Given initially 40mg of radium-226 (one of the decay products of uranium-238), determine,

(a) The amount that will be left after 500years.

(b) The number of α particles the radium will have emitted during this time, and

(c) The amount of kinetic energy that will have been released.

(d) Find the decay rate of the radium at the end of the 500 year.

Given data:

• mass of radium$m=40mg=40\times {10}^{-6}kg$
• The half-life of radium is$T_{\frac{1}{2}}=1599y$ because we are calculating the measures after 500 years of decay t = 500 y.

The nuclear reaction that involves theof Radium-226:

${}_{88}{}^{226}Ra\to {}_2{}^{4}He+{}_{86}{}^{222}Rn$

Now, let us find out the initial number of radium atoms$N_0$that would have been at the start of the decay,

$N_0$= initial mass of the radium/atomic mass of radium

$$\begin{gathered} N_0=\frac{40\times {10}^{-6}kg}{226u\times 1.66\times {10}^{-27}kg/u} \\ {N}_0=1.0662\times {10}^{20} \end{gathered}$$

Now, let us find the decay constant $\lambda$of radium-226 as:

$\begin{array}{rcl}\lambda & =& \frac{In2}{t_{\frac{1}{2}}}\\ & =& \frac{In2}{1599}\\ & =& 4.334\times {10}^{-4}\text{decays/year}\\ & =& 4.334\times {10}^{-4}\times \frac{1}{3.16\times {10}^7}\\ & =& 1.3715\times {10}^{-11}\text{decays/second}\\ & & \end{array}$

$\begin{array}{rcl}\lambda & =& 4.334\times {10}^{-4}\text{decay/year}\\ & =& 1.3715\times {10}^{-11}\text{decays/second}\\ & & \end{array}$

First, let’s find out the number of radium-226 atoms leftover after 500 years of decay:

$\begin{array}{rcl}N& =& N_0{e}^{-\lambda t}\\ & =& 1.0662\times {10}^{20}\times e^{-4.334\times {10}^{-4}\times 500}\\ & =& 1.066\times {10}^{20}\times e^{-0.2167}\\ & =& 8.583\times {10}^{19}\\ & & \end{array}$

Mass of the radium after 500 years,

$\begin{array}{rcl}M& =& N\times atomicmassofradium-226\\ & =& 8.583\times {10}^{20}\times 226u\times 1.66\times {10}^{-27}\\ M& =& 32.19\times {10}^{-6}kg\\ & & \end{array}$

Thus, the amount that will be left after 500years is $32.19\times {10}^{-6}kg$.

Mass of radium will decay as:

$\begin{array}{rcl}{M}_d& =& m-M\\ & =& 40mg-32.2mg\\ M_d& =& 7.8\times {10}^{-6}kg\\ & & \end{array}$

The number of radium atoms that are decayed in 500 years can be calculated as:

$\begin{array}{rcl}{N}_{500yrs}& =& \frac{M_d}{Z_{rad}}\\ & =& \frac{{\displaystyle 7.8\times {10}^{-6}}}{{\displaystyle 226u\times 1.66\times {10}^{-27}kg/u}}\\ N_{500yrs}& =& 2.07\times {10}^{19}\end{array}$

Since 1 atom of radium-226 gives 1 particle,

The number of particles released in the decay of radium226 over 500 years is:

$$\begin{gathered} N_{\alpha }=N_{500yrs} \\ {N}_{\alpha }=2.07\times {10}^{19} \end{gathered}$$

Thus, the number of α particles the radium emitted during this time is $2.07\times {10}^{19}$.

We had already find out the formula:

$$\begin{gathered} Q=\left(m_i-m_f\right)c^2 \\ =(22.6.0254024u-222.101754u-4.002603u)\left(931.4\frac{MeV}{u}\right) \\ =4.868MeV \end{gathered}$$

The above kinetic energy is released if 1 atom of radium-226 decays, i.e. if 226u of radium decays. But since $7.8\times {10}^{-6}kg$of radium-226 is decayed, it releases kinetic energy $Q_{500yrs}$.

$\begin{array}{rcl}{Q}_{500yrs}& =& \frac{7.8\times {10}^{-6}kg\times 4.868MeV}{226u\times 1.66\times {10}^{-27}kg/u}\\ & =& 1.011\times {10}^{20}MeV\\ & =& 1.6\times {10}^{7}J\\ & & \end{array}$

Thus, the amount of kinetic energy that will have been released is$1.6\times {10}^{7}J$ .

We know that:

$\begin{array}{rcl}R& =& \lambda N\\ & =& 8.583\times {10}^{19}\times 1.3715\times {10}^{-11}\\ R& =& 1.2\times {10}^9{s}^{-1}\\ & & \end{array}$

Here we substituted the value of $\lambda$in seconds because the decay rate is usually determined per second.

Thus, the decay rate of the radium at the end of the 500 year is$1.2\times {10}^9{s}^{-1}$ .