p43 054

background image

54.

(a) We recall that mc

2

= 0.511 MeV from Table 38-3, and note that the result of problem 3 in

Chapter 39 can be written as hc = 1240 MeV

·fm. Using Eq. 38-51 and Eq. 39-13, we obtain

λ

=

h

p

=

hc

K

2

+ 2Kmc

2

=

1240 MeV

·fm



(1.0 MeV)

2

+ 2(1.0 MeV)(0.511 MeV)

= 9.0

× 10

2

fm .

(b) r = r

0

A

1/3

= (1.2 fm)(150)

1/3

= 6.4 fm.

(c) Since λ

 r the electron cannot be confined in the nuclide. We recall from Chapters 40 and 41,

that at least λ/2 was needed in any particular direction, to support a standing wave in an “infinite
well.” A finite well is able to support slightly less than λ/2 (as one can infer from the ground state
wavefunction in Fig. 40-8), but in the present case λ/r is far too big to be supported.

(d) A strong case can be made on the basis of the remarks in part (c), above.


Document Outline


Wyszukiwarka

Podobne podstrony:
p44 054
P23 054
p36 054
054 KP ka8
P17 054
p43 069
03 2005 053 054
p43 004
p43 002
p43 062
p41 054
p19 054
054, Sztuka celnego strzelania
02 2005 054 056
P27 054
P26 054
p43 051
KLIR biuletyn 054

więcej podobnych podstron