Thorium (with half-life T1/2 = 1.913 yr. and atomic mass 228.0287 u) undergoes a

Question

Thorium (with half-life T1/2 = 1.913 yr. and atomic mass 228.0287 u) undergoes alpha decay and produces radium (atomic mass 224.0202 u) as a daughter nucleus. (Assume the alpha particle has atomic mass 4.002603 u.)

1. What is the decay constant of thorium? (Note that the answer must be in units of hrs-1.) =

2. 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). What is the change in temperature of the water after 1 hr.? T =

Explanation / Answer

1) The decay constant = 1/Tmean          where Tmean = T1/2/ln2

so = ln(2)/(1.913*365d/yr*24hr/d) = 4.14x10^-5/hr

2) Calculate m and energy for the whole 10 g:

amu0 = 228.028715, m0 = 0.010 kg
amu0 = 224.020186 + 4.002603 = 228.022789, m1 = 0.010 * 228.022789 / 228.028715 = 9.99974E-03 kg.
m = 2.6E-7 kg
Energy E = m*c^2 = 2.34E10 J
Now calc. E for 1st 1.0 h:
E (1.0 hr) = 2.34E10 * (1 - 0.5^(1.0/(698*24))) = 968201.32 J
Specific heat of water Qs = 4181.3 J/kg-K
T = E/(m*Qs) = 968201.32/(3.8*4181.3) = 60.94 C.

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