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Badger
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Homework Statement
The radium isotope [tex]^{223}{\rm Ra}[/tex], an alpha emitter, has a half-life of 11.43 days. You happen to have a 1.30 g cube of [tex]^{223}{\rm Ra}[/tex], so you decide to use it to boil water for tea. You fill a well-insulated container with 340 mL of water at [tex]16.0^\circ {\rm C}[/tex] and drop in the cube of radium.
Homework Equations
Alpha Decay:
X (A, Z) yields Y (A-4, Z-2) + alpha + energy
Energy = K_alpha = (M_x - M_y - M_He)c^2
Nuclear Decay and Half-life
N = N_0 e ^(-r/t) = N_0 e ^(-t/T) = N_0 (1/2)^(t)/(t/2)
T = 1/r
Heat Transformation?
Q = E_th = McT
The Attempt at a Solution
I think is a heat transformation problem except I learned that 20+ chapters ago, last semester and need some advice.
Since it's alpha decayer Radium (223, 88) becomes Rodium (219, 86) + alpha + energy. Loses 2 protons and 2 neutrons.
E = (mass_Ra - mass_Rn - mass_He) * 931.5 MeV/u
E = (223.018499 u - 219.009477 u - 4.002602) * 931.5 MeV/u
E = 5.98 MeV
Change in T = T = 84 Kelvin or Celsius
Mass of Water = .998g/cm^3 * 340 mL * 1kg/1000g = .33592 kg
Q = E_th = McT = (.33592 kg)(4190 J/kg K)(84 K) = 118230.4032 J
Am I going about this the right way? This chapter is about radioactive decay and quantum physics. I can't seem to make the connection to alpha decay and heating water.