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Chapter 2 Solutions
Prob. 2.
(a&b) Sketch a vacuum tube device. Graph photocurrent I versus retarding voltage V for
several light intensities.
Note that Vo remains same for all intensities.
(c) Find retarding potential.
o
λ=2440Å=0.244μm =4.09eV
1.24eV μm 1.24eV μm V = hν - Φ = - = - 4.09eV = 5.08eV - 4.09eV 1eV λ(μm) 0.244μm
Prob. 2.
Show third Bohr postulate equates to integer number of DeBroglie waves fitting within
circumference of a Bohr circular orbit.
2 2 2 2 o n (^2 ) o 2 2 2 2 2 2 2 2 2 o o n n n (^2 2 2 2 2 2 ) B n n 2 2 2 2 2 n
n
4 n q mv r = and = and p = mvr mq 4 π r r
4 n n 4 π r n r n r = = = = mq mr q mr mv m v r
m v r = n
mvr = n
p = n is the third Bohr postulate
I
V V o
light
intensity
I
V V o
light
intensity
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Prob. 2.
(a) Find generic equation for Lyman, Balmer, and Paschen series.
4 4
2 2 2 2 2 2 2 2 o 1 o 2 4 2 2 4 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 o 1 2 o 1 2 2 2 2 2 2 3 2 2 o 1 2 o 1 2 4 2 2 4 2 2 2 1 2 1 -12 (^) F 2 m
c mq mq ΔE = = - λ 32π n 32π n
c mq (n - n ) mq (n - n ) = = λ 32 n n π 8 n n
8 n n c 8 ε c (^) n n = = mq (n - n ) mq n - n
h
h
h
h h h
(^34 3 8) m 2 2 s 1 2 -31 -19 4 2 2 2 1 2 2 2 2 8 1 2 1 2 2 2 2 2 2 1 2 1
1
10 Js) 2.998 10 (^) n n
9.11 10 kg (1.60 10 C) n - n
n n n n = 9.11 10 m = 9. n - n n - n
n =1 for Lyman, 2 for Balmer, and 3 for Paschen
Å
(b) Plot wavelength versus n for Lyman, Balmer, and Paschen series.
n n^2 n^2-1 n^2/(n^2-1) 911n^2/(n^2-1) n n^2 n^2-9 9n^2/(n^2-9) 9119n^2/(n^2-9) 2 4 3 1.33 1215 4 16 7 20.57 18741 3 9 8 1.13 1025 5 25 16 14.06 12811 4 16 15 1.07 972 6 36 27 12.00 10932 5 25 24 1.04 949 7 49 40 11.03 10044 8 64 55 10.47 9541 9 81 72 10.13 9224 10 100 91 9.89 9010
n n^2 n^2-4 4n^2/(n^2-4) 9114n^2/(n^2-4) 3 9 5 7.20 6559 4 16 12 5.33 4859 5 25 21 4.76 4338 6 36 32 4.50 4100 7 49 45 4.36 3968
BALMER LIMIT 3644Ǻ
PASCHEN LIMIT 8199Ǻ
BALMER SERIES
LYMAN SERIES PASCHEN SERIES
LYMAN LIMIT 911Ǻ
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A) For a wavefunction (x), we know
Ρ = (x) (x)dx = 1
0 c = 0 Ρ = (x) (x)dx = c dx Ρ= (x) cannot be a wave function c 0
^
B) For 5 x 8 , (^) (x)has two values, C and 3.5C. For c 0 , (^) (x)is not a function
and for c = 0:
Ρ = (x) (x)dx = 0 (x)
cannot be a wave function.
C)
iC x-2 2 x 5 3 (x)= iC x-10 5 x 10 5
(^5 2 ) *^2
c c Ρ = (x) (x)dx = x-2 dx + x-10 dx 9 25
2 2 3 5 3 10 2 5
2 2
c c = (x-2) + (x-10) 3×9 3×
27 125 8c = c + = 27 3×25 3
2 8c Ρ = 1 =1 c=0.612 (x) can be a wave function 3
Since (x) = 0for (^) x 2 and x 10 , the potential energy should be infinite in these two
regions.
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Prob. 2.
A particle is described in 1D by a wavefunction:
Ψ = Be
-2x for x ≥0 and Ce
+4x for x<0, and B and C are real constants. Calculate B and C to make
Ψ a valid wavefunction. Where is the particle most likely to be?
A valid wavefunction must be continuous, and normalized.
For(0) = C = B
To normalize ,
2
dx = 1
0 2 8x 2 -4x
C e dx + C e dx = 1
2 0 8x 2 -4x 0
C 1
e + C e 1 8 4
^
2 2 C C 8
Prob. 2.
The electron wavefunction is Ce
ikx between x=2 and 22 cm, and zero everywhere else. What is
the value of C? What is the probability of finding the electron between x=0 and 4 cm?
ikx = Ce
22
2
dx = C (20) = 1 C = cm 20
4 2 2
0
Probability = dx = 2 = 20 10
Prob. 2.
Find the probability of finding an electron at x<0. Is the probability of finding an electron at
x>0 zero or non-zero? Is the classical probability of finding an electron at x>6 zero or non?
The energy barrier at x=0 is infinite; so, there is zero probability of finding an electron at
x<0 (|ψ|
2 =0). However, it is possible for electrons to tunnel through the barrier at 5<x<6;
so, the probability of finding an electron at x>6 would be quantum mechanically greater
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Prob. 2.
Find the uncertainty in position (Δx) and momentum (Δρ).
L -2πjEt/ *
0 L L
0 0 L L 2 * 2 2 2
0 0
2 πx Ψ(x,t) = sin e and dx = 1 L L
2 x x = x dx = x sin dx = 0.5L (from problem note) L L
2 x x = x dx = x sin dx = 0.28L (from problem note) L L
Δx = x
h
2 2 2 2
- x = 0.28L - (0.5L) = 0.17L
p = 0. 4 π Δx L
h h
Prob. 2.
Calculate the first three energy levels for a 10Ǻ quantum well with infinite walls.
2 2 2 -34 2 2 -20 2 n (^2 31 9 )
20 1
2
3
n π (6.63 10 ) E = = n = 6.03 10 n 2 m L 8 9.11 10 (10 )
E = 6.03 10 J = 0.377eV
E = 4 0.377eV = 1.508eV
E = 9 0.377eV = 3.393eV
Prob. 2.
Show schematic of atom with 1s
2 2s
2 2p
4 and atomic weight 21. Comment on its reactivity.
This atom is chemically reactive because
the outer 2p shell is not full. It will tend
to try to add two electrons to that outer
shell.
= proton
= neturon
= electron
= proton
= neturon
= electron
nucleus with 8 protons and 13 neutrons
2 electrons in 1s
2 electrons in 2s
4 electrons in 2p
= proton
= neturon
= electron
= proton
= neturon
= electron
nucleus with 8 protons and 13 neutrons
2 electrons in 1s
2 electrons in 2s
4 electrons in 2p
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