(a) (i) A radioactive sample produces 1536 decays per second at one point in an

Question

(a) (i) A radioactive sample produces 1536 decays per second at one point in an experiment and 55.888 years later produces 320 decays per second. What is the half-life of this sample? (ii) What is the decay (in y-1)? (iii) What is the mean lifetime of the material? (b) What fraction of a sample of phosphrous-32 is left after 64.26 days? (the half-life of phosphorous-32 is 14.28 days) (c) What period of time is required for a sample of radioacive sulfur-35 to lose 59.75% of its activity? (The half-life of this sample is 87.4 days)

Explanation / Answer

a)

let the number of particles in the beginning be N0.

let the govering equation be

N(t)=N0*exp(-t/T)

where N(t)=number of particles remaining at any time t

T=a constant

then decay rate=dN(t)/dt=(-N0/T)*exp(-t/T)

so if we assume the point where it decays 1536 per seconds to be t=0,

then N0/T=1536

at t=55.888, dN/dt=320

then (N0/T)*exp(-55.888/T)=320

1536*exp(-55.888/T)=320

T=

then half life=ln(2)*T=24.696 years

iii)mean life time=1.44*half life time=35.5623 years

b)governing equation: N(t)=N0*exp(-t/T)

where N0=initial concentration

T=half life/ln(2)=20.6017 days

putting t=64.26 days, we get N(t)=0.0442*N0

then after 64.26 days, 4.42% of phosphrous 32 is left.

c)let at time t , it loses 59.75% of its activity.

then if activity at initial period is A0 and at any time t is A(t)

then A(t)=A0*exp(-t/T)

where T=half life/log(2)=126.0915 days

then given that A(t)=1-0.5975=0.4025

0.4025=1*exp(-t/126.0915)

t=114.751 days

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