photoelectric calculations
DESCRIPTION
Worked examples for calculations regarding the photoelectric effectTRANSCRIPT
Photoelectric Effect
Calculations
ENERGY PER PHOTONE=hf
E = hf
• E = Energy per photon• f = frequency of light• h = Plank’s constant = 6,63 x 10-34 J·s
Question 1
How much energy does each photon of UV light of wavelength 288 nm have?
Solution: Given and required
Required: EGiven: = 288 nmKnown: h = 6,63 x 10-34 J·s
Equation: E = hf
Solution: how to find f
Solution: finding f
E = hf
Given: = 288 nmKnown: nm = 10-9 mTherefore 288 nm = 288 x 10-9 m
Solution: finding f
E = hf
Given: = 288 nmKnown: nm = 10-9 mTherefore 288 nm = 288 x 10-9 ms-1 = = Hz
Solution: finding f
E = hf
f = 1,04 x 1015 Hz
Solution: finding E
E = hf
Required: EKnown: h = 6,63 x 10-34 J·s
f = 1,04 x 1015 Hz
Solution: finding E
E = hf = 6,63 x 10-34 J·s x 1,04 x 1015 Hz = 6,63 x 10-34 J·s x 1,04 x 1015 s-1
= 6,91 x 10-19 J
PHOTON
PHOTONcarries amount of
Energy = hf
PHOTONcarries amount of
Energy f
PHOTONcarries amount of
Energy = hf
PHOTONcarries amount of
Energy = hf
- - - - - - - -
give
s AL
L to
ON
E
inELECTRON
metal
in metal
PHOTONcarries amount of
Energy = hf
- - - - - - - -
give
s AL
L to
ON
E
ELECTRON
PHOTONcarries amount of
Energy = hf
- - - - - - - -
give
s AL
L to
ON
E
inELECTRON
metal
photoelectric emission-
- - - - - - - -
PHOTONcarries amount ofgi
ves
ALL
to O
NE
in
Energy = hf
photoelectric emission-
ELECTRON
type determinesneeds
for
W0 = hf0
metal
type determines
metal
- - - - - - - -
PHOTONcarries amount ofgi
ves
ALL
to O
NE
in
Energy = hf
-
needs
for
W0 = hf0
ELECTRON
photoelectric emission
- - - - - - - -
PHOTONcarries amount ofgi
ves
ALL
to O
NE
in
Energy = hf
photoelectric emission-
ELECTRON
type determinesneeds
for
W0 = hf0
metal
W0 = hf0
type determines
metal
in
carries amount of
Energy = hf
needs - - - - - - - -
give
s AL
L to
ON
E
-
ELECTRON
for
photoelectric emission
PHOTON
in
type determinesneeds
for
metal
- - - - - - - -
PHOTONcarries amount ofgi
ves
ALL
to O
NE
Energy = hf
photoelectric emission-
ELECTRON
W0 = hf0
W0 = hf0
in
type determinesneeds
for
metal
- - - - - - - -
carries amount ofgi
ves
ALL
to O
NE
photoelectric emission-
ELECTRON
Energy = hf
PHOTON
type determines- - - - - - - -
PHOTONcarries amount ofgi
ves
ALL
to O
NE
in
Energy = hf
photoelectric emission-
needs
for
W0 = hf0
ELECTRON
metal
type determines- - - - - - - -
PHOTONcarries amount ofgi
ves
ALL
to O
NE
in
Energy = hf
photoelectric emission-
needs
for
W0 = hf0
ELECTRON
metal
for surface electrons occurs with
KEmax= hf–hf0
type determines- - - - - - - -
PHOTONcarries amount ofgi
ves
ALL
to O
NE
in
Energy = hf
photoelectric emission-
needs
for
W0 = hf0
ELECTRON
metal
for surface electrons occurs with
KEmax= hf–hf0
KEmax= ½ mvmax
2=
type determines- - - - - - - -
PHOTONcarries amount ofgi
ves
ALL
to O
NE
in
Energy = hf
photoelectric emission-
needs
for
W0 = hf0
ELECTRON
metal
for surface electrons occurs with
KEmax= hf–hf0
KEmax= ½ mvmax
2=
energy photon gives
energy e- needs to break free from Zn
energy photon gives=6,91 x 10-19 J
energy e- needs to break free from Zn
energy photon gives=6,91 x 10-19 J
energy e- needs to break free from Zn
=6,91 x 10-19 J
energy photon gives=6,91 x 10-19 J
Zn’s work function, W0
=6,91 x 10-19 J
energy photon gives
energy e- needs to
break free
e- ‘s kinetic energy
E =hf
W0=hf0 KE
E =hf
W0=hf0 KE
E =hf
W0=hf0 KE
Metal 0 (nm) f0 (Hz) MoviePt 196 1,53 x 1015
Cu 263 1,14 x 1015
Zn 289 1,04 x 1015
Ca 427 7,03 x 1014
Na 539 5,57 x 1014
Na’s W0
Na’s W0
=3,38 x 10-19 J
Zn’s W0
=6,91 x 10-19 J
Na’s W0
=3,38 x 10-19 J
Question 2
How much energy does each photon of UV light of frequency 1,5 x 1015 Hz have?
Solution
E = hf = 6,63 x 10-34 J·s x 1,5 x 1015 Hz = 6,63 x 10-34 J·s x 1,5 x 1015 s-1
= 9,95 x 10-19 J
energy e- needs to break free from Zn
=6,91 x 10-19 J
energy photon gives=9,95 x 10-19 J
energy e- needs to break free from Zn
=6,91 x 10-19 J
energy photon gives
energy e- needs to break free from metal
e- ‘s kinetic energy
E=hf
energy e- needs to break free from metal
e- ‘s kinetic energy
E=hf
W0=hf0 e- ‘s kinetic energy
E=hf
W0=hf0 KE = E-W0 hf-hf0
Question 3
How much kinetic energy do electrons emitted from the surface of zinc have when light of frequency 1,5 x 1015 Hz is incident on the zinc?
Solution
KE = E-W0
= 9,95 x 10-19 J - 6,91 x 10-19 J = 3,03 x 10-19 J
E=hf=9,95 x 10-19 J
E=hf=9,95 x 10-19 J
W0=hf0
=6,91 x 10-19 Je- ‘s
kinetic energy
E=hf=9,95 x 10-19 J
W0=hf0
=6,91 x 10-19 JKE = hf-W0= 3,03 x 10-19 J
Question 4
At what maximum speed are electrons emitted from the surface of zinc have when light of frequency 1,5 x 1015 Hz is incident on the zinc?
Solution: v subject of formula
KEmax = ½ mvmax2
KEmax x = ½ mvmax2 x
= vmax2
= =
=
Solution: Known values
=
Known: m of 1 e- = 9,11 x 10-31 kgKnown: 3,03 x 10-19 J (from Question 3)
Solution
=
= 816 000 m•s-1
Question 4
Light of a certain frequency is incident on a piece of sodium. The maximum kinetic energy of the emitted electrons is 1,07x10-20 J. What is the frequency of the incident light?
Solution: making f the subject of the formula
KEmax = E - W0
KEmax = hf – hf0
KEmax + hf0= hf – hf0 + hf0
KEmax + hf0= hf
hf = KEmax + hf0
=
=
Solution: Known values
=
KEmax = 1,07x10-20 J
Na’s f0 = 5,57 x 1014 Hz
h = 6,63 x 10-34 J·s
Solution: Known values
=
f = 5,73 x 1014 Hz
energy photon gives
energy e- needs to
break free
e- ‘s kinetic energy
E =hf
W0=hf0 KE
E =hf
W0=hf0 KE
E =hf
W0=hf0 KE
E =hf
W0=hf0 KE
LIGHT
colour
has
photoelectric emission
determined by light’s
frequency
PHOTONS
consistsof
energy
carry amount of
determined by
determines if
occurs from a
=f0<f0 >f0
if if
type determines f0W
0
W0
for emission, each e- needs energy
has has >
is an amount of
metal
LIGHT
intensity
photoelectric emission
determined by
emission rate
PHOTONS
consists of
metal
occurs from a
of
affects rate that
has
LIGHT
intensity colou
r
has
photoelectric emission
determined by light’s
frequency
determined by
emission rate
PHOTONS
consists of
energy
carry amount of
determined by
metal
determines if
occurs from a
of
affects rate that
LIGHT
intensity
colour
has
photoelectric emission
determined by light’s
frequency
determined by
emission rate
PHOTONS
consists of
energy
carry amount of
determined by
metal
determines if
occurs from a
of
affects rate of