Study guide for chapter 28
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Study Guide for Chapter 28. Nuclear Chemistry. 1. Positron. A. - particle of charge +1 and mass equal to that of an electron. 2. Alpha particle. D. Emitted helium nucleus. 3. Beta particle. F. Energetic electron from a decomposed neutron. 4. Transuranium elements.

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Study Guide for Chapter 28

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Study guide for chapter 28

Study Guide for Chapter 28

Nuclear Chemistry


1 positron

1. Positron

  • A. - particle of charge +1 and mass equal to that of an electron.


2 alpha particle

2. Alpha particle

  • D. Emitted helium nucleus.


3 beta particle

3. Beta particle

  • F. Energetic electron from a decomposed neutron.


4 transuranium elements

4. Transuranium elements

  • E. Element with atomic number greater than 92


5 gamma radiation

5. Gamma radiation

  • B. High energy electromagnetic radiation


6 transmutation

6. transmutation

  • C. conversion of an atom of one element to an atom of another element


7 fission

7. fission

  • B. Splitting of nucleus into two similar – sized pieces.


8 fusion

8. Fusion

  • C. combination of two nuclei to form a large nucleus


9 radioisotope

9. Radioisotope

  • A. Element with unstable nucleus


10 what is the charge on an alpha particle

10. What is the charge on an alpha particle?

  • 42 He - +2

Lower number is the atomic number.

This represents the number of protons.

Each proton has a +1 charge.


How many neutrons are there in an alpha particle 4 2 he

How many neutrons are there in an alpha particle?42 He

Mass #

Atomic #

  • Neutrons = Mass # - Atomic #

Neutrons = 4 – 2 = 2


12 what is the change in the atomic number when an atom emits an alpha particle

12. What is the change in the atomic number when an atom emits an alpha particle.

This is an alpha particle 42 He

Atomic number = 2

So the atomic number would

be reduced by 2.


13 what is the change in atomic mass when an atom emits an alpha particle

13. What is the change in atomic mass when an atom emits an alpha particle.

This is an alpha particle 42 He

The mass number is 4, so the mass

Would be reduced by 4.


14 what is the change in the atomic number when an atom emits a beta particle

14. What is the change in the atomic number when an atom emits a beta particle.

This is an beta particle: 0-1 e

When a neutron loses a beta particle,

the neutron turns into a proton.

Thus increasing the atomic number

by 1.


15 what is the change in the atomic number when an atom emits gamma radiation

15. What is the change in the atomic number when an atom emits gamma radiation?

This is an gama particle 00 γ

So, the atomic number of the atom does not change.


16 what particle is emitted in alpha radiation

16. What particle is emitted in alpha radiation?

an alpha particle also known as a helium nucleus. 42 He


17 which symbol is used for an alpha particle

17. Which symbol is used for an alpha particle.

42 He


18 what is the minimum thickness needed to stop an alpha particle

18. What is the minimum thickness needed to stop an alpha particle.

  • A sheet of paper can stop an alpha particle.

  • Alpha particles are the weakest form of radiation.


19 what symbol is used for beta radiation

19. What symbol is used for beta radiation?

  • 0-1 e


20 what is the minimum thickness needed to stop a beta particle

20. What is the minimum thickness needed to stop a beta particle.

  • A sheet of aluminum foil.


21 what is the minimum thickness needed to stop gamma radiation

21. What is the minimum thickness needed to stop gamma radiation

  • Three inches of lead.


22 the most penetrating form of radiation is

22. The most penetrating form of radiation is -

  • gamma rays


23 which type of ionizing radiation can be blocked by clothing

23. Which type of ionizing radiation can be blocked by clothing?

  • Alpha particles


Study guide for chapter 28

24. If the half life of a radioactive material is 8 years, how many years will it take for one half of the original amount to decay.

  • 8 years


Study guide for chapter 28

25. A piece of wood found in an ancient burial mound contains only half as much carbon-14 as a piece of wood cut from a living tree growing nearby. If the half-life for carbon-14 is 5730 years, what is the approximate age of the ancient wood.


Study guide for chapter 28

This problem tells us that the ancient piece of wood has half the amount of Carbon-14 than a fresh sample. They also told us that the half life for carbon-14 is 5730 years.Therefore the wood must be 5730 years old.


26 after 42 days 2 g of phosphorous 32 has decayed to 25 g what is the half life phosphorous 32

26. After 42 days, 2 g of phosphorous-32 has decayed to .25 g. What is the half-life phosphorous-32.

The first thing we need to do is

to calculate the rate of decay. “K”


Study guide for chapter 28

Nt

N0

= -kt

ln

Amount at time “t” (time passed)

Time that passed

Amount at time 0

K = the decay rate constant


Study guide for chapter 28

.25 g

2 g

= -k(42 days)

ln

Amount at time “t” (time passed)

Time that passed

Amount at time 0

k = the decay rate constant


Study guide for chapter 28

.25 g

2 g

.25 g

2 g

= -k(42 days)

= -k(42 days)

ln

ln

Now do the math.

-2.08 = k(-42)

-42

-42

k = .0495


Study guide for chapter 28

0.693

k

0.693

.0495

= t1/2

= t1/2

  • Now we plug our rate constant “k” into our half-life formula.

Now do the math.

t1/2 = 14 days


27 if the half life of sodium 24 is 15 hours how much remains from a 10 0 g sample after 60 hours

27. If the half-life of sodium-24 is 15 hours, how much remains from a 10.0 g sample after 60 hours?

  • This problem is similar to the last problem.

  • Except this time we need to start with the half-life formula to figure out our rate constant “k”.


Study guide for chapter 28

0.693

k

0.693

15

= 15 hours

= k

  • We begin by plugging the half-life into the half-life equation.

Rearrange, and

do the math.

k= .0462


Study guide for chapter 28

Now we plug our givens and “k” into

this formula.

Nt

N0

= -kt

ln

Amount at time “t” (time passed)

Time that passed

Amount at time 0

K = the decay rate constant


Study guide for chapter 28

(?) g

10 g

= -.0462(60 hours)

ln

Amount at time “t” (time passed)

Amount at time 0

Time that passed

k = the decay rate constant


Study guide for chapter 28

Now do the math.

(?) g

10 g

(?) g

10 g

(?) g

10 g

= -.0462(60 hours)

= -2.772

= e-2.772

ln

ln

(?)g = e-2.772 (10 g)

(?) = .625 grams


28 what particle is needed to complete this nuclear reaction

28. What particle is needed to complete this nuclear reaction?

22286Rn21884Po + _____

To do these problems we have to make

sure that the superscripts on both sides

of the arrow add up to the same number.

We then need to make sure that the

subscripts on both sides of the arrow

add up the same number.


28 what particle is needed to complete this nuclear reaction1

28. What particle is needed to complete this nuclear reaction?

22286Rn21884Po + _____

So for the superscripts:

222 = 218 + ?

? = 4

And for the subscripts

? = 2

86 = 84 + ?


28 what particle is needed to complete this nuclear reaction2

28. What particle is needed to complete this nuclear reaction?

22286Rn21884Po + _____

Which radioactive particle has a

mass of 4 and a +2 charge?

? = 4

Alpha particle = 42He

? = 2


29 what particle is needed to complete this nuclear reaction

29. What particle is needed to complete this nuclear reaction?

5525Mn + 21H ____ + 2 10 N

To do these problems we have to make

sure that the superscripts on both sides

of the arrow add up to the same number.

We then need to make sure that the

subscripts on both sides of the arrow

add up the same number.


29 what particle is needed to complete this nuclear reaction1

29. What particle is needed to complete this nuclear reaction?

5525Mn + 21H ____ + 2 10 N

This means there are two neutrons. If you like you could rewrite the equation like this:

5525Mn + 21H _?_ + 10 N + 10 N


Study guide for chapter 28

5525Mn + 21H _?_ + 10 N + 10 N

So for the superscripts:

55 + 2 = ? + 1 + 1

? = 55

And for the subscripts

? = 26

25 + 1 = ? + 0 + 0


Study guide for chapter 28

So which element has an atomic number

of 24?

55

?

26

? = Fe or Iron


30 what particle is needed to complete this nuclear reaction

30. What particle is needed to complete this nuclear reaction?

147N + ____146C + 11H

To do these problems we have to make

sure that the superscripts on both sides

of the arrow add up to the same number.

We then need to make sure that the

subscripts on both sides of the arrow

add up the same number.


30 what particle is needed to complete this nuclear reaction1

30. What particle is needed to complete this nuclear reaction?

147N + ____146C + 11H

So for the superscripts:

14 + ? = 14 + 1

? = 1

And for the subscripts

? = 0

7 + ? = 6 + 1


Study guide for chapter 28

So which radioactive particle has a mass

of 1 and a charge of 0?

1

?

0

1

n

? = a neutron

0


31 what element will polonium 214 become when it loses an alpha particle

31. What element will polonium-214 become when it loses an alpha particle?

21484Po 42He + _?_

So for the superscripts:

214 = 4 + ?

? = 210

And for the subscripts

? = 82

84 = 2 + ?


Study guide for chapter 28

So which element has an atomic number

of 82?

210

?

82

? = Pb or Lead


32 to what does plutonium 239 decay when it loses an alpha particle

32. To what does plutonium-239 decay when it loses an alpha particle?

23994Pu 42He + _?_

So for the superscripts:

239 = 4 + ?

? = 235

And for the subscripts

? = 92

94 = 2 + 92


Study guide for chapter 28

So what element has an atomic number

of 92?

235

?

92

235

? = Uranium

U

92


Above what atomic number are all atoms radioactive

Above what atomic number are all atoms radioactive?

82


What type of radioactive decay will occur in an atom of an atomic number 87 and atomic mass of 224

What type of radioactive decay will occur in an atom of an atomic number 87 and atomic mass of 224.

Alpha Decay


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