416
7 Nuclear Physics – I
7.80 For electron z = −1. The given integral is easily evaluated assuming that this
refractive index μ is independent of frequency. Integrating between the limits
ν 1 and ν 2 .
−dW/dl = (4 π
2 e
2
/c
2 )(1 − 1/β
2
μ
2 )[(ν 2
2
− ν 1
2 )/2]
(1)
Calling the average photon frequency as
ν = 1 / 2 (ν 1 + ν 2 )
( 2 )
the average number of photons emitted per second is given by
N = 1/ h ¯
v = (4 π
2 e
2
/ hc)(1 − 1/β
2
μ
2 )(ν 2 − ν 1 )
= (2 π /137)(1 − 1/ β
2
μ
2 )(1/ λ 2 − 1/ λ 1 )
( 3 )
where λ 1 = c/ν 1 and λ 2 = c/ν 2 are the vacuum wavelengths and μ is the
average index of refraction over the wavelength interval λ 2 = 4,000 ˚
A to
λ 1 = 8,000 ˚
A. Substituting β μ = 0.9 × 1.33 = 1.197,
λ 1 = 8×10
−5 cm and λ 2 = 4×10
−5 cm, in (3) we find the number of photons
emitted per cm, N = 173.
7.3.8 Nuclear Resonance
7.81 The condition for resonance fluorescence is
ΔE γ /E γ = v/c
Put Δ E γ = Γ
Γ = vE γ /c = (1 cm/s/3 × 10
10 cm/s) × (129 keV) = 4.3 × 10
−6 eV
The mean lifetime
τ = /Δ E γ = /Γ = 1.05×10
−34
/1.6×10
−19
×4.3×10
−6
= 1.5×10
−10 s
7.82 Energy conservation gives
h ν = ΔW − E R
(1)
Momentum conservation gives
P R = h ν /c
(2)
Therefore E R = P R
2
/2M = (h ν)
2
/2Mc
2
(3)
Eliminating E R between (1) and (3), we get a quadratic equation in hν.
(h ν)
2
/2Mc
2
+ h ν − ΔW = 0
Which has the solution
h ν = −Mc
2
+ Mc
2 (1 + 2ΔW/Mc
2 )
1/2
Expanding the radical binomially and retaining up to second power of ΔW ,
and simplifying
h ν = ΔW (1 − ΔW/Mc
2 )
ν = (ΔW/ h)(1 − ΔW/Mc
2 )
7.83 The root mean square velocity of
137 Cs atoms
√
>= v = (3kT /m)
1/2
Substituting k = 1.38 × 10
−23 J K
−1
= 0.8625 × 10
−10 MeV K
−1
T = 288 K, mc
2
= 137 × 931.5 = 1.276 × 10
5 , we find v/c = 7.64 × 10
−7
7 Nuclear Physics – I
7.80 For electron z = −1. The given integral is easily evaluated assuming that this
refractive index μ is independent of frequency. Integrating between the limits
ν 1 and ν 2 .
−dW/dl = (4 π
2 e
2
/c
2 )(1 − 1/β
2
μ
2 )[(ν 2
2
− ν 1
2 )/2]
(1)
Calling the average photon frequency as
ν = 1 / 2 (ν 1 + ν 2 )
( 2 )
the average number of photons emitted per second is given by
N = 1/ h ¯
v = (4 π
2 e
2
/ hc)(1 − 1/β
2
μ
2 )(ν 2 − ν 1 )
= (2 π /137)(1 − 1/ β
2
μ
2 )(1/ λ 2 − 1/ λ 1 )
( 3 )
where λ 1 = c/ν 1 and λ 2 = c/ν 2 are the vacuum wavelengths and μ is the
average index of refraction over the wavelength interval λ 2 = 4,000 ˚
A to
λ 1 = 8,000 ˚
A. Substituting β μ = 0.9 × 1.33 = 1.197,
λ 1 = 8×10
−5 cm and λ 2 = 4×10
−5 cm, in (3) we find the number of photons
emitted per cm, N = 173.
7.3.8 Nuclear Resonance
7.81 The condition for resonance fluorescence is
ΔE γ /E γ = v/c
Put Δ E γ = Γ
Γ = vE γ /c = (1 cm/s/3 × 10
10 cm/s) × (129 keV) = 4.3 × 10
−6 eV
The mean lifetime
τ = /Δ E γ = /Γ = 1.05×10
−34
/1.6×10
−19
×4.3×10
−6
= 1.5×10
−10 s
7.82 Energy conservation gives
h ν = ΔW − E R
(1)
Momentum conservation gives
P R = h ν /c
(2)
Therefore E R = P R
2
/2M = (h ν)
2
/2Mc
2
(3)
Eliminating E R between (1) and (3), we get a quadratic equation in hν.
(h ν)
2
/2Mc
2
+ h ν − ΔW = 0
Which has the solution
h ν = −Mc
2
+ Mc
2 (1 + 2ΔW/Mc
2 )
1/2
Expanding the radical binomially and retaining up to second power of ΔW ,
and simplifying
h ν = ΔW (1 − ΔW/Mc
2 )
ν = (ΔW/ h)(1 − ΔW/Mc
2 )
7.83 The root mean square velocity of
137 Cs atoms
√
>= v = (3kT /m)
1/2
Substituting k = 1.38 × 10
−23 J K
−1
= 0.8625 × 10
−10 MeV K
−1
T = 288 K, mc
2
= 137 × 931.5 = 1.276 × 10
5 , we find v/c = 7.64 × 10
−7