0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# In a Uranium ore the ratio of $$U^{238}$$ nuclei to $$Pb^{206}$$ nuclei is $$\eta = 2.8$$. Evaluate the age of the ores, assuming all the lead $$Pb^{206}$$ to be a final decay product of the uranium series. The half life of $$U^{238}$$ nuclei is $$4.5 \times 10^{9}$$ years.

Solution
Verified by Toppr

#### What this implies is that in the time since the ore was formed, $$\dfrac{\eta}{1 + \eta} U^{238}$$ nuclei have remained undecayed. Thus $$\dfrac{\eta}{1 + \eta} = e^{-t \times \dfrac{ln\ 2}{T_{1/2}}}$$ Taking logarithm both side , we get or $$t = T_{\dfrac{1}{2}} \dfrac{ln \dfrac{1 + \eta}{\eta}}{ln\ 2}$$ Substituting $$T_{\dfrac{1}{2}} = 4.5 \times 10^{9}$$ years , $$\eta = 2.8$$we get $$t = 1.98 \times 10^{9}$$ years

Was this answer helpful?
0
Similar Questions
Q1
In a Uranium ore the ratio of $$U^{238}$$ nuclei to $$Pb^{206}$$ nuclei is $$\eta = 2.8$$. Evaluate the age of the ores, assuming all the lead $$Pb^{206}$$ to be a final decay product of the uranium series. The half life of $$U^{238}$$ nuclei is $$4.5 \times 10^{9}$$ years.
View Solution
Q2
In a uranium ore the ratio of U238 to Pb206 nuclei is 1:15. Evalute the age of ore, assuming Pb206 to be the final, initially only uranium was present (given ratio is of number of nuclei).
(Half life= 4.5×109 years)
View Solution
Q3
In an ore containing uranium the ratio of U238 to Pb206 nuclei in 3. Calculate the age of ore, assuming that all the lead present in the ore is the final stable product of U238. t1/2 for U238=4.5×109year.
View Solution
Q4
In an ore containing uranium, the ratio of U238 to Pb206 is 3. Calculate the age of the ore, assuming that all the leas present in the ore is the final stable product of U238. Take the half-life of U238 to be 4.5×109 yr.
View Solution
Q5
92U238 by successive radioactive decays changes to 82Pb206. A sample of uranium ore was analyzed and found to contain 1.0 g of U238 and 0.1 g of Pb206. Assuming that all the Pb206 had accumulated due to decay of U238, find out the age of the ore. (Half life of U238=4.5×109years)
View Solution