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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 237
�e values of k is therefore
k = p
ħ
= +0.01 kgms−1
1.0546 × 10−34 J s
= +9.5 × 1031 m−1
where 1 J = 1 kgm2 s−2 is used.�e wavefunction is then, with x is measured
in metres,
ψ(x) = eikx = e+i(9.5×10
31 m−1)x
E7D.3(b) �e energy levels of a particle in a box are given by [7D.6–263], En = n2h2/8mL2,
where n is the quantum number. With the mass equal to that of the electron
and the length as 1.5 nm, the energies are
En =
n2h2
8mL2
= n2 × (6.6261 × 10−34 J s)2
8 × (9.1094 × 10−31 kg) × (1.5 × 10−9 m)2
= n2 × (2.67... × 10−20 J)
To convert to kJmol−1, multiply through by Avogadro’s constant and divide by
1000.
= n2 × (2.67... × 10−20 J) × 6.0221 × 10
23mol−1
1000
= n2 × (16.1... kJmol−1)
To convert to electronvolts divide through by the elementary charge
= n2 × (2.67... × 10−20 J) × 1
1.6022 × 10−19 C
= n2 × (0.167... eV)
To convert to reciprocal centimetres, divide by hc, with c in cm s−1
= n2 × (2.67... × 10−20 J)
(6.6261 × 10−34 J s) × (2.9979 × 1010 cms−1)
= n2 × (1.34... × 103 cm−1)
�e energy separation between two levels with quantum numbers n1 and n2 is
∆E(n1 , n2) = En2 − En1 = (n22 − n21) × (2.67... × 10−20 J)
�e values in the other units are found by using the appropriate value of the
constant, computed above.
(i) ∆E(2, 3) = 1.3 × 10−19 J , 81 kJmol−1 , 0.84 eV , or 6.7 × 103 cm−1
(ii) ∆E(6, 7) = 3.5 × 10−19 J , 2.1 × 102 kJmol−1 , 2.2 eV ,
or 1.8 × 104 cm−1