Preparing for the Unified State Exam in Physics

Write in the comments which other problem you would like to see solved.

How many neutrons the nucleus contains

Let us look at an example.

A polonium atom ²¹²₈₄Po contains 84 protons. This is immediately visible from the subscript.

It also contains 128 neutrons.

How did I find that out - I simply subtracted 84 from 212.

212 - 84 = 128.

A 17

Problem statement:

A boron atom ⁸₅B contains:

Answer options

1) 8 protons 5 neutrons and 13 electrons

2) 8 protons 13 neutrons and 8 electrons

3) 5 protons 3 neutrons and 5 electrons

4) 5 protons, 8 neutrons and 13 electrons

How to solve

To solve such problems, you need to understand what atoms consist of.

1) The simplest fact, which is easy to remember: in neutral atoms the number of protons and electrons is the same.

That is, if the problem statement does not say that the atom has a charge (i.e. it is not a simple atom but an ion), we assume that there are as many electrons as protons.

2) The subscript is the number of protons.

3) The superscript is the sum of protons and neutrons

4) As additional information, you can keep in mind that in light nuclei there are fewer neutrons than protons; as the mass grows, the number of neutrons catches up with the number of protons, and in heavy nuclei there can be far more neutrons.

Example

A boron atom ⁸₅B contains:

1) 8 protons 5 neutrons and 13 electrons

2) 8 protons 13 neutrons and 8 electrons

3) 5 protons 3 neutrons and 5 electrons

4) 5 protons, 8 neutrons and 13 electrons

⁸₅B

The subscript is the number of protons (see point 2). So the number of protons in this atom is 5

Nothing is said about the charge of the atom, so it is neutral.

Consequently, there are as many electrons as protons - there are five of them too.

This knowledge alone is enough to choose answer number 3.

To check, we see that 5 protons plus 3 neutrons makes 8. That is, the superscript must equal 8. It checks out.

Answer: 3.

A 18

Problem statement:

A graph is given of the change in the number of nuclei of a radioactive isotope in a test tube over time. The half-life of this isotope -

First let us state the problem and the solution without a specific graph. Then we will go through an example.

Answer options

1) 1 month 2) 2 months 3) 3 months 4) 4 months or any other time interval.

How to solve such problems in the general case

1) The half-life is the time during which the number of nuclei decreases by half.

2) We look at the graph. We look for the axis along which the number of nuclei is plotted.

3) We look for any convenient section in which the number has changed by half.

4) We mark two points - the first is the point at which the number is larger. The second is the point at which the number of nuclei is half of that at the first.

5) We remember that every point of the graph has two coordinates: the number of nuclei and time.

6) After we have marked the points, we look at which time values correspond to them.

7) We subtract the smaller time from the larger one - that is the half-life.

Note: it is quite possible that several convenient pairs of points can be chosen on the graph; if the graph is correct, the answer will not change.

How to solve this particular example

Problem statement: A graph is given of the change in the number of nuclei of a radioactive isotope in a test tube over time. The half-life of this isotope -

1) 1 month 2) 2 months 3) 3 months 4) 4 months

Graph:

2) We look at the graph. We look for the axis along which the number of nuclei is plotted.

In our case this is the ordinate axis. It is underlined in orange.

3) We look for any section in which the number has changed by half.

For example, let us choose the section in which the number of nuclei decreased from 6 to 3.

To do this, let us draw two straight lines. The red one will correspond to the number 6 * N¹⁸ and the green one to 3 * N¹⁸

Now we need to find and mark the points at which these lines cross our graph.

Note: If you are solving a problem not about radioactive decay but about something else, there can be many such points - you need to choose two adjacent ones. In our case there will be only two of them.

4) We mark two points - the first is the point, at which the number is larger.

The second is the point, at which the number of nuclei is half of that at the first.

5) We remember that every point of the graph has two coordinates: the number of nuclei and time.

6) After we have marked the points, we look at which time values correspond to them.

In our case these are 0 months and 1 month

7) We subtract the smaller time from the larger one - that is the half-life.

1 month - 0 months = 1 month

Answer: 1 month

A 19

Problem statement:

Radioactive polonium ²¹⁶₈₄Po, having undergone one α-decay and two electron β-decays, turned into the isotope

Answer options

1) lead ²¹²₈₂Pb

2) polonium ²¹²₈₄Po

3) bismuth ²¹²₈₃Bi

4) thallium ²⁰⁸₈₁Po

How to solve it without going into the essence at all

To put it quite simply: alpha decay subtracts 4 from the superscript and 2 from the subscript. It adds nothing anywhere.

Electron beta decay does not touch the superscript but adds 1 to the subscript.

The letter depends on the subscript; if the subscript has not changed, the element has remained itself and the letter will not change.

216 - 4 = 212

84 - 2 + 1*2 = 84

Getting to the essence of what is happening

1) An alpha particle is two protons together with two neutrons.

That is, the total number of particles (hadrons) is four.

It can be written with the following formula ⁴₂α₂


2) α-decay: alpha decay is a process in which an alpha particle flies out of the nucleus.

That is, the nucleus has two protons and two neutrons fewer.


3) The subscript is the number of protons in the nucleus. ²¹⁶₈₄Po, that is, in our nucleus there are 84 protons


4) The superscript is the sum of protons and neutrons in the nucleus. ²¹⁶₈₄Po in our nucleus there are 84 protons and, consequently, 216 - 84 = 132 neutrons.


5) One α-decay means minus 4 from the superscript and minus 2 from the subscript.


6) electron β-decay or β^- decay is a process in which a neutron turns into a proton and emits another pair of particles that do not remain in the nucleus.


7) β^- decay does not change the superscript but increases the subscript by 1

Solving our example

²¹⁶₈₄ minus an alpha particle is ²¹²₈₂

that is, this is lead ²¹²₈₂Pb

Two electron beta decays increase the subscript by 2.

We get ²¹²₈₄ and this is polonium ²¹²₈₄Po

Article author: Andrei Olegovich