In this problem we are asked to write the nuclear equation for the transmutation of uranium isotope $^{234}_{92}$U into a thorium isotope $^{230}_{90}$Th and an $alpha$-particle.

$$
^{234}_{92}textrm{U}rightarrow ^{230}_{90}textrm{Th}+^{4}_{2}alpha
$$

$$
^{234}_{92}textrm{U}rightarrow ^{230}_{90}textrm{Th}+^{4}_{2}alpha
$$

In this problem we are asked to write the nuclear equation for the transmutation of thorium isotope $^{230}_{90}$Th into a radium isotope $^{226}_{88}$Th.

$$
^{230}_{90}textrm{Th}rightarrow ^{226}_{88}textrm{Ra}+^{4}_{2}alpha
$$

$$
^{230}_{90}textrm{Th}rightarrow ^{226}_{88}textrm{Ra}+^{4}_{2}alpha
$$

In this problem we are asked to write the nuclear equation for the transmutation of the radium isotope $^{226}_{88}$Ra into a radon isotope $^{222}_{86}$Rn and an $alpha$-particle.

$$
^{226}_{88}textrm{Ra}rightarrow ^{222}_{86}textrm{Rn}+^{4}_{2}alpha
$$

$$
^{226}_{88}textrm{Ra}rightarrow ^{222}_{86}textrm{Rn}+^{4}_{2}alpha
$$

In this problem we are asked to write the nuclear equation for the transmutation of a radioactive lead isotope $^{214}_{82}$Pb into a bismuth isotope $^{214}_{83}$Bi. This transmutation is accompanied by the emission of an electron and an antineutrino.

$$
^{214}_{82}textrm{Pb}rightarrow ^{214}_{83}textrm{Bi}+e^-+hatnu
$$

$$
^{214}_{82}textrm{Pb}rightarrow ^{214}_{83}textrm{Bi}+e^-+hatnu
$$

In this problem, we are asked to write the nuclear equation for $beta$-decay of a radioactive carbon isotope $^{14}_{6}$Pb into a nitrogen atom $^{14}_{7}$N. This decay is accompanied by the emission of an electron and an antineutrino.

$$
^{14}_{6}textrm{C}rightarrow ^{14}_{7}textrm{N}+e^-+hatnu
$$

$$
^{14}_{6}textrm{C}rightarrow ^{14}_{7}textrm{N}+e^-+hatnu
$$

center{a) The reaction is given as follows} [^{14}_{6}textrm{C}rightarrow ^A_ZX+_{textrm{-1}}^{textrm{ 0}}e+^0_0hatnu]
and we have that
[A=14-0-0=14]
[Z=6+1=7]
So the element we are looking for is $boxed{^{14}_{7}textrm{N}}$.

center{b) The second reaction is given as follows} [^{55}_{24}textrm{C}rightarrow ^A_ZX+_{textrm{-1}}^{textrm{ 0}}e+^0_0hatnu]
which makes that
[A=55-0-0=55]
[Z=24+1=25]
So the element we are looking for is $boxed{^{55}_{25}textrm{Mn}}$.

$$
textrm{a) } ^{14}_{7}textrm{N}
$$

$$
textrm{b) } ^{55}_{25}textrm{Mn}
$$

In this problem, we are asked to write the nuclear equation for the transmutation of seaborgium isotope $^{263}_{106}$Sg into a rutherfordium isotope $^{259}_{104}$Rf and an $alpha$-particle.

$$
^{263}_{106}textrm{Sg}rightarrow ^{259}_{104}textrm{Rf}+^{4}_{2}alpha
$$

$$
^{263}_{106}textrm{Sg}rightarrow ^{259}_{104}textrm{Rf}+^{4}_{2}alpha
$$

In this problem we have a proton collision with a nitrogen isotope $^{15}_{7}$N which forms another element and an alpha particle. We should find the product element.

In order to do se, we have to write down this nuclear reaction first. We have that

$$
^{15}_{7}textrm{N}+^{1}_{1}textrm{H} rightarrow ^{A}_{Z}textrm{X}+ ^4_2alpha
$$

center{Finally we get that }
[A=15+1-4=12]
[Z=7+1-2=6]
so the element we are looking for is $boxed{^{12}_6textrm{C}}$.

$$
^{15}_{7}textrm{N}+^{1}_{1}textrm{H} rightarrow ^{12}_{6}textrm{C}+ ^4_2alpha
$$

In this problem we are asked to find the child nuclei during $beta$-decays of given parent nuclei. We have to know that $beta$-decay is given as

$$
^{A}_{Z}textrm{X}rightarrow^{A}_{Z+1}textrm{Y}+ {^{textrm{
0}}_{textrm{-1}}}e+^0_0hatnu
$$

So we see that the atomic number of the child nuclei increases by 1 while the mass number remains the same.

a) In the case of $^{210}_{80}$Pb we have that the isotope that matches above description is $^{210}_{81}$Tl and the equation is

$$
^{210}_{80}textrm{Pb}rightarrow^{210}_{81}textrm{Tl}+ {^{textrm{
0}}_{textrm{-1}}}e+^0_0hatnu
$$

b) In the case of $^{210}_{83}$Bi we have that the isotope that matches above description is $^{210}_{84}$Po and the equation is

$$
^{210}_{83}textrm{Bi}rightarrow^{210}_{84}textrm{Po}+ {^{textrm{
0}}_{textrm{-1}}}e+^0_0hatnu
$$

c) In the case of $^{234}_{90}$Th we have that the isotope that matches above description is $^{234}_{91}$Pa and the equation is

$$
^{234}_{90}textrm{Th}rightarrow^{234}_{91}textrm{Pa}+ {^{textrm{
0}}_{textrm{-1}}}e+^0_0hatnu
$$

d) In the case of $^{239}_{93}$Np we have that the isotope that matches above description is $^{239}_{94}$Pu and the equation is

$$
^{239}_{93}textrm{Np}rightarrow^{239}_{94}textrm{Pu}+ {^{textrm{
0}}_{textrm{-1}}}e+^0_0hatnu
$$

$$
textrm{a) }^{210}_{80}textrm{Pb}rightarrow^{210}_{81}textrm{Tl}+ {^{textrm{
0}}_{textrm{-1}}}e+^0_0hatnu
$$

$$
textrm{b) }^{210}_{83}textrm{Bi}rightarrow^{210}_{84}textrm{Po}+ {^{textrm{
0}}_{textrm{-1}}}e+^0_0hatnu
$$

$$
textrm{c) }^{234}_{90}textrm{Th}rightarrow^{234}_{91}textrm{Pa}+ {^{textrm{
0}}_{textrm{-1}}}e+^0_0hatnu
$$

$$
textrm{d) }^{239}_{93}textrm{Np}rightarrow^{239}_{94}textrm{Pu}+ {^{textrm{
0}}_{textrm{-1}}}e+^0_0hatnu
$$

We can solve this problem in a very simple way by applying the radioactive decay formula which tells us that

$$
m=m_0cdot(frac{1}{2})^{frac{T}{T_{1/2}}}
$$

$$
m=0.25textrm{ g}
$$

In this problem we are given four grams of $^{238}_{93}$Np for which we should calculate the remaining mass after 8 days.

We can solve this problem in a very simple way by applying the half-life formula which tells us that

$$
m=m_0cdot(frac{1}{2})^{frac{T}{T_{1/2}}}
$$

$$
m=0.25textrm{ g}
$$

In this problem we are given a sample of polonium isotope of a given activity for which we should calculate the activity after $approx$275 days.

We can solve this problem in a very simple way by applying the radioactive decay formula which tells us that

$$
A=A_0cdot(frac{1}{2})^{frac{T}{T_{1/2}}}
$$

Now, having in mind that halflife of the given isotope is 138 days we can plug in the given values to have that

$$
A=2times 10^6cdot(frac{1}{2})^{frac{275}{138}}=2times 10^6cdot frac{1}{4}=boxed{0.5 times 10^6textrm{ Bq}}
$$

$$
A=0.5 times 10^6textrm{ Bq}
$$

Since the activity is proportional to the glow this problem can be solved in a very simple way by applying the radioactive decay formula which tells us that

$$
A=A_0cdot(frac{1}{2})^{frac{T}{T_{1/2}}}
$$

Now, having in mind that the halflife of the tritium is 12.3 years we can plug in the given values to have that

$$
A=A_0(frac{1}{2})^{frac{6}{12.3}}=boxed{0.71 A_0}
$$

So the brightness is 0.71 of the original.