merzbacher 3rd problems 1 and 2

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  • 8/8/2019 Merzbacher 3rd Problems 1 and 2

    1/2

    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

    2/2

    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