Lecture 2 Atomic Theory II Tutorial - AP...

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Lecture 2 Atomic Theory II Tutorial 1) A photon travels with a frequency of 8.57 x 10 14 Hz. a. Find the wavelength of this photon in nm? 8 14 1 7 2 3 10 m/s 8.57 10 s 3.50 10 m 3.50 10 nm c c λ ν λ ν λ λ λ = = × = × = × = × b. Find the energy contained by a single photon. Put your answer in kJ. 34 14 1 19 16 = = (6.63x10 Js) (8.57 x 10 s ) = 5.68 x 10 J = 5.68 x 10 kJ E h E E E ν 2) An electron drops from n = 5 to n = 3 in a hydrogen atom. a. Find the energy contained by the photon that is released during this transition. 18 5 2 20 5 2.178 10 J 5 8.712 10 J E E × = =− × 18 3 2 19 3 2.178 10 J 3 2.420 10 J E E × = =− × 3 5 19 20 19 = = 2.420 10 J ( 8.712 10 J) = 1.549 10 J E E E E E Δ Δ × −− × Δ × b. Find the frequency of the electromagnetic radiation that is released. 19 14 1 34 1.549 10 J 2.34 10 E s 6.63 10 Js E h h ν ν × = = = × = × c. At what speed does this electromagnetic radiation travel? (3x10 8 m/s) The speed of electromagnetic waves (light) is a constant. www.apchemsolutions.com © 2009, 2008 AP Chem Solutions. All rights reserved. 1

Transcript of Lecture 2 Atomic Theory II Tutorial - AP...

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Lecture 2 Atomic Theory II

Tutorial

1) A photon travels with a frequency of 8.57 x 1014 Hz.

a. Find the wavelength of this photon in nm?

8

1 4 1

7

2

3 1 0 m /s8.5 7 10 s3 .5 0 1 0 m3 .5 0 1 0 n m

ccλ ν

λν

λ

λ

λ

=

=

×=

×= ×

= ×

b. Find the energy contained by a single photon. Put your answer in kJ.

34 14 1

19

16

= = (6.63x10 Js) (8.57 x 10 s ) = 5.68 x 10 J = 5.68 x 10 kJ

E hEEE

ν− −

2) An electron drops from n = 5 to n = 3 in a hydrogen atom.

a. Find the energy contained by the photon that is released during this transition.

18

5 2

205

2.178 10 J5

8.712 10 J

E

E

− ×=

= − ×

18

3 2

193

2.178 10 J3

2.420 10 J

E

E

− ×=

= − ×

3 5

19 20

19

=

= 2.420 10 J ( 8.712 10 J)= 1.549 10 J

E E E

EE

− −

Δ −

Δ − × − − ×

Δ − ×

b. Find the frequency of the electromagnetic radiation that is released.

1914 1

34

1.549 10 J 2.34 10E s

6.63 10 Js

E h

h

ν

ν−

−−

×= = = ×

=

×

c. At what speed does this electromagnetic radiation travel?

(3x108 m/s) The speed of electromagnetic waves (light) is a constant.

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d. What is the wavelength of this electromagnetic radiation?

8

1 4 1

6

3

3 .0 0 1 0 m /s2 .3 4 1 0 s

1 .2 8 1 0 m1 .2 8 1 0 n m

ccλ ν

λν

=

=

λ

λ

λ

×=

×= ×

×

=

3) An electron is traveling with a velocity of 3.92 x 106 m/s. (mass e- = 9.11 x 10-31 kg)

a. What is its wavelength in nm?

347 2

34

6.63 10 Js 1.86 10 m 1.86 10 nm(9.11 10 kg)(3.92

hmv

λ−

−−

×= = = × = ×

× 6 /s)10 m×

b. What is its frequency?

8

-7

1 5 1

3 1 0 m /s1 .8 6 × 1 0 m1 .6 1 1 0 s

ccλ ν=

=νλ

ν

ν −×

×

=

=

4) Find the momentum of an electron with a wavelength of 232 nm.

3427

9

6.63 10 Js 2.86 10 kg m/s232 10 m

mvh h

mv

ρ

λρ

hρλ

−−

=

= = = × ⋅ii

= =

5) How many possible orientations exist in a p-sublevel?

The angular momentum quantum number (l) for a p-sublevel is 1. The magnetic

uantum numbers range from - l to + l . For a p-sublevel the magnetic quantum ientations in the

-sublevel.

qnumbers are -1, 0, and +1. Thus, there are THREE possible orp

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6) What is the lowest principal quantum number that can have p-orbitals?

n-1 = l n = l +1

n = 2

n = 1 + 1

7) If n=5: a. What are the possible angular momentum numbers?

The angular momentum numbers range from 0 to n-1. If n = 5 the possible angular momentum numbers are:

0, 1, 2, 3, and 4.

b. What do these numbers represent? The angular momentum numbers rep r type of orbital. 0 is an s-orbital, is a p-orbital, 2 is a d-orbital, and 3 is an f-orbital.

resent the shape o1

8) If l = 3:

-3, -2, -1, 0, 1, 2, and 3

a. What are the possible magnetic quantum numbers?

These numbers define th e-dimensional space. In this ase, there are seven different orientations (an f-sublevel).

b. What do these numbers represent?

e orientation of the orbital in threc

9) What are the possible angular momentum, magnetic, and electron spin quantum

= 0 (An s-orbital is defined by l = 0) m =ms= + ½

numbers associated with a single electron in the s-orbital of n = 2.

l - l to + l , so m = 0 (there is only one orientation for an s-orbital)

or – ½ (the one electron could be spin up or spin down.)

omentum, magnetic, and electron spin quantum .

= 3 (l = 3 defines an f-orbital) m =for f-orms= + ½ or – ½ (the one electron could be spin up or spin down.)

10) What are the possible angular mnumbers associated with a single electron in an f-orbital of n = 5

l

- l to + l , so m = -3, -2, -1, 0, +1, +2, or +3 (there are seven possible orientations bitals)

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