Question

Strontium-90 is a dangerous by-product of atomic testing because it mimics the action of calcium in the body. It decay in two B-emissions to give zirconium-90 (nuclear mass = 89.8824 g). a) write a balanced nuclear equation for the overall decat of Sr-90. b) calculate ∆m in grams when one mole of Sr-90 decays to Zr-90.
c) How much energy (in KJ) is given off by the decay of 6.50 mg of Sr-90?

Answer 1

a) At the first step of $\beta$-emission, the $^{90}Sr$ goes to produce $^{90}Y$ and releases an electron and some amount of energy

$^{90}Sr \rightarrow ^{90}Y + \bar{e} + \bar{\nu}_e$

At the second step of $\beta$-emission, $^{90}Y$ as a not stable isotope decay to a stable $^{90}Zr$

$^{90}Y \rightarrow ^{90}Zr + \bar{e} + \bar{\nu}_e$

The overall decay then will look like

$^{90}Sr \rightarrow ^{90}Zr + 2\bar{e} +2 \bar{\nu}_e$

b) The difference in atomic mass (a.u.) can be found by taking a table values of atomic masses of each isotopes by neglecting a mass of electron:

$M_A(^{90}Sr) = 89.907738 a.u\\ M_A(^{90}Zr) = 89.904703 a.u$

$\Delta M_A = M_A(^{90}Zr)-M_A(^{90}Sr) = \\=89.904703 - 89.907738= -0.003035 a.u$

By looking at the decay equation from a), the number of moles is equivalent, so one mole of $^{90}Sr$ produces one mole of $^{90}Zr$. So, to calculate the mass difference, the atomic units have to be converted to kilograms. The conversion factor is:

$1a.u = 0.0166054 \times 10^{-25} kg\\ \Delta m = -0.003035 \times 0.0166054 \times 10^{-25} = 0.504 \times 10^{-29} kg$

c) The energy which is given off by the decay is:

$E=(\Delta m)c^2$

The 6.5 mg have to be converted to a.u first.

$\Delta m=6.5 mg = 3.91492 \times 10^{21} a.u$

Using $1 a.u = 931.5MeV/c^2$ , we obtain the simpler way of the energy calculation:

$E= (3.91492 \times 10^{21} a.u)(931.5MeV/c^)c^2 = 3.64675 \times 10^{24} MeV$

Converting to kJ:

$E=3.64675 \times 10^{24} MeV \times 1.60218\times 10^{-16} = 584.27373 \times 10^{6} kJ$