1. A surface of some metal was illuminated first by light with   0.35 and
then by light with   0.54  . It was found that maximum velocity of the
electrons differ by factor   2 . Find the work function of the metal.
Solution:
hc / 1  We
  2 => We  hc( 2 / 2  1 / 1 ) /( 2  1) = 1.9 eV
hc / 2  We
.2
‫ העלנו את‬. m1 ‫ ואורך גל בו העוצמה מקסימלית‬T1 K  ‫נתון גוף שחור עם טמפרטורה‬
‫ עקב כך אורך גל בו העוצמה מקסימלית קטן פי‬,‫ נוספות‬400K-‫הטמפרטורה של הגוף ב‬
.2
T1 , T2 , m1 , m 2 ‫ מצא את‬.‫א‬
.‫ מצא בכמה תשתנה הממוצע של הפוטונים במהוד‬.‫ב‬
‫פיתרון‬
.‫א‬
T2  T1  400  T2  400  T1
m 
hc
: ‫לפי‬
5kT
hc

5kT1 T2
T
2  m1 

 2  2  T2  2T1  400  T1  2T1  T1  400K 
hc
m 2
T1
T1
5kT2
T2  2T1  800k 
hc
6.626  10 34  3  10 8
m1 

 7.197  10 6  7197nm 
 23
5kT1 5  1.381  10  400
m 2
hc
6.626  10 34  3  10 8


 3.598  10 6  3598nm 
 23
5kT1 5  1.381  10  800
.‫ב‬
3
3
3
1
1
1
 kT 
 kT 
k
n  n2  n1  2.4   2   2 3  2.4   1   2 3  2.4     2 3  T23  T13  
  c
    c
    c
3
 1.381  10 23 
1
 
 2.4  
 800 3  400 3   20.5  10 6  448  10 6  918.4  1012
34 
2
8 3
1
.
05

10

  3  10 
3.
Amplitude of the electromagnetic wave with angular frequency
  3.6 x1015 s 1 is modulated with angular frequency   6 x1014 s 1 so that
electrical field E  a(1  cos t ) cos t . Find maximum kinetic energy
knocked out of the Li surface by this electromagnetic wave if the work
function of Li is We = 2.39 eV.
Solution: The wave is a superposition of the harmonic waves with
frequencies  and    , with photon energies   2.37eV , (   )  2.77eV
and (   )  1.98eV , respectively. Hence maximum kinetic energy
K= (   ) -We= 0.38 eV.
4. (a) Spectrum of the Sun radiation is close to that of the black body for which
the maximum emission is achieved at   0.48 . Find the mass lost by sun
per unite time due to this radiation. Find during what time the mass of the Sun
will decrease by 1% if mass and the radius of the sun are M=2x1030 kg and
R=6.95x108 m, respectively.
2.9 x10 3 mK
 6042 K , emitted mass per unit time
Solution: T 

M  4R 2T 4 /c2 = 5x109 kg/s. The time for which M decreases by 1% is
0.01M / M  1011 years.
(b) assuming that absorption efficiencies of the Sun and Earth are equal to 1
estimate the steady state temperature of the earth. The distance to the Sun is
Re-s=1.5x1011 m.
2
2
R
4
2 rearth / Re  s
2
 Te4 4rearth
Solution: Ts 4R
, hence Te  Ts
=291 K
4
2 Re s