Question

nuclear equations
20.write the equations showing that when protactinium - 229 undergoes two alpha decays, francium - 221 is formed. show each step and then combine into one overall step.
21.write the equations showing the daughter isotopes of each step as po - 210 undergoes two alpha decays, followed by a beta decay, followed by another alpha decay. show each step and then combine into one overall step.

Explanation:

Step1: Define alpha - decay

An alpha - particle is a helium nucleus with mass number 4 and atomic number 2, represented as $_{2}^{4}He$. When a nucleus undergoes alpha - decay, its mass number decreases by 4 and its atomic number decreases by 2.

Step2: First alpha - decay of protactinium - 229

Protactinium has atomic number 91. For the first alpha - decay of $_{91}^{229}Pa$:
$_{91}^{229}Pa
ightarrow_{89}^{225}Ac + _{2}^{4}He$
Here, the mass number of Pa (229) decreases by 4 to give 225 for actinium (Ac), and the atomic number of Pa (91) decreases by 2 to give 89 for Ac.

Step3: Second alpha - decay of the daughter nucleus

The daughter nucleus $_{89}^{225}Ac$ undergoes another alpha - decay:
$_{89}^{225}Ac
ightarrow_{87}^{221}Fr+_{2}^{4}He$

Step4: Overall equation for protactinium - 229 decay

Combining the two steps, we get the overall equation:
$_{91}^{229}Pa
ightarrow_{87}^{221}Fr + 2_{2}^{4}He$

Step5: First alpha - decay of polonium - 210

Polonium has atomic number 84. For the first alpha - decay of $_{84}^{210}Po$:
$_{84}^{210}Po
ightarrow_{82}^{206}Pb+_{2}^{4}He$

Step6: Second alpha - decay of the daughter nucleus

The daughter nucleus $_{82}^{206}Pb$ undergoes another alpha - decay:
$_{82}^{206}Pb
ightarrow_{80}^{202}Hg + _{2}^{4}He$

Step7: Beta - decay of the daughter nucleus

In beta - decay, a neutron in the nucleus is converted into a proton and an electron (beta - particle, $_{ - 1}^{0}e$). For the beta - decay of $_{80}^{202}Hg$:
$_{80}^{202}Hg
ightarrow_{81}^{202}Tl+_{ - 1}^{0}e$

Step8: Fourth alpha - decay of the daughter nucleus

The daughter nucleus $_{81}^{202}Tl$ undergoes an alpha - decay:
$_{81}^{202}Tl
ightarrow_{79}^{198}Au+_{2}^{4}He$

Step9: Overall equation for polonium - 210 decay

Combining all the steps:
$_{84}^{210}Po
ightarrow_{79}^{198}Au+3_{2}^{4}He + _{ - 1}^{0}e$

Answer:

For protactinium - 229:
Individual steps:
$_{91}^{229}Pa
ightarrow_{89}^{225}Ac + _{2}^{4}He$
$_{89}^{225}Ac
ightarrow_{87}^{221}Fr+_{2}^{4}He$
Overall: $_{91}^{229}Pa
ightarrow_{87}^{221}Fr + 2_{2}^{4}He$
For polonium - 210:
Individual steps:
$_{84}^{210}Po
ightarrow_{82}^{206}Pb+_{2}^{4}He$
$_{82}^{206}Pb
ightarrow_{80}^{202}Hg + _{2}^{4}He$
$_{80}^{202}Hg
ightarrow_{81}^{202}Tl+_{ - 1}^{0}e$
$_{81}^{202}Tl
ightarrow_{79}^{198}Au+_{2}^{4}He$
Overall: $_{84}^{210}Po
ightarrow_{79}^{198}Au+3_{2}^{4}He + _{ - 1}^{0}e$