102
2 Quantum Mechanics – I
cP γ = E γ = 67.5 MeV
λ =
h
p
= 2π
c
cp
=
(2π)(197.3 MeV.fm)
67.5 MeV
= 18.36 fm = 1.836 × 10
−14 m.
2.2 λ =
150
V
1/2
=
150
54
1/2
= 1.667 ˚
A
2.3 λ =
h
p
=
2πc
(2mc 2 T ) 1/2 = (2π) ×
197.3 MeV − 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
2.4 λ =
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)
m n v
2
2
=
p
2
2m n
=
3kT
2
=
3
2
× 1.38 × 10
−23
×
300
1.6 × 10 −19
= 0.0388 eV
cC p =
√ 2m n c
2
· E n
1/2 = (2 × 940 × 106 × 0.0388)
1/2
= 8,541 eV
λ =
h
p
=
hc
cp
= 3.9 × 10
−15 (eV − s) × 3 × 10
8 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) Δ p x .Δx =
Put the uncertainty in momentum equal to the momentum itself, Δ p x = p
cp =
c
Δx
=
197.3 MeV − fm
1.0 fm
= 197.3 MeV
E = (c
2 p
2
+ m
2 c
4 )
1/2
= [(197.3)
2
+ (940)
2 ]
1/2
= 960.48 MeV
Kinetic energy T = E − mc
2
= 960.5 − 940 = 20.5 MeV
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