For this question, we're told that  of a radioactive compound decays in  days. We're asked to determine the half-life for this compound.

The first thing we need to do is write an expression for radioactive decay. Remembering that radioactive decay processes follow a first-order reaction rate, we can use the following expression.

We can further rewrite this expression as follows.

Also, recall that in the beginning we are starting with  of the compound. After  of it has decayed, we will be left with  of the compound at time . Furthermore, in order to solve for the half life, we will first need to find the rate constant, .

Now that we have the rate constant for the decay reaction, we're equipped with what we need in order to calculate the half-life.

To solve for the half-life, we can derive an expression using one already shown above.

Now, using the above expression, we can plug in the value for the rate constant that we calculated in order to solve for the half-life.

Recall that beta decay occurs when the nucleus of the parent atom emits an electron. We can then write the following equation to illustrate the beta decay of thorium.

Make sure that the atomic numbers and masses add up to the same on both sides of the equation.

In order to solve this problem we first must setup an equation representing the decay of  particles. We know that an  particle is equivalent to , so we can write our radioactive decay equation as:

where  is the unknown element

 is the atomic number

and  is the mass number. We can create two equations. One to solve for the mass number and one to solve for the atomic number.

Let's start with the atomic number:

Since we have an atomic number of  on the reactants side of the equation we can set that equal to  (the combined atomic number of three  particles + .

As an equation this is written as:

This simplifies to . This means that is the atomic number, and we can identify the element by looking at the periodic table for element number  as the atomic number identifies the element. The element is Platinum also written as .

Now we must solve for the mass number similarly:

Following the instructions above except for the mass number we create an equation with one unknown to solve for the mass number of  produced by this decay.:

therefore 

This means our final answer is .

This can be double-checked by plugging the mass number and atomic number into the decay equation and simplifying it.

In order to determine the decay process that occurs we should check if the mass number charges or not in the decay. Since the mass number doesn't change we can eliminate any answer with   decay in it.

Now we can write out our decay equation. Since the atomic numbers aren't written in the question we must find them on the periodic table, where elements are ordered by their atomic number. In the case of Thallium, we find that it is element number  and in the case of Polonium we find that it is element .

So our equation is as follows:

where  is the element/particle name

 is the mass number

and  is the atomic number.

Since the mass number is the same on both sides of the equation we know that it is equal to .

Now we see that the atomic number increases by , which is the equivalent of a  on the product side of the equation. The only particle in the answer choices that is capable of having a negative atomic number value is a  particle, which is written as . In order to produce an atomic number change of  there must be   decays, which is the right answer.

Let's first look at the cases and determine which ones can maintain both the same atomic and mass number. We know that  decay produces a  nucleus so this cannot be the right answer as both the mass number and atomic number change with  decay.

As for  decay an electron is lost , so the atomic number increases, so therefore this cannot be the right answer.

In electron capture, an electron is gained so the atomic number decreases, so this cannot be the right answer either.

Positron emission means a positron  is lost, so this is not only equivalent to electron capture, but the atomic number changes by decreasing meaning this cannot be the right answer.

Therefore the right answer is gamma  decay, as in gamma decay the nuclei comes to a stable state and in order to do this must release ONLY energy, so neither the atomic number nor mass number change.

In order to begin determination of the mass number of phosphorus, we must first write out the equation and then solve for the mass number of phosphorus.

We know there is  present and it is bombarded by an  particle. An alpha particle is equivalent to . Next by reading the results  of the bombardment we can see that    , which are our reactants, produce  neutrons and an isotope of phosphorus. A neutron is an uncharged particle that has a "Z" value is equal to , and its atomic mass number is equal to . This gives us   since there are two neutrons. Next, we must look at the periodic table to find the atomic number of Phosphorus, which is . Since we must calculate its mass number, for now we mark it as unknown '', so for phosphorus we have . This means our products are 

Now let's write out the reaction that occurs and all of its reactants and products, so we can solve for X:

Now we can solve for the unknown mass number of phosphorus by adding up the mass numbers on each side of the equation and solving for the unknown.

On the reactants side we have  (from Al)  (from the alpha particle), and we set that equal to the products sum of mass numbers which are  (from the neutrons) .

This leaves us: 

Therefore the mass number of phosphorus is . This answer can be checked by plugging it into the bombardment equation and seeing if a true result is obtained.

In an electron capture, an electron is absorbed meaning the atomic number of the product is reduced by 1. Therefore we can write out our radioactive decay equation as:

Where  is the element name of the product

is the mass number of the product

and  is the atomic number of the product.

Since the mass number doesn't change at all . Since we are capturing an electron the atomic number is reduced by , therefore it becomes . Element  is , therefore   is the right answer.

In  decay a  nucleus is emitted, and 2  decays can be written as  so knowing this we can now write out the decay equation.

Where  is the element name of the product.

 is the atomic number of the product.

and  is the mass number of the product.

We can quickly determine the atomic number/element name by making an equation with an unknown with all the atomic numbers. This would be:

therefore the atomic number is  and the element polonium.

We can do the same for mass numbers to get the equation:

, so the final answer is .

Recall what an alpha particle is: .

Now, write the equation for the alpha decay of polonium-214. The polonium will be emitting an alpha particle.

Make sure that the masses and the atomic numbers on both sides of the equation will add up.

This is a question regarding radioactive decay, specifically alpha decay. An alpha particle has the same identity as the nucleus of a  atom, or . To solve this problem, we must perform simple addition and subtraction.

, and . So, the two new numbers become  and , eliminating one of the answer choices. Next, we must remember that the bottom number, in this case , signifies atomic number. Atomic number defines the identity of the element. So, we must find the element with the atomic number , which is .