Question1: [Light Harvesting Process]
How many photons does it take to split one water molecule during the photosynthesis process? In which part of the chloroplast does the Calvin cycle occur?
Only two photons are necessary. One for each electron harvested (water has two hydrogen atoms whose nuclei, the H + ions, fulfill important roles. The split off oxygen combines with another oxygen atom to form an O
2 molecule).
The binding energy of oxygen atom and a hydrogen atom is about 492 kJ/mol.
So the bond energy for single bond is
E b

492

10
3
J
6 .
022

10 23
/ mol
/ mol

8 .
17

10

19
J

5 .
1 e V
Energy of a photon of 430 nm(say, blue light) is
E photon
 hc
430

10

9 m

4 .
62

10

19
J

2 .
9 e V
Hence, the two photons indeed do have enough energy to break the bond. Of course, the energy is not directly transferred to the bond.
The Calvin cycle occurs in the stroma of the chloroplast.

Question2: [Atomic Energy Level]
From Quantum Mechanics, we know that energy levels of atom are discrete.
(a) If an atom would have three different excited levels, how many different wavelengths can the absorbed and emitted photons have?
T
13
T
23
There are 6 possible transitions as shown in the figure.
Remark: We assume that no selection rule here and transitions between all pairs of levels are allowed.
T
01
T
02
T
12
T
03
(b) The energy-level scheme for the hypothetical one-electron element Searsium is shown in Figure below. The potential energy is taken to be zero for an electron at an infinite distance from the nucleus.
(i) How much energy (in eV) does it takes to ionize an electron from the ground level?
Negative energy means a bound system
E ionize

E


E g

0

(

20 eV )

20 eV

(ii)An 18-eV photon is absorbed by a Sersium atom in its ground level. As the atom returns to its ground level, what possible energies can the emitted photons have? Assume that there can be transitions between all pairs of levels.
After the ground state atom absorbed the photon, it will
“jumps” to n=4 state. It can returns to ground state via followings:
(iii) What will happen if a photon with an energy of 8eV strikes a Searsium atom in its ground level?Why?
There is no energy level 8eV higher than ground state. Therefore the electron will not absorb the 8eV photon.

Question3: [Absorption Spectra of Chlorophyll]
We claim that the wavelength of a photon must be exactly match the energy difference between the states of an atom or molecule. Then why are the peaks in Figure 12.10 (see text book) so broad? Shouldn’t they be very sharp?
Molecules do not stand exactly still. Due to temperature alone, they will be vibrate and thus move towards or away from the light source. There may also be rotational or other energies that come into play.
Consequently, although the energy to transition from ground state to an excited state is exactly given, but amount of energy that needs to be added to effect that transition is not.
The energy levels of molecule is broader (with a finite width) mainly due to the extra degrees of freedom from rotational and vibration modes.

Question4: [Leaves Color and Absorption]
(i)What color would leaves have if they could also absorb green light?
If green light also absorbed, leaves would absorb most of the visible light. Hence they would appear blackish.
However, if the absorption peaks are fairly sharp, then the colors between green and red
(the color near yellow) and the colors between green and blue (near turquoise) would shine through.
(ii)What color would leaves have if they could only absorb green light?
If leaves only absorb green light, they will let red and green light pass. By additive mixing, red and blue together once obtain magenta. Thus, the leaves appeared to be in purple color.

Question5: [Sun or Stars light as Blackbody Radiation]
The brightest star in the sky is Sirius, the Dog Star. It is actually a binary system for twostars, the smaller one (Sirius B) being a white dwarf. Special analysis of Sirius B indicates that its surface temperature is 24,000K and that it radiates energy at a total rate of 1.0 x10 25 W. Assume that it behaves like an ideal blackbody.
(i)What is the total radiated intensity of Sirius B?
Intensity is given by the Stefan-Boltzmann law
I
 
T
4 
( 5 .
67

10

8
)( 24 , 000 )
4

1 .
9

10
10
W / m 2
(ii)What is the peak intensity wavelength? Is this wavelength visible to humans?
To find peak intensity, we use the Wien’s law.

 m
T
 m

2 .
90

10

3 mK

120 nm
This wavelength is shorter than 400nm. Thus we cannot see this light.
(iii) What is the radius of Sirius B? Express your answer in km and as a fraction of our sun’s radius.
The power radiated is given
P

R

IA
Sirius


I

4

2
R
Sirius
6510 km

As a ratio of sun’s radius, we have
R
Sirius
R
Sun

6 .
51

10
6 m
6 .
96

10
9 m

0 .
0093 Roughly 1%

(iv)Which star radiates more total energy per second, the hot Sirius B or the (relatively) cool sun with a surface temperature of 5800K? To find out, calculate the ratio of the total power radiated by our sun to the power radiated by Sirius B.
Using Stefan-Boltzmann law, we have
P
Sun
P
Sirius


A
4 sun
T sun

A
4 sirius
T sirius



R
Sirius
R
Sun


2 

T sun
T sirius


4

39
Even though the absolute surface temperature of Sirius B is about 4 times that of our sun, Sirius B radiates only 1/39 times that of our sun. The reason is that Sirius
B is much smaller than our sun.

Question6: [Young Double Slit Experiment]
Consider Figure 13.21 in the textbook. If we have the following settings: d = 2mm, L =
80cm and P = 2cm.
(i)What is the wavelength of our source?
 
Pd
L


( 2 cm )( 2 mm )
80 cm
50
 m
(ii) Is the radiation from the source visible? If so, what is its color?
No, the radiation from the source is not in between the range of visible light
(400nm-700nm). 400nm is violet/blue and 700nm is the red light.

Slit separation = d;
Distance on screen = x
Angle of incidence = θ x
= m λ  bright
= (m + ½)λ  dark m = 0, 1, 2, … sin θ ~ tanθ ~ θ ~ x/L, small θ
L

θ
θ
θ
x/L,
θ

Question7: [Double Slit Interference]
When green light ( λ =505nm) passes through a pair of double slits, the interference pattern
(a) is observed. When light of a different color passes through the same pair of slits, the pattern (b) is observed. (i) Is the wavelength of the second color greater than or less than
505nm?
The angular separation is given by sin( θ )=m λ /d, which is proportional to the wavelength of the light. Figure (b) have smaller θ , thus λ
2
<505nm.
(ii) Find the wavelength of the second color.
From figure, 4 th order minima (m
1
=4) of green light coincides with the 5 th order maxima (m
2
=5) of 2 nd color.
For minima, For maxima,

Question8: [Speed of Light]
(i)What is the speed of light in Diamond?
The refractive index of Diamond is given by 2.42
(from text book page 288).
c n
 v
 v

3 .
0

10
8 m

2 .
42
1 .
24

10
8 m
/
/ s s
(ii) How long will it take a light pulse to travel from Singapore to Vancouver (a distance of about 12,838km) in an optical fiber made from Plexiglas?
 c

Sin

1

 n n
2
1



Sin

1

41 .
47 o
1
1 .
51
S
 t
 vt ; v

3 .
0

10
8 m / s
1 .
51

1 .
986

12 , 838 km
1 .
986

10
8 m / s

0 .
065 s

10
8 m / s
In general, the time taken is longer than this since the light path is not parallel to the optical fiber, but goes through many total internal reflections.

Question9: [Reflection of Light]
A small vertical mirror hangs on the wall, 1.40m above the floor. Sunlight strikes the mirror, and the reflected beam forms a spot on the floor 2.50m from the wall. Later in the day, you notice that the spot has moved to a point 3.75m from the wall. (a) Were your two observations made in the morning or in the afternoon? Explain.
θ i1
θ r1
θ i2 a) For the 2 nd observation, since the distance from the wall is longer, we see that the incident/ reflection angle must be smaller (in green). Thus, the sun is at lower latitude (local frame) for 2 nd case (setting sun). The observation is made in the afternoon.
1.40m
(b) What was the change in the Sun’s angle of elevation between your two observations?
2.50m
3.75m

Question10: [Refraction of Light and Snell’s Law]
The observer, as shown, is positioned so that the far edge of the bottom of the empty glass is just visible. When the glass is filled to the top with water, the center of the bottom of the glass is just visible to the observer. Find the height, H, of the glass, given that its width isW = 6.2cm.
Consider water-air interface,
θ r
θ r
θ r
θ i
W/2
In both case, the refracted angle is the same, since the eye position is fixed.

Question11: [Electromagnetic Wave]
Microwaves can cook food (and hence your brain as well). Yet we can see inside through a metal sheet with small holes. Why can the light pass through but apparently not the microwaves?
Although both microwave and visible light consist of photons and are part of the electromagnetic wave, their wavelengths are rather different . For instance, green light is around 520nm while the wavelength of a typical microwave is 12.24cm (frequency of
2.45Ghz).
The door of a microwave oven has a glass panel with a conduction perforated sheet whose holes are a few mm in diameter. Compared to the wavelength of the visible photons, the holes are huge and hence visible light photons can easily pass through.
However, this hole size is considered tiny for the microwave photons and hence they cannot pass through .
Microwave oven make use of Faraday’s cage concept.
That’s why we are safe to use the oven and will not get our brain cooked!
For more details: http://factoidz.com/why-microwave-ovenscan-have-see-through-windows/

Question12: [Diversity]
How many different polymers with a length of six monomers can be made if one has twelve different types of monomers available?
If reading the sequence backward and forward is considered different (i.e. AAB ≠ BAA), then the number of different polymers
N is given by the number of types to the power of the length of the sequence.
N

 
6
Question13: [Biopolymers]
(i) Speculate as to why herbivores can eat grass and properly digest it while human beings cannot.
Both human beings and herbivores can break α linkages since the bond is not that strong.
β -linkages, however, are too strong for the mechanisms available to mammals to break. This is even so in the case of herbivores like cows and horses. Herbivore mammals solve this problem with the help of symbiotic bacteria that live in their digestive systems.
The bacteria produce enzymes which break down the cellulose into simple sugars that are then used by the mammal as food.

Question13: [Biopolymers]
(ii) How many side chains do the amino acids found in human beings have?
The differences between amino acids lie in their side chains. Since there are 20 different types of amino acids in human beings, there are 20 different types of side chains.
(iii)What is the type of bond that holds the pair A-T together in a strand of DNA?
The type of bond that holds pairs of nucleotides such as the A-T, G-C pairs is the hydrogen bond (relatively weak bond and easy to “zip and unzip” the DNA during the DNA replication).
(iv) Name two key differences between DNA and RNA.
The base thymine(T) in DNA is replaced by the base uracil(U) in RNA. RNA can have strong enzymatic properties while DNA does not. DNA generally only occurs double stranded while RNA can be both single and double stranded and furthermore fold in a myriad of ways. The sugar in DNA is deoxyribose while the sugar in RNA is ribose.

Question14: [DNA]
(i) If a species contains 23% A in its DNA, what is the percentage of guanine it would contain?
In DNA, we have base pairing as A-T and GC. So, percentage of Guanine is given by (1-2x0.23)/2= 0.27 or
27%.
(ii) Can we say that life is “programmed” by a random sequence of A, G, C & T in a chain of deoxyriboses and phosphates linked by phosphatediester bonds?Why?
No. For the functionality, it should be a given sequence rather than a random sequence.