<h3>
Answer:</h3>
Lead-205 (Pb-205)
<h3>
Explanation:</h3>
<u>We are given;</u>
We are supposed to identify its product after an alpha decay;
- Polonium-209 has a mass number of 209 and an atomic number of 84.
- When an element undergoes an alpha decay, the mass number decreases by 4 while the atomic number decreases by 2.
- Therefore, when Po-209 undergoes alpha decay it results to the formation of a product with a mass number of 205 and atomic number of 82.
- The product from this decay is Pb-205, because Pb-205 has a mass number of 205 and atomic number 82.
- The equation for the decay is;
²⁰⁹₈₄Po → ²⁰⁵₈₂Pb + ⁴₂He
- Note; An alpha particle is represented by a helium nucleus, ⁴₂He.
Answer : The activation energy of the reaction is, 
Solution :
The relation between the rate constant the activation energy is,
![\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7BK_2%7D%7BK_1%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial rate constant = 
= final rate constant = 
= initial temperature = 
= final temperature = 
R = gas constant = 8.314 kJ/moleK
Ea = activation energy
Now put all the given values in the above formula, we get the activation energy.
![\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7B8.75%5Ctimes%2010%5E%7B-3%7DL%2Fmole%5Ctext%7B%20s%7D%7D%7B4.55%5Ctimes%2010%5E%7B-5%7DL%2Fmole%5Ctext%7B%20s%7D%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20%288.314kJ%2FmoleK%29%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7B468K%7D-%5Cfrac%7B1%7D%7B531K%7D%5D)

Therefore, the activation energy of the reaction is, 
Heat of vaporization of water will be required as water is already at it's boiling point thus heat required will be 540*10=5400 cal