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,...

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