An undisturbed sediment
core containing varves from
the deposition zone of a
deep lake
The varves can be used to
calculate dates along the
core profile
Paleolimnology--Pollen stratigraphy in lake sediment cores
Cores can be dated with radioisotopes
137Cs
(half-life 30 yr) is
found in fallout from
bomb tests
The most commonly used isotope is 210Pb, half-life 22 yr
The Uranium 238 decay series
238U
226Ra
222Rn
218Po
210Pb
The littoral zone, what determines its outer boundary?
The transparency of lake water is measured by its
extinction coeficient
The extinction coefficient k increases with:
•the concentration of organic matter (colour) of the
water
•the amount of suspended matter
eg, phytoplankton, fine suspended particles, eg
clay
Light extinction --Light enters from above and its intensity (I) is sharply
attenuated with depth (z)—absorption by water or solute molecules or
scattered by particles
Section 10.6
Iz
z 50%
z 10%
z 1%
Iz  I 0e  kz where I is light intensity, and z  depth,
k is the extinction coefficien t in fraction/m
This equation can be rearranged to
give (take ln of both sides)
Photic ln I 0  ln Iz
zone k 
z
What fraction of light remains by depth z?
Iz
 e  kz
I0
What fraction of light is absorbed every z m
1  e  kz
z
Page 144 in text
In general Photosynthesis exceeds respiration above the 1% light level
and rooted plants can grow down to about the 10% light level
Iz
z 50%
z 10%
At what depth z1% will the light intensity be 1% of I 0
Iz
 e  kz so 0.01  e  kz and  ln 0.01  kz...
Photic I 0
zone  z  4.6 ,
k
for the depth at which light intensity is 10% of I 0
2.3
,
k
for the 50% light level
z
z 1%
z
z
Page 144 in text
0.69
k
Consider t he following example problem
If the light intensity at 2 m depth is 50% of the surface intensity
solve for k , the extinction coefficien t
Since
Iz
Iz
Iz  I 0e  kz , then
 e  kz and  0.5
I0
I0
then
e  kz  0.5
then take the natural log (ln) of both sides
 kz  ln 0.5 or kz  0.69
since in this problem z  2, we have
2k  0.69, or
k  0.345
This means that 34.5% of the incident light is absorbed in each m of depth
What fraction of the incident light woul d reach 4m?
I4
 e ( 40.345)  0.25
I0
25% of the light woul d reach 4 m
What would be the depth of the photic zone?
z  4.6 / k
z  4.6 / 0.345  13.3m