A radioactive sample of U238 decay to Pb through a process for which half life is 4-5 - 109 years- The ratio of number of nuclei of Pb to U238 after a time of 1-5 - 109 years- Given 21-3 - 1-26

The weight based ratio of $U^{238}$ and $P b^{226}$ in a sample of rock is 4 : 3. If the half life of $U^{238}$ is $4.5 \times 10^{9}$ years, then the age of rock is:

Q. The weight based ratio of

U^{238}

and

P b^{226}

in a sample of rock in

4 : 3

. If the half life of

U^{238}

4.5 \times 10^{9}

years then the age of rock is

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A radioactive sample of U238 decay to Pb through a process for which half life is 4.5×109 years. The ratio of number of nuclei of Pb to U238 after a time of 1.5×109 years. (Given 21/3 = 1.26)

The correct option is A 0.26Here [NN0=(12)n=(12)1/3]where n= Number of half lives [=13][NN0=11.26] [NUNPb+NU=11.26] [NPb=0.26 NU][NPbNU=0.26]

A radioactive sample of $U^{238}$ decay to $P b$ through a process for which half life is $4.5 \times 10^{9}$ years. The ratio of number of nuclei of $P b$ to $U^{238}$ after a time of $1.5 \times 10^{9}$ years.
(Given 21/3 = 1.26)

The correct option is A 0.26
Here $[\frac{N}{N_{0}} = {(\frac{1}{2})}^{n} = {(\frac{1}{2})}^{1 / 3}]$
where $n =$ Number of half lives $[= \frac{1}{3}]$
$[\frac{N}{N_{0}} = \frac{1}{1.26}]$ $[\frac{N_{U}}{N_{P b} + N_{U}} = \frac{1}{1.26}]$ $[N_{P b} = 0.26 N_{U}]$
$[\frac{N_{P b}}{N_{U}} = 0.26]$