merzbacher 3rd problems 1 and 2
TRANSCRIPT
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8/8/2019 Merzbacher 3rd Problems 1 and 2
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Exercise 1.1 : Calculate the quantized energy levels of a linear harmonic oscillator of angular frequency in the old quantum theory.
~ Using basic algebra and Plank's relation between energy and frequency we get:
E = h = h2
= 2 E = n E n=12
n
Potential Energy: V = k x2
2= m
2
x2
2x2= 2 V
k = 2 V
m 2x= 2 V m 2
Kinetic Energy: K = mv2
2
Classical
& p= m v T = p2
2 m
New
Total Energy: E = V T =m 2 x2
2p 2
2 mp2
2m= E
m 2 x2
2p2= 2 m E m2 2 x2
Solved for momentum: p = 2 m E m2 2 x2 & position: q= 2 V m 2 = 2m 2 m2 x2
2= x
Old Quantum Theory: p j d q j= n j h qq
p j d q j= qq
p j d q j q q p j d q j= n j h
if q is a distance traveled in a potential, x or r can be q because they are also distances.
2 qq
p j d q j = n j h = 2 qq 2 m E m2 2 x2 d x = 2
2 V m 2 2 V m 2 2 m E m2 2 x2 d x
Setting V = E 4 0 2 E m 2 2 m E m2 2 x2 d x can be pluged into my TI-89 to get: 2 E
2 E = n h E =n h2
= n
E n= 12n for n= 1,2,3,4,...
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8/8/2019 Merzbacher 3rd Problems 1 and 2
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Exercise 1.2 : Assuming that the electron moves in a circular orbit in a coulomb field, derive theBalmer
= E
1n 1
2 1n 2
2
formula for the spectrum of hydrogenic atoms from the quantum
condition (1.2) pi d qi= ni h and the Bohr formula (1.1) E = h .
~ Once more we use the same set of equations as (Exercise 1.1)
= h2
= 2 E = n circumference C = 2 r and angular momentum L= m v r
pi d q i= ni h p i= 2 mT = m v qi= distance on circumference from0 to 2 r
02
pi d q = n i h = pi r 02
d q= mv 02 r
d q = 2 r m v
Centripetal Force: F = ma F c= mv2
r equal to the force keeping it in (Bore's orbit) F c= F e
Electrical Force: F e= q E = ee
4 0 r 2 =
e2
r 21
4 0
e 2
r 21
4 0= m v
2
r e2
4 0= r m v 2
p= m v & k = 14 0
&e2
4 0= r m v 2 mk e 2= r p 2 r = m k e
2
p 2p= mk e2r
E = n & p = mv & = 2 f = vr
E = n vr
E v
= n
r p = 2 E
v= 2
n r
m k e 2= 4 r n r
2
mr k e2= 4 n 2 r = 4n 2
m k e2& p=
2 n r
= 2 n mk e2
2 n 2= mk e
2
2 n
Energy from Work : W = F d T = F e r =e 2
r 2r
4 0= e
2
4 0 r
Total Energy: E = T V = e2
4 0 r p2
2 m =mk e 2
2 n
21
2 m e2
k m k e2
4 n 2 =m k 2 e4
82 n 2 m k 2 e 4
4 2 n 2 = 18
mk 2 e4
2 n 2
The final solution is based on an intuitive understanding of proportions...
We know: E = n & E = 18
m k 2 e4
2 n 2
E 1n2
E n= E 11n 2
= E
1n 1
2 1n2
2