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In an ore containing uranium, the ratio ...

In an ore containing uranium, the ratio of `U^(238)` to `Pb^(206)`nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ores are the final stable product of `U^(238)`. The half-life of `U^(238)` is `4.5 xx 10^(9)` years.

A

`1.25 xx 10^(6)` yrs

B

`1.87 xx 10^(9)` yrs

C

`1.48 xx 10^(9)` yrs

D

`1.85 xx 10^(7)` yrs

Text Solution

Verified by Experts

The correct Answer is:
B
The radioactive decay follows first-order kinetic, Here we take `N_(0) = C_(0) and N_(1) = C_(t)` and `lambda = k`.
The first order rate equation for radioactive decay is :
`lambdat = 2.303 log "" (N_(0))/(N), " where" lambda = (0.693)/(t_(1//2))`
`N_(0) ` = initial nuclei (at t =0), `N_(1)` = final nuclei (at t), `lambda` = decay constant or dinintegration constant.
`U^(238) to Pb^(206)`
`N_(0) " " x " " 0 " " rArr (N_(0))/(N_(t)) = (x)/(x-y)`
`N_(t) " "x-u" " y "`
Using `lambdat = 2.303 log "" (N_(0))/(N)(N_(t))" Given" (x-y)/(y) =3 " "rArr (x)/(x-y) = (4)/(3)`
`rArr t = (2.303)/(0.693) xx 4.5 xx 10^(9) log "" (4)/(3) rArr t = 1.85 xx 10^(9)` years
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