416 7 Nuclear Physics – I
7.80 For electronz=−1. The given integral is easily evaluated assuming that this
refractive indexμis independent of frequency. Integrating between the limits
ν 1 andν 2.
−dW/dl=(4π^2 e^2 /c^2 )(1− 1 /β^2 μ^2 )[(ν 22 −ν 12 )/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 /hv ̄=(4π^2 e^2 /hc)(1− 1 /β^2 μ^2 )(ν 2 −ν 1 )
=(2π/137)(1− 1 /β^2 μ^2 )(1/λ 2 − 1 /λ 1 )(3)
whereλ 1 =c/ν 1 andλ 2 =c/ν 2 are the vacuum wavelengths andμis the
average index of refraction over the wavelength intervalλ 2 = 4 , 000 Ato ̊
λ 1 = 8 , 000 A. Substituting ̊ βμ= 0. 9 × 1. 33 = 1 .197,
λ 1 = 8 × 10 −^5 cm andλ 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
ΔEγ/Eγ=v/c
PutΔEγ=Γ
Γ=vEγ/c=(1 cm/s/ 3 × 1010 cm/s)×(129 keV)= 4. 3 × 10 −^6 eV
The mean lifetime
τ=/ΔEγ=/Γ= 1. 05 × 10 −^34 / 1. 6 × 10 −^19 × 4. 3 × 10 −^6 = 1. 5 × 10 −^10 s
7.82 Energy conservation gives
hν=ΔW−ER (1)
Momentum conservation gives
PR=hν/c (2)
ThereforeER=PR^2 / 2 M=(hν)^2 / 2 Mc^2 (3)
EliminatingERbetween (1) and (3), we get a quadratic equation inhν.
(hν)^2 / 2 Mc^2 +hν−ΔW= 0
Which has the solution
hν=−Mc^2 +Mc^2 (1+ 2 ΔW/Mc^2 )^1 /^2
Expanding the radical binomially and retaining up to second power ofΔW,
and simplifying
hν=ΔW(1−ΔW/Mc^2 )
ν=(ΔW/h)(1−ΔW/Mc^2 )7.83 The root mean square velocity of√^137 Cs atoms
<v^2 >=v=(3kT/m)^1 /^2
Substitutingk= 1. 38 × 10 −^23 JK−^1 = 0. 8625 × 10 −^10 MeV K−^1
T=288 K,mc^2 = 137 × 931. 5 = 1. 276 × 105 , we findv/c= 7. 64 × 10 −^7