Physics 204 – Section 9
QUIZ 10
21 April 2014
Name:
1. (10 total points) When light with a wavelength of 100 nm is incident on a certain
metal surface, electrons are ejected with a maximum kinetic energy of 7.308 eV.
(a) (2 points) Find the energy of a photon with a wavelength of 100 nm.
Solution: The energy of a photon is
E = hf =
hc
(4.136 ⇥ 10 15 eV s)(3 ⇥ 108 m/s)
=
= 12.408 eV = 1.988 ⇥ 10 18 J.
l
100 ⇥ 10 9 m
(1)
(b) (3 points) Find the work function W0 for the metal.
Solution: For the photoelectron effect,
E = h f = W0 + KEmax ,
(2)
so the work function
W0 = E
KEmax = 12.408 eV
7.308 eV = 5.1 eV = 8.17 ⇥ 10
19
J.
(3)
(c) (2 points) Find the energy of a photon with a wavelength of 200 nm.
Solution: The energy of a photon with a wavelength of 200 nm is
E=
hc
(4.136 ⇥ 10 15 eV s)(3 ⇥ 108 m/s)
=
= 6.204 eV = 9.9399 ⇥ 10
l
200 ⇥ 10 9 m
19
J.
(4)
(d) (3 points) Find the maximum kinetic energy of the electrons if the wavelength
of the photons is increased to 200 nm
Solution: The maximum kinetic energy of the electrons will be
KEmax = E
E = hf =
hc
l
W0 = 6.204 eV
h f = KEmax + W0
l=
5.1 eV = 1.104 eV = 1.7688 ⇥ 10
h
p
(Dp x )(Dx )
h
4⇡
19
J.
(5)
h = 4.136 ⇥ 10
15
eV s
Physics 204 – Section 11
QUIZ 10
21 April 2014
Name:
7
1. (10 total points) An electron has a de Broglie wavelength of 5 ⇥ 10
m.
(a) (5 points) Find the momentum of the electron.
Solution: The de Broglie wavelength
h
h
6.626 ⇥ 10 34 J s
l=
=) p = =
= 1.3252 ⇥ 10
p
l
5 ⇥ 10 7 m
27
kg m/s.
(1)
(b) (5 points) What is the de Broglie wavelength if the electron’s momentum is
doubled?
Solution: Planck’s constant is constant (hence the name), so we can rearrange the equation for the de Broglie wavelength to h = lp. Therefore, for
any two electrons,
l1 p1 = l2 p2 ,
(2)
so
l2 = l1
E = hf =
hc
l
p1
= (5 ⇥ 10
p2
h f = KEmax + W0
l=
h
p
7
m)
1
= 2.5 ⇥ 10
2
(Dp x )(Dx )
h
4⇡
7
m.
(3)
h = 4.136 ⇥ 10
15
eV s
Physics 204 – Section 13
QUIZ 10
21 April 2014
Name:
1. (10 total points) When light with a wavelength of 200 nm strikes the surface of the
metal lithium, electrons are ejected with maximum a kinetic energy of 3.3 eV.
(a) (2 points) Find the energy of a photon with a wavelength of 200 nm.
Solution: The energy of those photons is
E = hf =
hc
(4.136 ⇥ 10 15 eV s)(3 ⇥ 108 m/s)
=
= 6.204 eV.
l
200 ⇥ 10 9 m
(1)
(b) (4 points) Find the work function for this metal.
Solution: Because h f = KEmax + W0 , the work function
W0 = h f
KEmax = 6.204 eV
3.3 eV = 2.904 eV.
(2)
(c) (4 points) Find the maximum wavelength that a photon can have and still be
able to eject an electron.
Solution: When a photon has the minimum energy necessary to eject an
electron, all of the energy will go to removing the electron from its atom
(the work function) leaving it with no kinetic energy. Therefore, h f = W0 .
Because f = c/l, this implies
hc
(4.136 ⇥ 10 15 eV s)(3 ⇥ 108 m/s)
l=
=
= 427.273 nm
W0
2.904 eV
E = hf
h f = KEmax + W0
l=
h
p
(Dp x )(Dx )
h
4⇡
h = 4.136 ⇥ 10
(3)
15
eV s
Physics 204 – Section 20
QUIZ 10
21 April 2014
Name:
1. (10 total points) The uncertainty of an electron’s position is Dx = 10
size of an atom).
10 m
(about the
(a) (5 points) Find the minimum uncertainty in momentum.
Solution: The Heisenberg Uncertainty Principle says that
h
,
4⇡
DxDp
(1)
which implies
Dpmin =
h
6.626 ⇥ 10 34 J s
=
= 5.2728 ⇥ 10
4⇡Dx
4⇡(10 10 m)
25
kg m/s.
(2)
(b) (5 points) Find the minimum uncertainty in the electron’s velocity.
Solution: Because p = mv, the uncertainty in momentum is related to the
uncertainty in velocity by Dp = mDv. Therefore, the minimum uncertainty
in the electron’s velocity
Dvmin =
E = hf =
hc
l
Dpmin
5.2728 ⇥ 10
=
me
9.1 ⇥ 10
h f = KEmax + W0
l=
h
p
25
kg m/s
= 579 429 m/s.
kg
31
(Dp x )(Dx )
h
4⇡
(3)
h = 4.136 ⇥ 10
15
eV s