91Ó°ÊÓ

Given data:

• mass of radium$m=40mg=40×{10}^{-6}kg$
• 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:

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

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

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

Now, let us find the decay constant $\mathrm{λ}$of radium-226 as:

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

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

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

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

Mass of the radium after 500 years,

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

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

Mass of radium will decay as:

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

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

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

Since 1 atom of radium-226 gives 1 particle,

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

$$\begin{gathered} N_{\mathrm{Î\pm }}=N_{500yrs} \\ {N}_{\mathrm{Î\pm }}=2.07×{10}^{19} \end{gathered}$$

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

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 $7.8×{10}^{-6}kg$of radium-226 is decayed, it releases kinetic energy $Q_{500yrs}$.

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

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

We know that:

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

Here we substituted the value of $\mathrm{λ}$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$1.2×{10}^9{s}^{-1}$ .