Question

The gold nucleus Au(198,79) undergoes alpha decay with a half-life of 2.70 days.

A. )  Find how many neutrons and protons are in the daughter nucleus produced by this decay.

B.) Find the activity in Bq (decays/sec) of a sample that contains 5.0 grams of Au(198,79) nuclei.

C.) A 2.5 kg suitcase at an airport absorbs 2.0 × 10−3 J/kg from 100 keV x-rays. How many photons are absorbed?

Answer 1

A) protons are 79-2 = 77

neutrons are (194-77)= 117

B) activity is dN/dt = -lamda*N

lamda is the decay constant = 0.693/T1/2 = 0.693/(2.7*24*60*60) = 2.97*10^-6 s^-1

N 5 grms

then dN/dt = -lamda*N

N is the no.of atoms = (5/198)6.02*10^23 = 1.52*10^22 nuclei

then activity dN/dt = -lamda*N = 2.97*10^-6*1.52*10^22 = 4.52*10^16 decays/sec

C) total energy absorbed is 2*2.5*10^-3 J

E= 5*10^-3 J

but E = n*h*f

h*f = 100*1000*1.6*10^-19 = 1.6*10^-14 J

no.of photons are n = E/(h*f) = (5*10^-3)/(1.6*10^-14) = 3.125*10^11 photons