OCN400$Problem$Set$#3Key$$ Winter$2015$Due:$9:30a.m ...1/2#to#minutes#...
Transcript of OCN400$Problem$Set$#3Key$$ Winter$2015$Due:$9:30a.m ...1/2#to#minutes#...
OCN 400 Problem Set #3 Key Winter 2015 Due: 9:30 a.m., Monday , February 2 1. Secular Equilibrium a) 232Th90 decays by alpha (α) decay. This thorium isotope has an atomic number (protons) of 90 and an atomic weight (protons + neutrons) of 232. The decay product is radium. What is the atomic number and atomic weight of this radium isotope? (5)
𝐓𝐡𝟗𝟎𝟐𝟑𝟐 → 𝛂𝟐𝟒 + 𝐑𝐚𝟖𝟖
𝟐𝟐𝟖 𝐑𝐚𝟖𝟖
𝟐𝟐𝟖 , with an atomic weight of 228. 1pt for Ra 1pt for 228 1 pt for 88 (implied 1 pt for alpha decay) b) 232Th90 has a half-‐life of 1.4*1010 years. The radium isotope it decays to has a half life of 5.75 years. If there was none of this radium initially present, how long would the sample have to sit for secular equilibrium to be reached? (5) Five half lives of the daughter or 5 x 5.75 years = 28.75 years c) If the activity of 232Th and this Ra isotope were equal (secular equilibrium), what would be the ratio of their molal concentrations? (5) For secular equilibrium, the activity of the daughter must equal activity of the parent 𝐀 𝐓𝐡𝟐𝟑𝟐 = 𝐀 𝐑𝐚𝟐𝟐𝟖 A= -‐λN λ = 0.693/t½ −𝛌 𝐓𝐡𝟐𝟑𝟐 𝐍 𝐓𝐡𝟐𝟑𝟐 = −𝛌 𝐑𝐚𝟐𝟐𝟖 𝐍 𝐑𝐚𝟐𝟐𝟖 𝐍 𝐓𝐡𝟐𝟑𝟐
𝐍 𝐑𝐚𝟐𝟐𝟖=𝛌 𝐑𝐚𝟐𝟐𝟖
𝛌 𝐓𝐡𝟐𝟑𝟐
=𝟎.𝟔𝟗𝟑/𝐭½ 𝐑𝐚𝟐𝟐𝟖
𝟎.𝟔𝟗𝟑/𝐭½ 𝐓𝐡𝟐𝟑𝟐
=𝐭½ 𝐓𝐡𝟐𝟑𝟐
𝐭½ 𝐑𝐚𝟐𝟐𝟖=𝟏.𝟒×𝟏𝟎𝟏𝟎𝐲𝟓.𝟕𝟓 𝐲
= 𝟎.𝟐𝟒×𝟏𝟎𝟏𝟎
1 pt Setting activities equal 1 pt Plug in concentration times rate of decay 1 pt Knowing to solve for ratio 1 pt Using half life equation to get the ratio d) Thorium isotopes are attached to particles. Radium isotopes are more soluble (like calcium and barium) but not totally conservative. Suggest a possible geochemical application for this Th-‐Ra isotope pair. (5) 232Th is insoluble. It comes into the ocean on particles. Any solubilized in the ocean also becomes attached to particles. Because its half-‐life is so long it is mostly removed to the sediments. There it decays to 228Ra. Because 228Ra is much more soluble it can diffuse back to the overlying water. Thus there is excess 228Ra in the water column near the sediment water interface. The Ra is transported away by mixing. Thus 228Ra can be used to calculate the eddy diffusion coefficient due to mixing away from the bottom boundary. Because it’s half live is somewhat long it can be transported some distance (~ 1 km) before decaying. Also acceptable: anything that includes a use for both ions Chemistry of elements Half lives of elements How these can be combined to make a useful tool -‐2 for only considering 1 2. Radioactive Decay – units – Fukushima contamination a) Calculate the activity (in dpm/liter and Bq/liter) of 40K and 238U in seawater. 40K provides most of the natural background radiation in the ocean. (10) Useful information: Hint: 1 Becquerel (Bq) = 1 dps = 0.016 dpm The concentrations of natural potassium and uranium in seawater are [K]=400ppm;
[U]=3ppb [Hint: you will first need to calculate the atoms/liter for K and U, The first step is to calculate the concentration in moles then convert to atoms]. From a pass out on day 1 we know the following isotope abundances: 40K = 0.0117% of total K 238U = 99.27% of total U Their half-‐lives are: 40K = 1.25 x 109 yrs 238U = 4.47 x 109 yrs
You can use the equation A= λN (lecture 6, slide 6) to calculate activity. To do this, you will need to calculate both N (using the given concentration, the density of seawater, and the % of the relevant isotope) and λ (using the given half live, converted into minutes).
For 40K:
Calculate N (Keep in mind that only 0.0117% of all K in ocean is 40K) 0.000117 * (400mg/1kg) (1.025kg/L)(1g/1000mg)(1mol/40g)(6.02x1023 atoms/mol) = 7.22 x 1017 atoms 40K/L
Calculate λ in min-‐1: Start by converting t1/2 to minutes t1/2 = 1.25 x 109 years(365 days/year)(24hrs/day)(60min/hr) = 6.57 x 1014 min t1/2 = 0.693/λ (equation from lecture 6, slide 9), ààλ = 0.693/t1/2 λ = -‐.693/6.57 x 1014 minutes = 1.05 x 10-‐15 min-‐1
Substitute calculated values of N and λ into equation for A: A = λN = (1.05 x 10-‐15 min-‐1)(7.22 x 1017) = 762 dpm/L
Convert to Bq/L:
1110 dpm/L (1min/60 sec) = 18.4 dps/L = 12.7 Bq/L For 238U:
Follow same procedure as above, using values for U instead of K
N = 7.70 x 1015 atoms/L λ = 2.95 x 10-‐16 min-‐1 A = 2.27 dpm/L = 0.0378 Bq/L
Each element worth 5 pts Same mistake for both, then only take off 1 total through the problem Key points:
Converting t1/2 to minutes Getting concentration to atoms Multiplying by elemental abundance Determining b) In June 2013 surface waters off the west coast of North America had measured cesium isotope contamination from Fukushima. The measured Cs-‐137 reached 0.0009 Bq/ltr. How does this relate to the 40K background radiation? Should we be alarmed about this radioactivity? (5) (0.0009 Bq/L)/12.7 Bq/L = 0.000071 = 0.0071% of background 40K radiation. This is probably not a big deal... 3/5 for correct qualitative answer 5/5 for correct qualitative answer + calculation 5/5 for correct quantitative answer 3. Stable Isotopes – delta notation The carbon isotope ratio (13C/12C) in a sample of water is 0.010947. a. Calculate δ13CPDB for this sample.(5) Given 13C/12C sample = 0.010947 13C/12C PDB = 0.011237 (lecture 5, slide 14)
𝛅 𝐂𝟏𝟑 =
𝑪𝟏𝟑
𝑪𝟏𝟐𝒔𝒂𝒎𝒑𝒍𝒆
𝑪𝟏𝟑
𝑪𝟏𝟐𝑷𝑫𝑩
− 𝟏 ×𝟏𝟎𝟎𝟎
𝛅 𝐂𝟏𝟑 = 𝟎.𝟎𝟏𝟎𝟗𝟒𝟕𝟎.𝟎𝟏𝟏𝟐𝟑𝟕 − 𝟏 ×𝟏𝟎𝟎𝟎
𝛅 𝐂𝟏𝟑 = 𝟐𝟓.𝟖𝟎𝟖‰
3 pts identifying correct equation -‐1 wrong value for std
b. Was this sample more likely coal or a marine carbonate? (5) coal c. Explain why δ13C of atmospheric CO2 is growing more negative with time. (5) Burning of fossil fuels since the beginning of the industrial revolution – fossil fuels (coal, petroleum, natural gas) are created from buried organic matter, and therefore have an carbon isotopic composition similar to organic matter in the modern environment (~ -‐25‰). This is much lighter than average atmospheric CO2 (~ -‐ 7‰
at the beginning of the 20th century), and is thus decreasing the δ13C value of atmospheric CO2 (<-‐8‰ today). 5 pts: Input of fossil fuels is depleted carbon source; OM being burned is depleted as compared to atmosphere 4. Power of stable Isotopes – Mass Balances -‐ Methane Cycle We looked at one version of the global methane mass balance in PS #1. The values of the reservoirs and fluxes in this problem are slightly different. In this problem will use observations of two carbon isotopes to help determine the major sources and sinks of atmospheric methane. The total burden of methane in the atmosphere is 4680 Teragrams CH4 (Tg; 1 Tg = 1012 grams). This is equivalent to an average tropospheric atmospheric concentration of 1.7 ppm. Based on the kinetics of the reaction of methane with OH radical (the major sink for methane in the atmosphere), the rate of loss of methane from the atmosphere is 520 Tg CH4/year. a. What is the residence time (or atmospheric lifetime) of methane in the atmosphere? (5) τ = (mass CH4 in atmosphere)/(loss CH4 from atmosphere) τ = (4680 Tg)/(520 Tg/yr) τ = 9 years 3 pts correct residence time equation 2 pts correct values The Table below gives information on the isotopic composition of methane sources, together with the isotopic composition of methane consumed by reaction with OH. Note values for both 13C and 14C (two isotopes of C) are given. The 14C/12C ratio (divided by the standard’s ratio) is zero for methane derived from fossil fuel burning, since they were formed millions of years ago (long enough for all of the 14C to decay). Biospheric sources all
have 14C/12C ratios equal to 1.0 (i.e. the standard for 14C is defined so that the ratio of biospheric sources equals the standard). Since the residence time of methane in the atmosphere is short compared to the half-‐life of 14C/12C , we may disregard radioactive decay of 14C/12C and use it as we would a stable isotope. Let: W = CH4 derived from anaerobic bacteria in wetlands, ruminants, rice paddies, and termites, which are all similar isotopically; B = CH4 formed during biomass burning F = CH4 formed from fossil fuel sources (natural gas leaks, coal mines)
Source 13C
14C/12C ratio
W -‐65‰ 1.00 B -‐20‰ 1.00
F -‐40‰ 0.00
Methane Loss -‐57‰ 0.80 The total sources (W+B+F) add up to 520Tg CH4/year (with the isotopic signature equal to the methane loss signature) Assume CH4 is in steady state in the atmosphere (i.e. the sum of sources = loss rate) b. write the mass balance equation for CH4 (source = loss = 520Tg CH4/year) (5) W + B + F = Methane loss or W+ B + F = 520Tg/yr -‐1 for missing source/sink c. write the mass balance equation for 13CH4 (5) 13C/12CW(W) + 13C/12CB(B) + 13C/12CF(F) = 13C/12CML(ML) èè Need to convert given δ13C values to 13C/12CW in order to solve this equation. You can plug in known value of PDB to equation from 1a, or you can cancel out RPDB since it will appear in every term of the equation.
𝛅 𝐂𝟏𝟑 =
𝑪𝟏𝟑
𝑪𝟏𝟐𝒔𝒂𝒎𝒑𝒍𝒆
𝑪𝟏𝟑
𝑪𝟏𝟐𝑷𝑫𝑩
− 𝟏 ×𝟏𝟎𝟎𝟎
𝑹𝒔𝒂𝒎𝒑𝒍𝒆 =𝑪𝟏𝟑
𝑪𝟏𝟐𝒔𝒂𝒎𝒑𝒍𝒆
=𝛅 𝐂𝟏𝟑
𝟏𝟎𝟎𝟎+ 𝟏𝑪𝟏𝟑
𝑪𝟏𝟐𝑷𝑫𝑩
𝑪𝟏𝟑
𝑪𝟏𝟐𝑾=
−𝟔𝟓‰𝟏𝟎𝟎𝟎 + 𝟏 𝑪𝟏𝟑
𝑪𝟏𝟐𝑷𝑫𝑩
(13C/12C)W = 0.935*(13C/12C)PDB
𝑪𝟏𝟑
𝑪𝟏𝟐𝑩=
−𝟐𝟎‰𝟏𝟎𝟎𝟎 + 𝟏 𝑪𝟏𝟑
𝑪𝟏𝟐𝑷𝑫𝑩
(13C/12C)B = 0.980*(13C/12C)PDB
𝑪𝟏𝟑
𝑪𝟏𝟐𝑭=
−𝟒𝟎‰𝟏𝟎𝟎𝟎 + 𝟏 𝑪𝟏𝟑
𝑪𝟏𝟐𝑷𝑫𝑩
(13C/12C)B = 0.960* (13C/12C)PDB
𝑪𝟏𝟑
𝑪𝟏𝟐𝒎𝒆𝒕𝒉𝒂𝒏𝒆 𝒍𝒐𝒔𝒔
=−𝟓𝟕‰𝟏𝟎𝟎𝟎 + 𝟏 𝑪𝟏𝟑
𝑪𝟏𝟐𝑷𝑫𝑩
(13C/12C)methane loss = 0.943* (13C/12C)PDB
Plug these values back into initial equation: 0.935*(13C/12CPDB)*W + 0.980*(13C/12CPDB)*B + 0.960*(13C/12CPDB)*F = 0.943*(13C/12CPDB)*ML Cancel out 13C/12CPDB from each term: 0.935*W + 0.980*B + 0.960*F = 0.943*520 Tg/yr = 490.36 Tg/yr 5 for Mass balance equation can just have the R values – don’t necessarily need to calculate the fractions.
d. write the mass balance equation for 14CH4 (5) 1*W +1*B +0*F = 0.8*520 Tg/yr W + B = 416 Tg/yr Use R values or solve for them e. Using these three equations, you should be able to solve for the three unknowns (W, B and F). What are their values? What is the relative importance of the fossil fuel flux? (10) Hint: First solve the 14C mass balance (4d) for the sum of W + B. Then go to the first equation (4b) to solve for F. Now you have 2 equations and 2 unknowns, much easier to solve. From equation 4d, you know that W + B = 416 Tg/yr Substitute into equation 4b: 416Tg/yr+ F = 520 Tg/yr F = 520 Tg/yr – 416 Tg/yr F= 104 Tg/yr Solve equation 4d for B in terms of W: B= 416 Tg/yr -‐ W Substitute this expression and your solution for F into equation 4d: 0.935*W + 0.980(416Tg/yr – W) + 0.960*(104 Tg/yr) = 490.36 Tg/yr Solve for W: -‐0.045*W = -‐17.16 W = 381.33 Tg/yr Finally, substitute this value for W back into equation 4d and solve for B: 381.33 Tg/yr + B = 416 Tg/yr B = 34.67 Tg/yr To solve for the relative importance of the fossil fuel flux, determine what percent of the total atmospheric CH4 flux is from fossil fuels: (104 Tg/yr)/(520 Tg/yr) = 0.2 = 20% **Rounding allows for a fairly large spread, so just because values differ doesn’t mean math is wrong -‐Solve F -‐Solve B -‐Solve W -‐Solve final calculation