Question

1. write a balanced equation for each nuclear change.

a. carbon - 14 emits an alpha particle.
b. hydrogen - 2 and hydrogen - 3 react together to form a larger element.
c. carbon - 14 decays by beta emission.
d. nitrogen - 14 captures an alpha particle.
e. polonium - 208 emits an alpha particle.
f. polonium - 214 emits a particle to become lead - 210.

Explanation:

Step1: Recall alpha - decay formula

For alpha - decay, the general form is $_{Z}^{A}X
ightarrow_{2}^{4}He + _{Z - 2}^{A - 4}Y$.

Step2: Recall beta - decay formula

For beta - decay ($\beta^-$), the general form is $_{Z}^{A}X
ightarrow_{- 1}^{0}e+_{Z + 1}^{A}Y$.

Step3: Recall nuclear fusion formula

For the fusion of two hydrogen isotopes, we consider conservation of mass - number and atomic - number.

a.

Carbon - 14 ($_{6}^{14}C$) emits an alpha particle ($_{2}^{4}He$).
$_{6}^{14}C
ightarrow_{2}^{4}He+_{4}^{10}Be$

b.

Hydrogen - 2 ($_{1}^{2}H$) and hydrogen - 3 ($_{1}^{3}H$) react. This is a fusion reaction.
$_{1}^{2}H+_{1}^{3}H
ightarrow_{2}^{4}He+_{0}^{1}n$

c.

Carbon - 14 ($_{6}^{14}C$) decays by beta emission.
$_{6}^{14}C
ightarrow_{-1}^{0}e + _{7}^{14}N$

d.

Nitrogen - 14 ($_{7}^{14}N$) captures an alpha particle ($_{2}^{4}He$).
$_{7}^{14}N+_{2}^{4}He
ightarrow_{9}^{18}F$

e.

Polonium - 208 ($_{84}^{208}Po$) emits an alpha particle.
$_{84}^{208}Po
ightarrow_{2}^{4}He+_{82}^{204}Pb$

f.

Polonium - 214 ($_{84}^{214}Po$) becomes lead - 210 ($_{82}^{210}Pb$). Since the mass - number decreases by 4 and the atomic - number decreases by 2, it emits an alpha particle.
$_{84}^{214}Po
ightarrow_{2}^{4}He+_{82}^{210}Pb$

Answer:

a. $_{6}^{14}C
ightarrow_{2}^{4}He+_{4}^{10}Be$
b. $_{1}^{2}H+_{1}^{3}H
ightarrow_{2}^{4}He+_{0}^{1}n$
c. $_{6}^{14}C
ightarrow_{-1}^{0}e + _{7}^{14}N$
d. $_{7}^{14}N+_{2}^{4}He
ightarrow_{9}^{18}F$
e. $_{84}^{208}Po
ightarrow_{2}^{4}He+_{82}^{204}Pb$
f. $_{84}^{214}Po
ightarrow_{2}^{4}He+_{82}^{210}Pb$