Loading in 5 sec....

Study Guide for Chapter 28PowerPoint Presentation

Study Guide for Chapter 28

- 54 Views
- Uploaded on
- Presentation posted in: General

Study Guide for Chapter 28

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Study Guide for Chapter 28

Nuclear Chemistry

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

- D. Emitted helium nucleus.

- F. Energetic electron from a decomposed neutron.

- E. Element with atomic number greater than 92

- B. High energy electromagnetic radiation

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

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

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

- A. Element with unstable nucleus

- 42 He - +2

Lower number is the atomic number.

This represents the number of protons.

Each proton has a +1 charge.

Mass #

Atomic #

- Neutrons = Mass # - Atomic #

Neutrons = 4 – 2 = 2

This is an alpha particle 42 He

Atomic number = 2

So the atomic number would

be reduced by 2.

This is an alpha particle 42 He

The mass number is 4, so the mass

Would be reduced by 4.

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.

This is an gama particle 00 γ

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

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

42 He

- A sheet of paper can stop an alpha particle.
- Alpha particles are the weakest form of radiation.

- 0-1 e

- A sheet of aluminum foil.

- Three inches of lead.

- gamma rays

- Alpha particles

- 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.

The first thing we need to do is

to calculate the rate of decay. “K”

Nt

N0

= -kt

ln

Amount at time “t” (time passed)

Time that passed

Amount at time 0

K = the decay rate constant

.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

.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

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

- 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

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

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

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.

(?) 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

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.

22286Rn21884Po + _____

So for the superscripts:

222 = 218 + ?

? = 4

And for the subscripts

? = 2

86 = 84 + ?

22286Rn21884Po + _____

Which radioactive particle has a

mass of 4 and a +2 charge?

? = 4

Alpha particle = 42He

? = 2

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.

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

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

So for the superscripts:

55 + 2 = ? + 1 + 1

? = 55

And for the subscripts

? = 26

25 + 1 = ? + 0 + 0

So which element has an atomic number

of 24?

55

?

26

? = Fe or Iron

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.

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

of 1 and a charge of 0?

1

?

0

1

n

? = a neutron

0

21484Po 42He + _?_

So for the superscripts:

214 = 4 + ?

? = 210

And for the subscripts

? = 82

84 = 2 + ?

So which element has an atomic number

of 82?

210

?

82

? = Pb or Lead

23994Pu 42He + _?_

So for the superscripts:

239 = 4 + ?

? = 235

And for the subscripts

? = 92

94 = 2 + 92

So what element has an atomic number

of 92?

235

?

92

235

? = Uranium

U

92

82

Alpha Decay