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Session #3: Homework Solutions

Problem #1
From a standard radio dial, determine the maximum and minimum wavelengths
( λmax and λmin ) for broadcasts on the
(a) AM band
(b) FM band
Solution

c = νλ, ∴ λmin =

AM

λmin =

FM

λmin =

c
c
; λmax =
νmax
νmin

3 x 108m / s
1600 x 103Hz
3 x 108
108 x 106

= 188 m

= 2
...
02 x 1023 )

(a) ν (frequency) =

c
3 x 108 m/s
=
= 7
...
41 m

(b) ν (wavenumber) =

(c) λ = 408 x 10-9m x

1
1
=
= 2
...
63 x 10-34 x 7
...
02 x 1023 J/mole
= 2
...
09 x 1014 s–1) emitted by activated sodium,
determine:
(a) the wavelength ( λ ) in [m]
(b) the wave number ( ν ) in [cm–1]
(c) the total energy (in kJ) associated with 1 mole of photons
Solution
(a) The equation relating ν and λ is c = νλ where c is the speed of light = 3
...

λ =

c
3
...
89 x 10-7m
14
-1
ν
5
...
89 x 10–7 m x 100cm/m = 5
...
70 x 104 cm-1
-5
λ
5
...
We need to multiply the energy obtained by
Avogadro’s number to get the energy per mole of photons
...
62 x 10–34 J
...
09 x 1014 s-1

E = hν = (6
...
s) x (5
...
37 x 10-19 J per photon

This is the energy in one photon
...
023 x 1023 photons ⎞
E ⋅ NAv = (3
...
03 x 105 J per mole of photons
we get the energy per mole of photons
...
03 x 102 kJ
1000 J

2
...
The energy needed is 2
...
02 x 1023 electrons)
...
e
...


Ip
hν

This device should be called a phototube rather than a
photodiode – a solar cell is a photodiode
...
6 x 10-19 J
Erad = hν = (hc)/λ
The question is: below what threshold energy (h ν ) will a photon no longer be able to
generate a photoelectron?
2
...
02 x 10

photoelectrons

= 3
...
57 x 10-19

=

6
...
57 x 10-19

= 5
...
7102 x 10–5 cm, emitted by excited lithium atoms,
calculate:
(a) the frequency ( ν ) in s–1;
(b) the wave number ( ν ) in cm–1;
(c) the wavelength ( λ ) in nm;
(d) the total energy (in Joules) associated with 1 mole photons of the indicated
wavelength
...


For:

λ = 6
...
7102 x 10-7m

ν =

(b)

2
...
7102 x 10-7 m

1
1
=
= 1
...
4903 x 104 cm−1
-7
λ
6
...
7102 x 10-5 cm x

(d) E =

= 4
...
4677 Hz

1 nm
10-7 cm

= 671
...
62 x 10-34 Js x 2
...
7102 x 10-7 m

= 2
...
78 x 105 J/mole photons
Problem #6

Calculate the “Bohr radius” for He+
...
Correspondingly, the electron
energy (Eel) is given as:

Eel = -

Z2

me4

n2

8h2 ε o2

and the electronic orbit (rn):

h2εo

rn =

n2
Z

rn =

n2
a
Z o

πme2

For He+ (Z=2), r1 =

1
0
...
264Å
2
2

Problem #7

(a) Determine the atomic weight of He++ from the values of its constituents
...
(There is only one natural 42He
isotope
...
)
(a) The mass of the constituents (2p + 2n) is given as:
2p =
2 x 1
...
6952056 x 10-24 g
The atomic weight (calculated) in amu is given as:
6
...
660565 x 10-24 g

/ amu

He = 4
...
00260 (amu)
...
92841 x 10–2 amu, corresponding to 4
...

This mass defect appears as nuclear bond energy:
ΔE = 4
...
3765 x 10-12 J/atom

= 2
...
928 x 10-5 kg/mole = 0

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