Question

1. A caesium-137 source was manufactured in October 1977. At that time, it had an activity of 35.0 milliCuries. How many gamma rays per second does this source emit presently?

2. Calculate the rest mass energy of the electron (m_ec² ) in keV and show that α is 1.295

3. What is the largest and smallest possible energy for a Compton scattered caesium-137 gamma ray? For the scattered gamma rays of largest and smallest possible energies, what is the speed of the corresponding recoil electron, expressed as a fraction of the speed of light?

Answer 1

1. given Cs -137
half life, t' = 30.17 y
Ao = 35 mCi = 1.295*10^9 Bq
t = 40 years, 4 months = 40.3333 y

hence
A = Ao*e^(-t*ln(2)/t')
A = 5.12663*10^8 per second

2. rest mass energy of electron in keV = E
E = (9.1*10^-31)*c^2/1.6*10^-19 = 511875 eV = 511.875 keV

3. largest possible energy of C137 gamma ray, El = 662 keV
KE of electron = 0

smallest possibel energy , Es = hc/lambda'
lambda' - lambda = (h/me*c)*2
lambda' = lambda + 2h/me*c
now,
662*10^6*1.6*10^-19 = hc/lambda
lambda = 1.88916*10^-15 m
hence
lambda' = 4.888*10^-12 m
hence
Es = 255.838 keV

KE of electron = E - Es = 406.16141 eV