=I(0) exp(εMx) , where I(x)= intensity
of output light, I(0)= intensity of input light, x = path length, M is the
concentration of the absorber, and ε= molar absorption coefficient, For
a two level system, why I(x)/I(0) < 1? Under what condition, the
I(x)/I(0)>1 . In reality, how it works?(12)
I ┌┐ I(x)
0 ││
────→ ││────→
└┘
──→ｘ←──
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