91Ó°ÊÓ

Given,

• Mass of polonium-210$=10mg=10×{10}^{-6}kg$
• Mass of water$m_w=500g=500×{10}^{-3}kg$
• Half-life of polonium$t_{\frac{1}{2}}=140days=3360hours$
• Time period t = 1 hour

${}_{84}{}^{210}Po→{}_2{}^{4}He+{}_{82}{}^{206}Pb$

Now, let us find out the initial number of polonium atoms localid="1658423854044" $N_0$that would have been at the start of the decay

$$\begin{gathered} N_0=\frac{m_{p0}}{Z_p} \\ =\frac{{\displaystyle 10×{10}^{-6}kg}}{{\displaystyle 209.982848u×1.66×{10}^{-27}kg/u}} \\ {N}_0=2.86885×{10}^{19} \end{gathered}$$

Let us find the decay constant$\mathrm{λ}$ of polonium-210:

$\begin{array}{rcl}\mathrm{λ}& =& \frac{In2}{t_{\frac{1}{2}}}\\ & =& \frac{In2}{3360}\\ & =& 2.062×{10}^{-4}\text{decays/year}\\ & & \end{array}$

As we know:

$N=N_0{e}^{-\mathrm{λ}t}$

Substituting the values and we get,

$\begin{array}{rcl}N& =& 2.86885×{10}^{19}×e^{-2.062×{10}^{-4}}\\ & =& 2.86825×{10}^{19}\\ N& =& 8.583×{10}^{19}\\ & & \end{array}$

We have to find out the amount of kinetic energy released in the decay:

$\begin{array}{rcl}Q& =& (m_i-m_f)c^2\\ & =& (209.982873673-205.97446527-4.002603254)931.5MeV\\ & =& 5.407MeV\\ & & \end{array}$

Let us find out the amount of polonium-210 nuclei decayed as:

$\begin{array}{rcl}{\text{N}}_d& =& N_0-N\\ & =& 0.60×{10}^{14}\\ & & \end{array}$

One polonium-210 nuclei releases 5.407 MeV of energy, then $N_d$nuclei releases:

$\begin{array}{rcl}{Q}^{\text{'}}& =& N_d×Q\\ & =& 0.060×{10}^{14}×5.407MeV\\ & =& 3.244×{10}^{14}MeV\\ & =& 5190J\\ & & \end{array}$

Specific heat capacity of water$c=4200J/kg·C$

$\begin{array}{rcl}\mathrm{Δ}T& =& \frac{Q^{\text{'}}}{Cm}\\ \mathrm{Δ}T& =& \frac{5190}{4200×0.5}\\ \mathrm{Δ}T& =& 2.47°C\\ & & \end{array}$

Therefore, change in temperature$\mathrm{Δ}T=2.47°C$