Radioactivity of Thorium

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Homework Statement


Thorium (with half-life T1/2 = 1.913 yr. and atomic mass 228.028715 u) undergoes alpha decay and produces radium (atomic mass 224.020186 u) as a daughter nucleus. (Assume the alpha particle has atomic mass 4.002603 u.)

(a) What is the decay constant of thorium? (Note that the answer must be in units of hrs-1.)
Answer: 4.136E-5 hrs-1

The energy released from the decay of 10 g of thorium is used to heat 3.8 kg of water (assume all the energy released goes into the water).

(c) What is the change in temperature of the water after 1 hr.?


Homework Equations


E=mc2
N0= Mtotal/Mparticle
N=N0e^-λt
Q=mcT

The Attempt at a Solution


N0= Mtotal/Mparticle
N0=[(0.01kg)]/[(228.0287u)(1.66E-27kg)]
N0=2.64E22 atoms

N=N0e^-λt
N=(2.64E22 atoms)e^-(4.136E-5 hrs-1)(1hr)
N=2.642E22 atoms

mi=228.028715u

mf=224.020186u + 4.002603u
mf=228.022789u

m=mf-mi
m=228.022789u - 228.028715u
m=0.005926u

E=mc2
E=(0.005926u)[(931.5Mev/c2)/u]c2
E=5.52Mev

(5.52Mev)(1.602E-13J/Mev) = 8.843E-13 J

(8.843E-13J)(2.642E22 atoms)=2.336E10J

Q=mcT
2.336E10J=(3.8kg)(4186 J/kg-K)(T)
T=1468755.878 K = 1468482.878 C

Can someone help me catch my mistake? I'm not sure what I did wrong. Thank you.
 

Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
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(8.843E-13J)(2.642E22 atoms)=2.336E10J
This assumes all of the Th-228 decays.

One starts with No = 2.642E22 atoms, but how many atoms remain at the end of 1 hr? The difference No - N (1 hr) = the number of atoms that have decayed.

It's not that 10 g of Th-228 decayed, but a certain fraction of atoms in Th-228 have decayed, and that is related to the activity.
 
  • #3
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Thank you. I got it now.
 

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