Page 111 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:36 9in x 6in b985-ch08 Elementary Physical Chemistry

96 Elementary Physical Chemistry

Solution

2
2 −2
λ = h/p; E =1/2m ev ; p = (2m eE); J = kg m s

7
p = 2 × 9.11 × 10 −31 kg × 10 eV × 1.6 × 10 −19 J/eV
=1.707 × 10 −21 (kg J) = 1.707 × 10 −21 kg m s −1
1
λ =(6.626 × 10 −34 )J s /(1.707 × 10 −21 )kg m s −2
2 −2
=(6.626 × 10 −34 )kg m s s/(1.707 × 10 −21 )kg m s −1
=0.388 × 10 −12 m= 0.39 pm(pm = 10 −12 m)
Example 8.3. The uncertainty in speed of a particle weighing 1 g is
∆v =10 −6 ms −1 . What is the uncertainty in position?
Solution
2 −2
∆x∆p = h/4π =1.054 × 10 −34 kg m s s
2 −1
h/(4π∆p)= 5.267 × 10 −35 kg m s
∆x =
1.0 × 10 −3 kg × 1.0 × 10 −6 ms −1
−26
=5.3 × 10 m
3 1/2
Example 8.4. The ground-state wave-function of He + is ψ =(8/πa )
0
e −2r/a ,where r is the distance from the nucleus, and a 0 is the Bohr radius,
a 0 =52.9 pm. What is the probability of finding the electron within volume
3
element δv =1 pm ,
(a) around the nucleus;
(b) at a distance a 0 from the nucleus.
Solution

3
) δv.
The probability is ψψ δv =(8/πa )(e −2r/a 2
0
3
At nucleus r =0, Prob. =(8/πa ) δv
0
3
=(8/π)(52.9) −3 × (1) =1.72 × 10 −5
3
At r = a 0 , Prob. =(8/πa )(e −2a/a 2 −5 × e −4
) δv =1.72 × 10
0
=3.15 × 10 −7