1. (i) Find dispersion relation for the de Broglie wave corresponding to the relativistic
particle with the rest mass m0 and velocity v.
(ii) Find phase velocity of the wave and show that it is larger than the speed of
light
(iii) Find group velocity of the wave and show that it is equal to v.
Solution: (i) E 
m02 c 4  p 2 c 2 , E   , p  k =>   m02 c 4 /  2  k 2 c 2
(ii) v ph   / k  E / p , but p  Ev / c 2 hence v ph  c 2 / v  (c / v)c  c
d dE
2 pc 2
pc 2



(iii) v g 
, again we remind that
dk
dp 2 m02 c 4  p 2 c 2
E
pc 2 Evc2
 2 v
p  Ev / c and get v g 
E
c E
2. The main line of the Balmer series in emission of the atomic hydrogen, i. e. the line
corresponding to transition of the atom into the state with n = n1=2 from the nearest
state with n = n1+1=3, has a wavelength = 6.56 x 10-7m. In the emission of the
interstellar medium observed the series of emission of the same atomic hydrogen with
the wavelength of the main line = 48.8 cm. Find the corresponding quantum
number of the final state of the atom n2 (take into account that n2>>1).
2
Solution: Е ~ 1/n2 - 1/(n + 1)2 ~ 1/,
2/1 = [(n1)-2- (n1 + 1)-2]/[(n2)-2 - (n2 + 1)-2] ≈ (5/36)/[2n2/(n2)4];
n2 = [2 • 0.49 • 36)/(5 • 6,56 • 10-7)]1/3 ~ 220.
3. (i)Atomic hydrogen is illuminated by the ultraviolet radiation with wavelength
0
1000 A . Which lines will appear in the emission spectrum of hydrogen?
(ii) how many spectral lines emits the hydrogen atom excited to the level n?
3x1010 x6.626 x10 34
 12.42eV , h  Ry (1  1 / m 2 ) , hence
5
19
10 x1.6 x10
m  1 /(1  h / Ry )  3.4 , hence only m=2 and 3 can be excited. In this case the
lines 3->2, 3->1 and 2->1 will appear with the
0
0
ch
ch
wavelengths 32 
 6563 A , 21 
 1216 A ,
Ry (1 / 4  1 / 9)
Ry (1  1 / 4)
0
ch
31 
 1026 A
Ry (1  1 / 9)
(ii) there are n-1 transitions n->n-i (i=1, n-1); n-2 transitions from level n-2 …, 1
transition from level 2. Hence the total number of transitions N = 1+2+…+ (n-1) =
n(n-1)/2.
Solution: (i) h 
0
4. Excited atom emits photon during 0.01s . The wavelength is 6000 A . Find with
what accuracy can be determined the energy, wavelength and the location of the
photon.
Solution: E   / 2  3.6 x10 8 eV ,
 
E






E
=>
E

 


 10 4 A ;
hc 2 4c
2
2
0
E

, but p  E / c   / 2c => x  c  300cm
x 
2p
5. Inelastic collision of the hydrogen atom with another hydrogen atom at rest results
in photon emission by one of the atoms. What is the minimum kinetic energy of the
moving atom if both atoms are in the ground state before the collision?
Solution: minimum energy of excitation is Ry(1-1/22)=3/4Ry. Since Ry=13.6 eV is
much smaller than mc2 non relativistic consideration is valid.
From the momentum conservation we get mv  2mv1 => v1  v / 2 . The kinetic energy
changes by K  mv 2 / 2  2m(v / 2) 2 / 2  (1 / 2)mv 2 / 2  K / 2 , hence K/2>3/4Ry
=>K>3/2Ry.
6. Narrow beam of mono-energetic electrons falls on the surface of the monocrystal
aluminum by angle  = 300. The distance between the adjacent crystal planes parallel
to the surface is d=0.2 nm. The maximum of the mirror reflection was observed at the
accelerating voltage U0. Find U0 if the next maximum of the mirror reflection arouse
when the accelerating voltage was increased by =2.25.
Solution: 2d sin   n 
nh
2mU 0 e

(n  1)h
2mU 0e
. From the second equality it follows
that n  1 /(   1) . From the first equality we get
 2 2
U0 
 0.15keV
2med 2 sin 2  (   1) 2