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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

- E. Element with atomic number greater than 92

5. Gamma radiation

- B. High energy electromagnetic radiation

6. transmutation

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

7. fission

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

8. Fusion

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

9. Radioisotope

- A. Element with unstable nucleus

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?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.

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.

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.

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?

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? emits gamma radiation?

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

17. Which symbol is used for an alpha particle. emits gamma radiation?

42 He

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? particle.

- 0-1 e

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

- Three inches of lead.

22. The most penetrating form of radiation is - radiation

- gamma rays

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

- Alpha particles

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

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.

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.

The first thing we need to do is

to calculate the rate of decay. “K”

N g. What is the half-life phosphorous-32.t

N0

= -kt

ln

Amount at time “t” (time passed)

Time that passed

Amount at time 0

K = the decay rate constant

.25 g g. What is the half-life phosphorous-32.

2 g

= -k(42 days)

ln

Amount at time “t” (time passed)

Time that passed

Amount at time 0

k = the decay rate constant

.25 g g. What is the half-life phosphorous-32.

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

0.693 g. What is the half-life phosphorous-32.

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?

- 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”.

0.693 remains from a 10.0 g sample after 60 hours?

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

Now we plug our givens and “k” into remains from a 10.0 g sample after 60 hours?

this formula.

Nt

N0

= -kt

ln

Amount at time “t” (time passed)

Time that passed

Amount at time 0

K = the decay rate constant

(?) g remains from a 10.0 g sample after 60 hours?

10 g

= -.0462(60 hours)

ln

Amount at time “t” (time passed)

Amount at time 0

Time that passed

k = the decay rate constant

Now do the math. remains from a 10.0 g sample after 60 hours?

(?) 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?

22286Rn 21884Po + _____

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 reaction?

22286Rn 21884Po + _____

So for the superscripts:

222 = 218 + ?

? = 4

And for the subscripts

? = 2

86 = 84 + ?

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

22286Rn 21884Po + _____

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?

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 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

55 reaction?25Mn + 21H _?_ + 10 N + 10 N

So for the superscripts:

55 + 2 = ? + 1 + 1

? = 55

And for the subscripts

? = 26

25 + 1 = ? + 0 + 0

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 reaction?

147N + ____ 146C + 11H

So for the superscripts:

14 + ? = 14 + 1

? = 1

And for the subscripts

? = 0

7 + ? = 6 + 1

So which radioactive particle has a mass reaction?

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?

21484Po 42He + _?_

So for the superscripts:

214 = 4 + ?

? = 210

And for the subscripts

? = 82

84 = 2 + ?

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

Above what atomic number are all atoms radioactive? particle?

82

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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