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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 247
�e desired A′ appears only in (b) and (d). Solve these for A′ and set them
equal to each other.
A′ = e−ikW(CeκW − De−κW) = (κe−ikW/ik)(CeκW − De−κW)
Solve the resulting equation for C, and set it equal to the previously obtained
expression for C
C = (κ/ik + 1)De−2κW
κ/ik − 1
= (κ + ik)De−2κW
κ − ik
= 2Aik + D(κ − ik)
κ + ik
Solve this resulting equation for D in terms of A
(κ + ik)2e−2κW − (κ − ik)2
(κ − ik)(κ + ik)
D = 2Aik
κ + ik
and so
D = 2Aik(κ − ik)
(κ + ik)2e−2κL − (κ − ik)2
Substituting this expression back into the expression for C yields
C = 2Aik(κ + ik)e−2κW
(κ + ik)2e−2κW − (κ − ik)2
Substituting this into expression (b) for A′
A′ = 2Aike−ikW
(κ + ik)2e−2κL − (κ − ik)2
[(κ + ik)e−κW + (κ − ik)e−κL]
A′
A
= 4ikκe−ikWe−κW
(κ + ik)2e−2κL − (κ − ik)2
= 4ikκe−ikW
(κ + ik)2e−κL − (κ − ik)2eκW
�is leads to a transition probability of
T = ∣A
′
A
∣
2
= ( 4ikκe−ikW
(κ + ik)2e−κL − (κ − ik)2eκW )( −4ikκeikW
(κ − ik)2e−κL − (κ + ik)2eκW )
�e denominator, expanded separately is
(κ + ik)2(κ − ik)2e−2κW − (k − ik)4 − (k + ik)4 + (κ − ik)2(κ + ik)2e2κW
= (κ2 + k2)2(e2κW + e−2κW) − (κ2 − 2iκk − k2)2 − (κ2 + 2iκk − k2)2
= (κ4 + 2κ2k2 + k4)(e2κW + e−2κW) − (2κ4 − 12κ2k2 + 2k4)
�e term 12κ2k2 can be written as −4κ2k2 + 16k2κ2, which allows terms to be
collected
(κ4+2κ2k2+k4)(e2κW−2+e−2κW)+16κ2k2 = (κ2+k2)2(eκW−e−κW)2+16κ2k2
Hence, the probability is
T = 16k2κ2
(κ2 + k2)2(eκW − e−κW)2 + 16κ2k2