Isotope Geochemistry Questions and Answers

Q: Which one of the following stable isotopic ratio is used for estimation of palaeotemperature of sea water ?

Palaeotemperature of sea water is estimated from the \(^{18}\)O/\(^{16}\)O ratio.When calcium carbonate shells (of foraminifera, for example) precipitate from sea water, the fractionation of oxygen isotopes between water and carbonate depends on temperature: colder water gives carbonate richer in \(^{18}\)O. Measuring \(\delta^{18}\)O in fossil shells, corrected for the isotopic composition of the sea water, therefore gives the water temperature. This was Urey's discovery and underlies the marine oxygen isotope stages.The carbon ratio tracks productivity and the carbon cycle, \(^{87}\)Sr/\(^{86}\)Sr is used for stratigraphic dating and weathering history, and the nitrogen ratio traces nitrogen cycling and food webs.

Q: Isotopes are the elements having :

Isotopes have the same Z but different N and A.Z is the atomic number (number of protons), N the number of neutrons and A the mass number (Z + N). Isotopes are atoms of the same element, so they share Z, but they contain different numbers of neutrons and therefore have different mass numbers: \(^{12}\)C and \(^{14}\)C, or \(^{16}\)O and \(^{18}\)O.Atoms with the same N but different Z are isotones, atoms with the same A but different Z are isobars, and atoms with the same N, Z and A are simply identical nuclides.

Q: The half-life of \(^{238}\)U is 4468×10\(^{6}\) years and \(^{235}\)U is 704×10\(^{6}\) years. The ratio \(^{235}\)U/\(^{238}\)U in a metamorphosed sample is 0.007257. Given that the ratio was 0.4 at the time of formation, calculate the metamorphosed sample age.

The age is 4.8 × 10\(^{9}\) years.Both uranium isotopes decay, so their ratio R changes as \(R = R_0 e^{-(\lambda_{235} - \lambda_{238})t}\), with \(\lambda = \ln 2 / t_{1/2}\).\[ \begin{aligned} \lambda_{235} &= 0.693/7.04\times10^{8} = 9.85\times10^{-10} \\ \lambda_{238} &= 0.693/4.468\times10^{9} = 1.55\times10^{-10} \\ \Delta\lambda &= 8.29\times10^{-10}\ \text{yr}^{-1} \end{aligned} \]\[ \begin{aligned} t &= \frac{\ln(0.4/0.007257)}{\Delta\lambda} \\ &= \frac{4.01}{8.29\times10^{-10}} \\ &\approx 4.8\times10^{9}\ \text{years} \end{aligned} \]The other options have the same mantissa with the wrong power of ten. An age of 10\(^{6}\) or 10\(^{7}\) years would barely change the ratio, since both half-lives are hundreds of millions of years or more.

Q: Rb-Sr age based on the individual minerals often yield younger age than that of the whole rock age. The reason for this is :

The reason given by the key is that individual minerals have a lower closure temperature and Sr diffusion is random at the smaller scale.After a rock forms, a later heating event can make radiogenic \(^{87}\)Sr move out of Rb-rich minerals such as biotite into neighbouring minerals. The individual minerals close to Sr diffusion at relatively low temperatures, so their clocks are reset and record the later event, giving a younger age. But the Sr only moves over short distances, so a whole-rock sample of adequate size remains a closed system and still records the original age.The option claiming a higher closure temperature for minerals reverses the cause. The size of the sample does not by itself change an age, and '87Kb' (for \(^{87}\)Rb) content explains nothing about the age difference.

Q: Which one of the following methods is best suited for dating highly deformed and metamorphosed Archaean gneissic rocks containing multiple age components ?

The best method is single zircon dating.A strongly deformed and metamorphosed gneiss may contain zircons, or zones within one zircon, of several ages: inherited cores, magmatic growth and metamorphic rims. U-Pb dating of individual grains, or of spots within a grain by ion microprobe (SHRIMP) or laser ablation, can resolve each of those components, and zircon keeps its U-Pb record through high-grade metamorphism.Whole rock U-Pb and Pb-Pb dating mix all the components in the sample and give a single, often meaningless, average or a disturbed isochron. Carbon-14 dating works only on organic material younger than about 50,000 years and is useless for Archaean rocks.

Q: The half-life of radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. The estimated age of the artifact would be :

The artifact is about 1846 years old.Radioactive decay gives \(N = N_{0} e^{-\lambda t}\) with \(\lambda = \ln 2 / t_{1/2}\), so\[ \begin{aligned} t &= \frac{t_{1/2}}{\ln 2} \ln\frac{N_{0}}{N} \\ &= \frac{5730}{0.693} \times \ln\frac{100}{80} \\ &= 8267 \times 0.2231 \\ &\approx 1845\ \text{years} \end{aligned} \]That matches 1846 years to within rounding of the logarithms.Check: 80% remaining is much less decay than one half-life (50% remaining), so the age must be well under 5730 years, which rules out that option. 2303 and 3208 years would leave about 76% and 68% of the carbon-14, not 80%.

Q: 206Pb/238U versus 207Pb/235U ratios when plotted yield concordant ages, also indicate :

Concordant ages indicate No loss or gain of radiogenic lead.Uranium has two decay chains, 238U to 206Pb and 235U to 207Pb, with different half-lives. If a mineral such as zircon has been a closed system since it formed, both ratios give the same age, and the analysis plots on the concordia curve.Loss of lead (through an alteration or thermal event) or gain of lead moves the point off the curve, making it discordant, and a set of discordant points defines a discordia line. So concordance shows the opposite of a lead-loss or alteration event.

Q: Higher amounts of which of the following isotopes indicate plume related enriched mantle origin of early Deccan volcanic rocks.

The indicator is a high \(^{3}\)He/\(^{4}\)He ratio.Helium-3 is primordial, left over from Earth's formation, while helium-4 is continually produced by radioactive decay of uranium and thorium. Mid-ocean ridge basalts have about 8 times the atmospheric ratio, and lavas derived from deep, less degassed mantle plumes have much higher values, as at Hawaii and Iceland. High \(^{3}\)He/\(^{4}\)He in the early Deccan lavas and associated rocks is therefore taken as a fingerprint of the Reunion plume.Sr and Os isotope ratios are raised by crustal contamination as well as by enriched sources, so they are not specific to plumes, and \(^{87}\)Rb/\(^{86}\)Rb is not a meaningful isotope ratio.

Q: Pb isotopic data from southern hemisphere shows compositional similarity to EM type of mantle reservoir is known as :

The southern-hemisphere anomaly is called DUPAL.Stan Hart (1984) noticed that many oceanic basalts of a belt in the southern hemisphere, notably in the Indian and South Atlantic Oceans, have higher 208Pb/204Pb and 207Pb/204Pb than northern-hemisphere basalts, together with high 87Sr/86Sr, resembling the enriched mantle (EM) reservoirs. He named it after Dupre and Allegre, who first described it.HIMU is a separate end-member with very high U/Pb, FOZO is the common 'focal zone' composition, and NHRL is the Northern Hemisphere Reference Line against which the DUPAL excess is measured.

Q: Higher ratios of 87Sr/86Sr and lower values of epsilon Nd (εNd) are observed in continental flood basalts because of :

The signature comes from Assimilation of continental crust.Continental crust has high Rb/Sr and low Sm/Nd ratios, so over time it develops high 87Sr/86Sr and strongly negative εNd. When mantle-derived basalt melts and assimilates such crust on its way up, the magma inherits both features, which is why many continental flood basalts, including parts of the Deccan, show them.A deeper mantle source or entrained mantle material would give mantle-like values (lower Sr ratio, positive εNd), and the degree of partial melting changes trace element concentrations but not isotope ratios.

Which one of the following stable isotopic ratio is used for estimation of palaeotemperature of sea water ?

A.\(^{13}\)C/\(^{12}\)C
B.\(^{18}\)O/\(^{16}\)O
C.\(^{87}\)Sr/\(^{86}\)Sr
D.\(^{15}\)N/\(^{14}\)N

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

Palaeotemperature of sea water is estimated from the \(^{18}\)O/\(^{16}\)O ratio.

When calcium carbonate shells (of foraminifera, for example) precipitate from sea water, the fractionation of oxygen isotopes between water and carbonate depends on temperature: colder water gives carbonate richer in \(^{18}\)O. Measuring \(\delta^{18}\)O in fossil shells, corrected for the isotopic composition of the sea water, therefore gives the water temperature. This was Urey's discovery and underlies the marine oxygen isotope stages.

The carbon ratio tracks productivity and the carbon cycle, \(^{87}\)Sr/\(^{86}\)Sr is used for stratigraphic dating and weathering history, and the nitrogen ratio traces nitrogen cycling and food webs.

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Isotopes are the elements having :

A.Same N but different Z and A
B.Same Z but different N and A
C.Same A but different Z and N
D.Same N, Z and A

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

Isotopes have the same Z but different N and A.

Z is the atomic number (number of protons), N the number of neutrons and A the mass number (Z + N). Isotopes are atoms of the same element, so they share Z, but they contain different numbers of neutrons and therefore have different mass numbers: \(^{12}\)C and \(^{14}\)C, or \(^{16}\)O and \(^{18}\)O.

Atoms with the same N but different Z are isotones, atoms with the same A but different Z are isobars, and atoms with the same N, Z and A are simply identical nuclides.

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The half-life of \(^{238}\)U is 4468×10\(^{6}\) years and \(^{235}\)U is 704×10\(^{6}\) years. The ratio \(^{235}\)U/\(^{238}\)U in a metamorphosed sample is 0.007257. Given that the ratio was 0.4 at the time of formation, calculate the metamorphosed sample age.

A.4.8 × 10\(^{6}\) years
B.4.8 × 10\(^{9}\) years
C.4.8 × 10\(^{3}\) years
D.4.8 × 10\(^{7}\) years

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The age is 4.8 × 10\(^{9}\) years.

Both uranium isotopes decay, so their ratio R changes as \(R = R_0 e^{-(\lambda_{235} - \lambda_{238})t}\), with \(\lambda = \ln 2 / t_{1/2}\).

\[ \begin{aligned} \lambda_{235} &= 0.693/7.04\times10^{8} = 9.85\times10^{-10} \\ \lambda_{238} &= 0.693/4.468\times10^{9} = 1.55\times10^{-10} \\ \Delta\lambda &= 8.29\times10^{-10}\ \text{yr}^{-1} \end{aligned} \]

\[ \begin{aligned} t &= \frac{\ln(0.4/0.007257)}{\Delta\lambda} \\ &= \frac{4.01}{8.29\times10^{-10}} \\ &\approx 4.8\times10^{9}\ \text{years} \end{aligned} \]

The other options have the same mantissa with the wrong power of ten. An age of 10\(^{6}\) or 10\(^{7}\) years would barely change the ratio, since both half-lives are hundreds of millions of years or more.

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Rb-Sr age based on the individual minerals often yield younger age than that of the whole rock age. The reason for this is :

A.The whole rock gives enough sample to date the rock accurately
B.The whole rock contains more \(^{87}\)Kb than the individual minerals
C.Individual minerals have a higher closure temperature and diffusion of Sr atoms is not random at smaller scale (~ a few meters)
D.Individual minerals have a lower closure temperature and diffusion of Sr atoms is random at smaller (~ a few meters) scale

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The reason given by the key is that individual minerals have a lower closure temperature and Sr diffusion is random at the smaller scale.

After a rock forms, a later heating event can make radiogenic \(^{87}\)Sr move out of Rb-rich minerals such as biotite into neighbouring minerals. The individual minerals close to Sr diffusion at relatively low temperatures, so their clocks are reset and record the later event, giving a younger age. But the Sr only moves over short distances, so a whole-rock sample of adequate size remains a closed system and still records the original age.

The option claiming a higher closure temperature for minerals reverses the cause. The size of the sample does not by itself change an age, and '87Kb' (for \(^{87}\)Rb) content explains nothing about the age difference.

Discuss in Forum

Which one of the following methods is best suited for dating highly deformed and metamorphosed Archaean gneissic rocks containing multiple age components ?

A.Whole rock U-Pb dating
B.Single zircon dating
C.Whole rock Pb-Pb dating
D.Carbon dating

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The best method is single zircon dating.

A strongly deformed and metamorphosed gneiss may contain zircons, or zones within one zircon, of several ages: inherited cores, magmatic growth and metamorphic rims. U-Pb dating of individual grains, or of spots within a grain by ion microprobe (SHRIMP) or laser ablation, can resolve each of those components, and zircon keeps its U-Pb record through high-grade metamorphism.

Whole rock U-Pb and Pb-Pb dating mix all the components in the sample and give a single, often meaningless, average or a disturbed isochron. Carbon-14 dating works only on organic material younger than about 50,000 years and is useless for Archaean rocks.

Discuss in Forum

The half-life of radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. The estimated age of the artifact would be :

A.3208 years
B.5730 years
C.1846 years
D.2303 years

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The artifact is about 1846 years old.

Radioactive decay gives \(N = N_{0} e^{-\lambda t}\) with \(\lambda = \ln 2 / t_{1/2}\), so

\[ \begin{aligned} t &= \frac{t_{1/2}}{\ln 2} \ln\frac{N_{0}}{N} \\ &= \frac{5730}{0.693} \times \ln\frac{100}{80} \\ &= 8267 \times 0.2231 \\ &\approx 1845\ \text{years} \end{aligned} \]

That matches 1846 years to within rounding of the logarithms.

Check: 80% remaining is much less decay than one half-life (50% remaining), so the age must be well under 5730 years, which rules out that option. 2303 and 3208 years would leave about 76% and 68% of the carbon-14, not 80%.

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²⁰⁶Pb/²³⁸U versus ²⁰⁷Pb/²³⁵U ratios when plotted yield concordant ages, also indicate :

A.Pb loss event
B.Alteration event
C.Loss or gain of radiogenic lead
D.No loss or gain of radiogenic lead

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

Concordant ages indicate No loss or gain of radiogenic lead.

Uranium has two decay chains, ²³⁸U to ²⁰⁶Pb and ²³⁵U to ²⁰⁷Pb, with different half-lives. If a mineral such as zircon has been a closed system since it formed, both ratios give the same age, and the analysis plots on the concordia curve.

Loss of lead (through an alteration or thermal event) or gain of lead moves the point off the curve, making it discordant, and a set of discordant points defines a discordia line. So concordance shows the opposite of a lead-loss or alteration event.

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Higher amounts of which of the following isotopes indicate plume related enriched mantle origin of early Deccan volcanic rocks.

A.\(^{87}\)Sr/\(^{86}\)Sr
B.\(^{87}\)Rb/\(^{86}\)Rb
C.\(^{3}\)He/\(^{4}\)He
D.\(^{187}\)Os/\(^{188}\)Os

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The indicator is a high \(^{3}\)He/\(^{4}\)He ratio.

Helium-3 is primordial, left over from Earth's formation, while helium-4 is continually produced by radioactive decay of uranium and thorium. Mid-ocean ridge basalts have about 8 times the atmospheric ratio, and lavas derived from deep, less degassed mantle plumes have much higher values, as at Hawaii and Iceland. High \(^{3}\)He/\(^{4}\)He in the early Deccan lavas and associated rocks is therefore taken as a fingerprint of the Reunion plume.

Sr and Os isotope ratios are raised by crustal contamination as well as by enriched sources, so they are not specific to plumes, and \(^{87}\)Rb/\(^{86}\)Rb is not a meaningful isotope ratio.

Discuss in Forum

Pb isotopic data from southern hemisphere shows compositional similarity to EM type of mantle reservoir is known as :

A.HIMU
B.NHRL
C.FOZO
D.DUPAL

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The southern-hemisphere anomaly is called DUPAL.

Stan Hart (1984) noticed that many oceanic basalts of a belt in the southern hemisphere, notably in the Indian and South Atlantic Oceans, have higher ²⁰⁸Pb/²⁰⁴Pb and ²⁰⁷Pb/²⁰⁴Pb than northern-hemisphere basalts, together with high ⁸⁷Sr/⁸⁶Sr, resembling the enriched mantle (EM) reservoirs. He named it after Dupre and Allegre, who first described it.

HIMU is a separate end-member with very high U/Pb, FOZO is the common 'focal zone' composition, and NHRL is the Northern Hemisphere Reference Line against which the DUPAL excess is measured.

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10.

Higher ratios of ⁸⁷Sr/⁸⁶Sr and lower values of epsilon Nd (εNd) are observed in continental flood basalts because of :

A.Its derivation from deeper mantle source
B.Due to lower amounts of partial melting
C.Entrainment of mantle material
D.Assimilation of continental crust

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The signature comes from Assimilation of continental crust.

Continental crust has high Rb/Sr and low Sm/Nd ratios, so over time it develops high ⁸⁷Sr/⁸⁶Sr and strongly negative εNd. When mantle-derived basalt melts and assimilates such crust on its way up, the magma inherits both features, which is why many continental flood basalts, including parts of the Deccan, show them.

A deeper mantle source or entrained mantle material would give mantle-like values (lower Sr ratio, positive εNd), and the degree of partial melting changes trace element concentrations but not isotope ratios.

Discuss in Forum

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