NCEA Level 3 Physics (90522) 2010 — page 1 of 4
Assessment Schedule – 2010
Physics: Demonstrate understanding of atoms, photons and nuclei (90522)
Evidence Statement
Q
Evidence
ONE
(a)(i)
Circles are energy levels (of electrons) or
electron shells or orbits.
Not atoms. Not just shells
Lines represent electron transitions between
energy levels, or spectral lines ( not
electromagnetic spectrum), wavelengths
emitted (or absorbed)
1
Correct
identification of
energy levels and
electron transitions.
(ii)
Emission spectra explained by: a transition
from higher to a lower energy level causing
the emission of a photon of a discrete energy
level. Only certain energies are involved, so
the photons emitted have certain energies /
wavelengths – hence the lines in the spectrum.
Absorption spectra explained by: a transition
from lower to a higher energy levels caused
by the absorption of only specific photons
with the ‘right’ energies – hence the dark lines
in the spectrum.
The wavelength corresponds to the CHANGE
in energy level.
1
Describes emission
or absorption
spectra in terms of
electron transitions.
(b)
 1
1
 R 2  2 

S
L 
 1 1
1
 1.097  107  2  2 

2 5 
2
Correct method.
2
Correct E1 in J.
1
Achievement
Achievement with
Merit
1
Explains emission
OR absorption
spectra in terms of
electron transitions of
discrete energies.
2
Correct answer.
 2.304  106
  4.341 10Π7 Κm
(c)
En 
hcR
2
n
– 2.182  10-18 J
6.626  1034  3.00  108  1.097  107
E1  
Accept ONE
12
18
mistake, OR
E1  2.182  10
wrong energy
2.182  1018
calculation but

 13.6ΚeV
19
converted into eV
1.6  10
correctly.
13.6 eV
Achievement with
Excellence
1
must include that
wavelength
(energy released
or absorbed)is
directly
proportional to
Change in
Energy.
NCEA Level 3 Physics (90522) 2010 — page 2 of 4
(d)
2
 1 1
 1.097  10  2  2   8.2275  106

1 1 
OrΚusesΚvalueΚgiven,Κ122Κnm
hc
E  hf 
1
7
Converts  into J
3.66  1.6  10

 6.626  1034  3.00  108  8.2275  106
 1.635  1018 ΚJ
hf    EK
EK  hf  


EK  1.635  1018  3.66  1.6  1019
EK  1.04  1.6  10
18
19
 5.86  1019 ΚJ
OR frequency
calculated correctly.
f = 2.47  1015 Hz
2
2
Correct calculation of
photon energy.
Complete answer.
= 1.635  10
–18
J
= 1.04  10–18 J
= 1.05  10–18 J
Or in eV = 10 eV

ΚJ
OrΚusesΚvaluesΚgiven,122Κnm
hc
E  hf 


6.626  1034  3.00  108
1.22  109
 1.629  1018 ΚJ
hf    EK
EK  hf  


EK  1.629  1018  3.66  1.6  1019

EK  1.04  1018 ΚJ
Also accept 1.05  10–18 J
TWO
(a)
E  hf 
hc

hc 6.626  1034  3.00  108
 
E
4.97  1019
  4.00  107 Κm
2
Correct frequency
of
E  hf
2
400 nm
= 4.00  10–7 m
E
h
4.97  1019
f 

6.626  1034
 7.5  1014 ΚHz
(b)
(c)
Photo electric effect: Threshold frequency –
there is no emission of electrons when the
surface is illuminated with light below a
threshold frequency, however intense this
light is.
This suggests that the light is interacting with
the surface in discrete amounts of energy. If
the energy of this photon is too low then no
electron emission can occur.
Photo electric effect: No time delay at low
light intensity. There is no evidence the
energy is spread out at low energies because
electron emission begins immediately. This
agrees with a 1 photon = 1 electron emission
model.
Blackbody radiation answers are also
acceptable.
EK  eV  1.6  1019  10  103
 1.6  1015 ΚJ
Mentions
photoelectric effect
OR loosely
describes
Photoelectric effect.
2
Correct answer.
1
Mentions threshold
frequency or no time
delay in photo
electric effect.
1
Complete answer
explaining either
why threshold
frequency or no
time delay means
that light energy is
quantised.
NCEA Level 3 Physics (90522) 2010 — page 3 of 4
(d)
Ephoton  hf 

6.626  10
V
2
hc
34

 3.00  108
2.36  109
 8.423 10
17
2
Calculation of photon
energy.
2
Correct answer.
526 V or 527 V
= 8.423  10–17J
Ek 8.4231017

 526ΚV
e
1.6 1019
The electrons must have energy equal to (or
greater than) that of the photon (all the
electron energy can be released as a photon).
m  66.34446  1027
THREE
(a)
frequency
calculated
correctly.
f = 1.27  1017 Hz
1
Correct assumption.
Do not accept
work function is
negligible or
conservation of
energy.
2
Correct mass
deficit.
 66.34121 1027
2
Correct answer.
= 2.339  10–30 kg
= 2.1  10–13 J
Binding energy is the energy required to
completely separate the nucleons in a nucleus.
Ca must be at a lower energy, because energy
was given off when the reaction happened.
So Ca must have a greater binding energy.
Both nuclei have the same number of
nucleons, so the Ca has the greater binding
energy per nucleon.
OR
Fe is the most stable nucleus (A), and an
excellence level explanation to compliment
answer can gain full marks
1
Recognition that Ca
has a greater
binding energy.
OR correct
definition of
binding energy.
OR that the nuclei
have the same
number of
nucleons.
OR Fe
1
6.626  1034  3.00  108
2
Correct calculation.
1
Correct answer can
be used to
demonstrate
concept knowledge
 0.000911 1027
 2.339  1030 Κkg
E  mc 2

E  2.339  1030 3.00  108

2
 2.1 1013 ΚJ
(b)
(c)
E
hc


8.5  1013
 2.339  1013 ΚJ
This is more than the energy release in the
reaction so it can’t have come from it.
OR using Energy from part (a) to find
λ (9.45  10–13), and comparing λ (8.5 10–13),
Correct definition
of Binding energy
+ Recognition that
Ca has the greater
binding energy per
nucleon because
• more mass deficit
• Ca has less mass
• Ca is formed
spontaneously
• energy is given out.
1
Ca has more binding
energy, but same
number of nucleons,
so more binding
energy per
nucleon.
NCEA Level 3 Physics (90522) 2010 — page 4 of 4
Judgement Statement
Achievement
Achievement with
Merit
C1
2 A1
1 A1 + 2 M1
C2
3 A2
2 A2 + 3 M2
Achievement with Excellence
3 E,
including at least one from each criterion
+ 1 M1 and 1 M2