Physics · 5 Nuclear physics

5.2.3 Radioactive decay

Before you start

A nucleus contains protons and neutrons. The proton number $Z$ is the number of protons, the nucleon number $A$ is protons plus neutrons, so the number of neutrons is $A - Z$ (5.1.2).
Nuclide notation $_{}^{}_{Z}^{A} X$ : the nucleon number goes on top, the proton number at the bottom (5.1.2).
Isotopes of an element have the same number of protons but different numbers of neutrons (5.1.2).
Relative charges: proton $+ 1$ , neutron $0$ , electron $- 1$ (5.1.2).
An α-particle is two protons and two neutrons, a β-particle is a fast-moving electron, and γ-radiation is electromagnetic radiation (5.2.2).

Syllabus objectives covered (5)

Know that radioactive decay is a change in an unstable nucleus that can result in the emission of α-particles or β-particles and/or γ-radiation and know that these changes are spontaneous and random
State that during α-decay or β-decay, the nucleus changes to that of a different element
Know that isotopes of an element may be radioactive due to an excess of neutrons in the nucleus and/or the nucleus being too heavy
Describe the effect of α-decay, β-decay and γ-emissions on the nucleus, including an increase in stability and a reduction in the number of excess neutrons; the following change in the nucleus occurs during β-emission: neutron → proton + electron
Use decay equations, using nuclide notation, to show the emission of α-particles, β-particles and γ-radiation

Part 1 of 6

What radioactive decay is

Most nuclei never change. A few are unstable: their arrangement of protons and neutrons cannot last, and at some point it changes. That change is radioactive decay.
When an unstable nucleus decays, it gives out radiation. It can emit an α-particle or a β-particle, and γ-radiation can come out as well, either together with one of these or on its own. That is what the 'and/or' in the objective means.
Keep the location clear: the change happens in the nucleus. The electrons around the nucleus are not what decays, so a decay is not the same thing as an atom becoming an ion.
Spontaneous means the decay happens by itself. Nothing from outside has to cause it, and nothing you do from outside can trigger it or stop it. Heating the sample, cooling it, squeezing it or making the atoms part of a chemical compound does not make a nucleus decay.
Random means you cannot predict when a particular nucleus will decay, or which nucleus in a sample will decay next. Every nucleus of the same isotope has the same chance of decaying in the next second, but nothing tells you which one it will be.
You can see the randomness with a detector and counter. Put a detector near a source and count for 10 s, again and again. The counts are not all the same: they go up and down about an average value, because the number of nuclei that happen to decay in each 10 s changes.
The trap here is to treat the two words as one. 'Spontaneous' answers what causes it (nothing outside). 'Random' answers when (you cannot predict it for one nucleus). The objective asks for both, so learn both meanings.
Random does not mean that nothing can be predicted. In a very large number of nuclei, the fraction that decays in a given time is predictable. That idea is what half-life is built on, in 5.2.4.

Part 2 of 6

Why some isotopes are radioactive

Remember that isotopes of one element all have the same number of protons and differ only in their number of neutrons. An element can have some isotopes that are stable and some that are radioactive.
Carbon shows this. Carbon-12 has 6 protons and 6 neutrons, and carbon-13 has 6 protons and 7 neutrons: both are stable. Carbon-14 has 6 protons and 8 neutrons, and it is radioactive.
The only difference is the neutrons, so the neutrons must be the cause. Carbon-14 has too many neutrons for its 6 protons. The syllabus calls this an excess of neutrons.
Hydrogen is the same. Hydrogen-1 (no neutrons) and hydrogen-2 (one neutron) are stable, but hydrogen-3, with 1 proton and 2 neutrons, is radioactive because of its excess of neutrons.
The second reason is size. A very large nucleus, with a great many protons and neutrons, is unstable because it is too heavy. Every isotope of uranium (92 protons) is radioactive, for example.
The objective says 'and/or' again: a nucleus can be unstable because of an excess of neutrons, because it is too heavy, or for both reasons at once.
Each reason points to a way out. A nucleus with an excess of neutrons can become more stable by β-decay, which changes a neutron into a proton. A nucleus that is too heavy can become more stable by α-decay, which takes 4 nucleons away. The next sections show both.

Part 3 of 6

α-decay: the nucleus loses two protons and two neutrons

An α-particle is 2 protons and 2 neutrons held together. That is the same as a helium nucleus, so in an equation you write it as $_{}^{}_{2}^{4} He$ : nucleon number 4, proton number 2. You may also see it written $_{}^{}_{2}^{4} α$ .
When a nucleus emits an α-particle, those 4 nucleons leave it. So the nucleon number $A$ goes down by 4 and the proton number $Z$ goes down by 2.
Because $Z$ has changed, the new nucleus belongs to a different element. The element is decided by the proton number and nothing else, so any change in $Z$ means a new element.
Written in general, with X for the parent nucleus and Y for the new one:
$_{}^{}_{Z}^{A} X \overset{}{\to}_{}^{}_{Z - 2}^{A - 4} Y +_{}^{}_{2}^{4} He$
The new nucleus is smaller and lighter, and energy has been released (the α-particle leaves with kinetic energy). The nucleus is now more stable than before. This is how a nucleus that is too heavy becomes more stable.
Notice what α-decay does not do. It removes 2 neutrons and 2 protons, the same number of each, so the difference between the neutron number and the proton number stays the same. α-decay does not reduce the excess of neutrons. That is β-decay's job.

Part 4 of 6

β-decay: a neutron becomes a proton

In β-decay, one neutron inside the nucleus changes into a proton and an electron. This is the change the syllabus names, and you need to be able to state it exactly.
$neutron \to proton + electron$
The proton stays in the nucleus. The electron does not: it is emitted at high speed, and that fast electron is the β-particle.
This is the trap. The β-particle is not one of the electrons that orbit the nucleus. It is made inside the nucleus at the moment the neutron changes, and then it leaves.
Now count. The nucleus has one neutron fewer and one proton more. The total number of nucleons is the same, so $A$ is unchanged, and $Z$ goes up by 1.
As $Z$ has changed, this is also a different element: the one whose proton number is one higher than before.
This is the decay that reduces the excess of neutrons. There is one neutron fewer and one proton more, so the neutron number falls and comes closer to the proton number. The nucleus becomes more stable.
In an equation, the electron is written $_{}^{}_{- 1}^{0} e$ (or $_{}^{}_{- 1}^{0} β$ ). The top number is 0 because an electron is not a nucleon. The bottom number is $- 1$ because that is its charge. In general:
$_{}^{}_{Z}^{A} X \overset{}{\to}_{}^{}_{Z + 1}^{A} Y +_{}^{}_{- 1}^{0} e$
You can check the neutron change itself in the same notation. A neutron is $_{}^{}_{0}^{1} n$ and a proton is $_{}^{}_{1}^{1} p$ . The top numbers give $1 = 1 + 0$ and the bottom numbers give $0 = 1 + (- 1)$ , so it balances.
$_{}^{}_{0}^{1} n \overset{}{\to}_{}^{}_{1}^{1} p +_{}^{}_{- 1}^{0} e$

Part 5 of 6

γ-emission: energy leaves, nothing else changes

γ-radiation is electromagnetic radiation. It has no mass and no charge, so in an equation it is written $_{}^{}_{0}^{0} γ$ .
After an α- or β-decay, the new nucleus may be left with extra energy. It gets rid of that energy by emitting γ-radiation. This is why γ can come out together with an α- or β-particle.
Because a γ-ray carries away no protons and no neutrons, $A$ and $Z$ are both unchanged. The nucleus is the same element and the same isotope. It just has less energy, which makes it more stable.
For example, a technetium-99 nucleus with extra energy emits a γ-ray and is still technetium-99:
$_{}^{}_{43}^{99} Tc \overset{}{\to}_{}^{}_{43}^{99} Tc +_{}^{}_{0}^{0} γ$
So of the three, only α-decay and β-decay change the element. γ-emission on its own never does.
A nucleus made by a decay can still be unstable and decay again, so one decay can be followed by another. A chain like this goes on until it reaches a stable nucleus. Uranium-238, for example, goes through a chain of α- and β-decays that ends at lead-206, which is stable.

Part 6 of 6

Writing and checking decay equations

Every symbol in a decay equation carries two numbers: the nucleon number on top and the proton number (or the charge, for an electron) at the bottom. Write both numbers on every symbol, including the emitted particle.
Rule 1: the top numbers balance. The total nucleon number on the left equals the total on the right, because no nucleons are made or destroyed.
Rule 2: the bottom numbers balance. The total on the left equals the total on the right, because charge is not made or destroyed. This is why the $- 1$ on the β-particle matters: it balances the extra $+ 1$ in the new nucleus.
To find a missing nucleus, take the parent's numbers and subtract the particle's numbers: top minus top, bottom minus bottom.
To find a missing particle, work out what its top and bottom numbers must be to make both rules hold, then recognise it: 4 and 2 is an α-particle, 0 and $- 1$ is a β-particle, 0 and 0 is γ.
There is no periodic table in a Physics paper, so you cannot look up which element has a given proton number. The question gives you the symbol or the name of the new element; your job is to get the two numbers right.
A good way to picture the three decays is as moves on a grid, with proton number across and neutron number up. α moves the nucleus 2 to the left and 2 down. β moves it 1 to the right and 1 down. γ does not move it at all.
graph diagram — not drawn yet
A square grid with proton number Z on the horizontal axis and neutron number N on the vertical axis; a parent nucleus marked as a dot at (Z, N); an arrow labelled α from the dot to (Z − 2, N − 2); an arrow labelled β from the dot to (Z + 1, N − 1); and a label γ beside the parent dot reading 'no move'.
α takes away 2 protons and 2 neutrons, so it moves diagonally down to the left. β swaps one neutron for one proton, so it moves one step down and one step right. γ leaves the dot where it is.

Worked examples

Example 1. Radium-226 has proton number 88. It decays by emitting an α-particle and forms radon (Rn). Write the decay equation, and state why radon is a different element from radium.

Show the working

Write the parent nucleus in nuclide notation first, so both of its numbers are in front of you: $_{}^{}_{88}^{226} Ra$ .
Write the α-particle with both of its numbers: $_{}^{}_{2}^{4} He$ .
Top numbers must balance, so the radon nucleon number is what is left after taking 4 away.
$A = 226 - 4 = 222$
Bottom numbers must balance, so the radon proton number is what is left after taking 2 away.
$Z = 88 - 2 = 86$
Write the full equation.
$_{}^{}_{88}^{226} Ra \overset{}{\to}_{}^{}_{86}^{222} Rn +_{}^{}_{2}^{4} He$
Check it: on the right, $222 + 4 = 226$ and $86 + 2 = 88$ , the same as on the left.
Radon is a different element because its proton number (86) is not the same as radium's (88), and the proton number is what decides the element.

Answer: $_{}^{}_{88}^{226} Ra \overset{}{\to}_{}^{}_{86}^{222} Rn +_{}^{}_{2}^{4} He$ . The proton number has changed from 88 to 86, and the proton number decides which element a nucleus is.

Example 2. Strontium-90 ( $_{}^{}_{38}^{90} Sr$ ) decays by β-emission to yttrium (Y). Write the decay equation. Then give the numbers of protons and neutrons before and after the decay, and explain how the decay changes the stability of the nucleus.

Show the working

In β-decay one neutron becomes a proton and an electron, and the electron leaves. So $A$ does not change and $Z$ goes up by 1.
Yttrium therefore has nucleon number 90 and proton number $38 + 1 = 39$ .
Write the equation, with the β-particle as $_{}^{}_{- 1}^{0} e$ .
$_{}^{}_{38}^{90} Sr \overset{}{\to}_{}^{}_{39}^{90} Y +_{}^{}_{- 1}^{0} e$
Check the bottom numbers: $39 + (- 1) = 38$ . The top numbers: $90 + 0 = 90$ . Both balance.
Before: 38 protons, and $90 - 38 = 52$ neutrons.
After: 39 protons, and $90 - 39 = 51$ neutrons.
So the nucleus has one neutron fewer and one proton more. Its excess of neutrons has gone down, and the nucleus is more stable.

Answer: $_{}^{}_{38}^{90} Sr \overset{}{\to}_{}^{}_{39}^{90} Y +_{}^{}_{- 1}^{0} e$ . Before: 38 protons and 52 neutrons. After: 39 protons and 51 neutrons. A neutron has changed into a proton, so the excess of neutrons is reduced and the nucleus is more stable.

Example 3. Americium-241 decays to neptunium-237, emitting a particle X and γ-radiation: $_{}^{}_{95}^{241} Am \overset{}{\to}_{}^{}_{93}^{237} Np + X +_{}^{}_{0}^{0} γ$ . Identify X.

Show the working

Start with the γ-radiation. It has 0 on top and 0 at the bottom, so it changes neither total. You can ignore it for the balancing.
Top numbers: X must supply the nucleons that are missing on the right.
$241 - 237 = 4$
Bottom numbers: X must supply the charge that is missing on the right.
$95 - 93 = 2$
A particle with nucleon number 4 and proton number 2 is two protons and two neutrons: an α-particle, $_{}^{}_{2}^{4} He$ .
Notice that this one decay gives out an α-particle and γ-radiation. This is the 'and/or' in the definition of radioactive decay.

Answer: X is an α-particle, $_{}^{}_{2}^{4} He$ (nucleon number 4, proton number 2).

Example 4. A uranium-238 nucleus ( $_{}^{}_{92}^{238} U$ ) emits an α-particle and forms thorium (Th). The thorium nucleus emits a β-particle and forms protactinium (Pa). The protactinium nucleus then emits a β-particle. Determine the nucleon number and proton number of the final nucleus, and identify the element.

Show the working

Track the two numbers one decay at a time. Do not try to do all three in one go, because that is where an α gets counted as a β.
α-decay: $A$ down 4, $Z$ down 2. So thorium has $A = 238 - 4 = 234$ and $Z = 92 - 2 = 90$ .
First β-decay: $A$ unchanged, $Z$ up 1. So protactinium has $A = 234$ and $Z = 90 + 1 = 91$ .
Second β-decay: $A$ unchanged, $Z$ up 1 again. So the final nucleus has $A = 234$ and $Z = 91 + 1 = 92$ .
The proton number is 92 again, the same as at the start. The element is decided by the proton number, so the final nucleus is uranium.
It is a different isotope, though: it has 4 fewer nucleons, so $_{}^{}_{92}^{234} U$ rather than $_{}^{}_{92}^{238} U$ . Each single α- or β-decay changed the element; it is only the three together that brought the proton number back to 92.

Answer: Nucleon number 234, proton number 92: uranium-234, $_{}^{}_{92}^{234} U$ , an isotope of the element the chain started from.

Where people go wrong

Mistake: Writing the β-particle as $_{}^{}_{1}^{0} e$ , or making the proton number go down in β-decay.

Instead: The β-particle is $_{}^{}_{- 1}^{0} e$ , and the proton number of the nucleus goes up by 1.

A neutron turns into a proton, so the nucleus gains a proton. The bottom numbers still balance because $(Z + 1) + (- 1) = Z$ .

Mistake: Thinking the β-particle is one of the electrons that orbit the nucleus.

Instead: It is made inside the nucleus when a neutron changes into a proton and an electron, and it is then emitted.

Radioactive decay is a change in the nucleus. Losing an orbiting electron would only make the atom a positive ion; it would not change the proton number or the element.

Mistake: Changing the element symbol, or one of the numbers, after γ-emission.

Instead: After γ-emission the nucleus has the same nucleon number, the same proton number and the same symbol. Only its energy has changed.

γ-radiation has no mass and no charge, $_{}^{}_{0}^{0} γ$ , so it takes nothing away from either total.

Mistake: Saying that α-decay reduces the excess of neutrons.

Instead: α-decay removes 2 neutrons and 2 protons, so the difference between them does not change. It is β-decay that reduces the excess of neutrons.

β-decay changes a neutron into a proton: one fewer neutron and one more proton. α-decay makes a too-heavy nucleus lighter, which is a different effect.

Mistake: Using 'spontaneous' and 'random' as if they meant the same thing.

Instead: Spontaneous: nothing outside causes it, and conditions such as temperature do not trigger or stop it. Random: you cannot predict when a particular nucleus will decay, or which one decays next.

They answer two different questions, so a one-word answer for one does not cover the other.

Check yourself

State what is meant by saying that radioactive decay is random.
Answer
You cannot predict when a particular nucleus will decay, or which nucleus in a sample will decay next.
A radioactive sample is heated from $20$ °C to $200$ °C. Does the heating change how its nuclei decay? Give the word that describes this property of decay.
Answer
No. The nuclei go on decaying exactly as before. Decay is spontaneous: nothing from outside, including heating, makes a nucleus decay or stops it.
A nucleus emits a β-particle. State why the new nucleus is a different element.
Answer
Its proton number has gone up by 1, and the proton number decides the element.
Hydrogen-1 ( $_{}^{}_{1}^{1} H$ ) is stable but hydrogen-3 ( $_{}^{}_{1}^{3} H$ ) is radioactive. Suggest why.
Answer
Hydrogen-3 has 2 neutrons but only 1 proton. It has an excess of neutrons, so its nucleus is unstable.
State the change that happens in the nucleus during β-emission, and state which particle leaves the nucleus.
Answer
A neutron changes into a proton and an electron. The electron leaves as the β-particle; the proton stays in the nucleus.
Describe the effect of γ-emission on the nucleon number, the proton number and the stability of the nucleus.
Answer
The nucleon number and proton number do not change. The nucleus loses energy, so it becomes more stable.
Complete the equation by filling in the two boxes: $_{}^{}_{84}^{210} Po \overset{}{\to}_{}^{}_{□}^{□} Pb +_{}^{}_{2}^{4} He$ .
Answer
$_{}^{}_{82}^{206} Pb$ : nucleon number $210 - 4 = 206$ , proton number $84 - 2 = 82$ .
Iodine-131 ( $_{}^{}_{53}^{131} I$ ) decays by β-emission to xenon (Xe). Write the decay equation.
Answer
$_{}^{}_{53}^{131} I \overset{}{\to}_{}^{}_{54}^{131} Xe +_{}^{}_{- 1}^{0} e$

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Thorium-232 ( $_{}^{}_{90}^{232} Th$ ) is radioactive. It decays by emitting an α-particle and forms radium (Ra). The radium nucleus is also radioactive and decays by β-emission to form actinium (Ac).

(a) State what is meant by saying that the decay of thorium-232 is random. 1 mark No calculator

Mark scheme

B1 it cannot be predicted when a particular nucleus will decay / which nucleus will decay next

(b) Complete the decay equation for the α-decay of thorium-232 by filling in the three boxes: $_{}^{}_{90}^{232} Th \overset{}{\to}_{}^{}_{□}^{□} Ra + □$ 3 marks No calculator

Mark scheme

B1 nucleon number of Ra is 228
B1 proton number of Ra is 88
B1 the α-particle written as ⁴₂He or ⁴₂α, with both numbers

(c) Describe the change that happens inside the radium nucleus when it emits a β-particle. 2 marks No calculator

Mark scheme

B1 a neutron changes into a proton and an electron
B1 the electron is emitted (as the β-particle) and the proton stays in the nucleus

(d) Deduce the nucleon number and the proton number of the actinium nucleus that is formed. 2 marks No calculator

Mark scheme

B1 nucleon number 228 (unchanged); ecf from (b)
B1 proton number 89 (one more than radium); ecf from (b)

(e) State why actinium is a different element from radium. 1 mark No calculator

Mark scheme

B1 the proton number has changed / it has a different number of protons (89, not 88)

Question 2. Carbon-12 ( $_{}^{}_{6}^{12} C$ ) is stable. Carbon-14 ( $_{}^{}_{6}^{14} C$ ) is radioactive. It decays by β-emission to nitrogen-14 ( $_{}^{}_{7}^{14} N$ ), which is stable.

(a) Suggest why carbon-14 is radioactive when carbon-12 is not. 2 marks No calculator

Mark scheme

B1 carbon-14 has 8 neutrons and carbon-12 has 6 / carbon-14 has two more neutrons, with the same number of protons
B1 carbon-14 has an excess of neutrons (too many neutrons for its 6 protons), so its nucleus is unstable, owtte

(b) Describe the effect of this β-decay on the nucleus. 3 marks No calculator

Mark scheme

B1 the number of neutrons decreases by 1 / from 8 to 7
B1 the number of protons increases by 1 / from 6 to 7 (nucleon number unchanged at 14 not required); accept 'a neutron changes into a proton (and an electron)' for both of these first two marks
B1 the excess of neutrons is reduced / the nucleus becomes more stable

Question 3. A technetium-99 nucleus ( $_{}^{}_{43}^{99} Tc$ ) that has extra energy emits γ-radiation. A polonium-210 nucleus ( $_{}^{}_{84}^{210} Po$ ) emits an α-particle and forms lead (Pb).

(a) State the nucleon number and the proton number of the technetium nucleus after it has emitted the γ-radiation. 1 mark No calculator

Mark scheme

B1 nucleon number 99 and proton number 43 (both unchanged); both needed

(b) Compare the effect on the nucleus of the γ-emission by technetium-99 with the effect of the α-emission by polonium-210. 3 marks No calculator

Mark scheme

B1 similarity: both nuclei lose energy / both become more stable
B1 α-emission: nucleon number decreases by 4 (to 206) and proton number decreases by 2 (to 82)
B1 γ-emission: no change in nucleon number or proton number, so technetium stays the same element, while polonium becomes a different element (lead), ORA

How this is examined

Completing a decay equation in nuclide notation: finding the missing nucleus or the missing particle: a two- or three-mark 'complete the equation' question, with gaps for the nucleon number, the proton number or the emitted particle; the element symbol may already be printed, so that only the numbers are missing Top numbers that add up to the same total on both sides, and bottom numbers that do the same. The particle written with both numbers: α as $_{}^{}_{2}^{4} He$ (or $_{}^{}_{2}^{4} α$ ), β as $_{}^{}_{- 1}^{0} e$ (or $_{}^{}_{- 1}^{0} β$ ), γ as $_{}^{}_{0}^{0} γ$ . A β with $+ 1$ at the bottom, or a symbol with a number missing, is not correct.
What each emission does to the nucleon number and proton number, and why the nucleus ends up more stable: a table to complete for α, β and γ, or a two- or three-mark Describe or Compare about one or two named decays α: nucleon number down 4, proton number down 2. β: nucleon number unchanged, proton number up 1 (one neutron fewer). γ: neither changes; the nucleus loses energy. For stability, say 'the nucleus becomes more stable', and for β add 'the excess of neutrons is reduced'.
The change inside the nucleus during β-decay: a two-mark Describe or State, sometimes after a decay equation Two points: 'a neutron changes into a proton and an electron' and 'the electron is emitted (as the β-particle) and the proton stays in the nucleus'. 'An electron leaves the atom' with no neutron change does not describe the nucleus.
What 'spontaneous' and 'random' each mean for decay: a one- or two-mark State or Explain, which may refer to a count rate that goes up and down Random: 'you cannot predict when a particular nucleus will decay' or 'which nucleus will decay next'. Spontaneous: 'it happens without any outside cause', 'it is not affected by conditions such as temperature'. Each word needs its own meaning.
Why one isotope of an element is radioactive when another is stable: a one- or two-mark Suggest or Explain that gives you a stable and a radioactive isotope of the same element in nuclide notation The neutron count worked out from $A - Z$ and compared ('it has more neutrons, with the same number of protons'), then the reason: 'an excess of neutrons, so the nucleus is unstable'. For a very large nucleus, 'the nucleus is too heavy'.
Why the product of α- or β-decay is a different element: a one-mark State, after an equation 'The proton number has changed' (with the two values if you have them). γ-emission alone does not change the element.

Words to know

Radioactive decay: A change in an unstable nucleus, in which it emits an α-particle or a β-particle and/or γ-radiation.
Unstable nucleus: A nucleus whose arrangement of protons and neutrons cannot last, so at some point it decays.
Spontaneous: Happening without any outside cause. Conditions such as temperature do not make a nucleus decay or stop it decaying.
Random: Not predictable for a single nucleus: you cannot say when a particular nucleus will decay, or which one will decay next.
Excess of neutrons: Too many neutrons for the number of protons in the nucleus, which makes the nucleus unstable. β-decay reduces it.
α-decay: Emission of an α-particle, $_{}^{}_{2}^{4} He$ : the nucleon number falls by 4 and the proton number by 2.
β-decay: A neutron in the nucleus changes into a proton and an electron, and the electron is emitted, $_{}^{}_{- 1}^{0} e$ : the nucleon number is unchanged and the proton number rises by 1.
γ-emission: Emission of electromagnetic radiation, $_{}^{}_{0}^{0} γ$ , from a nucleus with extra energy: the nucleon number and proton number do not change.
Decay equation: An equation in nuclide notation for a decay, in which the top numbers and the bottom numbers each add up to the same total on both sides.

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