1000 Solved Problems in Modern Physics

102 2 Quantum Mechanics – I

cPγ=Eγ= 67 .5MeV

λ=

h

p

= 2 π

c

cp

=

(2π)(197.3MeV.fm)

67 .5MeV

= 18 .36 fm= 1. 836 × 10 −^14 m.

2.2λ=

(

150

V

) 1 / 2

=

(

150

54

) 1 / 2

= 1. 667 A ̊

2.3λ=

h

p

=

2 πc

(2mc^2 T)^1 /^2

=(2π)×

197 .3MeV−fm

(2×939 T−MeV)^1 /^2

= 28. 6 × 10 −^5 A ̊/T^1 /^2

where T is in MeV. If T is in eV,λ= 0. 286 A ̊/T^1 /^2

2.4λ=

h

p

=

6. 63 × 10 −^34 J−s

(2× 9. 1 × 10 −^31 × 1. 6 × 10 −^19 )^1 /^2

= 12. 286 × 10 −^10 m/V^1 /^2 =

(

151

V

) 1 / 2

.

2.5 (a)

mnv^2

2

=

p^2

2 mn

=

3 kT

2

=

(

3

2

)

× 1. 38 × 10 −^23 ×

300

1. 6 × 10 −^19

= 0 .0388 eV

cC p=

(√

2 mnc^2 ·En

) 1 / 2

=(2× 940 × 106 × 0 .0388)^1 /^2 = 8 ,541 eV

λ=

h

p

=

hc

cp

= 3. 9 × 10 −^15 (eV−s)× 3 × 108 m−s−^1 / 8 ,541 eV

= 1. 37 × 10 −^10 m= 1. 37 A ̊

Such neutrons can be diffracted by crystals as their deBroglie wavelength

is comparable with the interatomic distance in the crystal.

(b)Δpx.Δx=

Put the uncertainty in momentum equal to the momentum itself,Δpx=p

cp=

c

Δx

=

197 .3MeV−fm

1 .0fm

= 197 .3MeV

E=(c^2 p^2 +m^2 c^4 )^1 /^2 =[(197.3)^2 +(940)^2 ]^1 /^2 = 960 .48 MeV

Kinetic energyT=E−mc^2 = 960. 5 − 940 = 20 .5MeV

2.6 E^2 =c^2 p^2 +m^2 c^4

^2 ω^2 =c^2 ^2 k^2 +m^2 c^4

ω=

(

c^2 k^2 +

m^2 c^4

^2

) 1 / 2