1000 Solved Problems in Modern Physics

(Romina) #1

2.3 Solutions 109


Iron:EK= 13 .6(26−1)^2 = 8 ,500 eV
λK=

1 , 241

8 , 500

= 0 .146 nm= 1. 46 A ̊

2.25 The minimum wavelength of the photon will correspond to maximum fre-
quency which will be determined byE=hvmax


λmin=

c
vmax

=

hc
hvmax

=

hc
E

=

2 πc
E

=

2 π× 197 .3fm−MeV
30 × 10 −^3 MeV

= 4. 13 × 105 fm= 4. 13 A ̊

2.26λC=


hc
eV
h=

eVλC
c

=

1. 6 × 10 −^19 × 80 × 103 × 0. 15 × 10 −^10

3 × 108

= 6. 4 × 10 −^34 J−s

2.27λc=


hc
eV
h=

λceV
c

=

0. 247 × 10 −^10 × 1. 6 × 10 −^19 × 50 , 000

3 × 108

= 6. 59 × 10 −^34 J−s

2.28 According to Mosley’s law
1
λ


=A(Z−1)^2

1

λI

=A(26−1)^2

1

λCu

=A(29−1)^2

λCu
λI

=

252

282

= 0. 797 →λCu= 193 × 0. 797 = 153 .8pm

2.29 λK−λC=84 pm= 0. 84 A(1) ̊


1 , 200
(28−1)^2


12. 4

V

= 0. 84 (2)

whereλC=

hc
eV

=

12. 4

V

(V is in kV) (3)

Solving forVin (2),V= 15 .4kV

2.30 TheLαline is produced due to transitionn= 3 →n=2. For then=2 shell
the quantum numbers arel=0orl=1 andj=l±^12 , the energy states
being^2 S 1 / 2 ,^2 P 1 / 2 ,^2 P 3 / 2 .Forn=3 shell the energy states are^3 S 1 / 2 ,^3 P 1 / 2 ,


(^3) P 3 / 2 , (^3) d 3 / 2 , (^3) d 5 / 2
The allowed transitions are
(^3) S
1 / 2 →
(^2) P
1 / 2 ,
(^3) S
1 / 2 →
(^2) P
3 / 2
(^3) P
1 / 2 →
(^2) S
1 / 2 ,
(^3) P
3 / 2 →
(^2) S
1 / 2

Free download pdf