You visited us 1 times! Enjoying our articles? Unlock Full Access!

Surface Energy

The weight ba...

Question

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:

9.0 \times 10^{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.3 \times 10^{9} y e a r s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.5 \times 10^{9} y e a r s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.78 \times 10^{9} y e a r s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $3.78 \times 10^{9} y e a r s$
$t = \frac{l n (\frac{A_{0}}{A_{t}})}{k} = \frac{l n (\frac{\frac{4}{238} + \frac{3}{226}}{\frac{4}{238}})}{l n 2} t_{\frac{1}{2}}$
$k = \frac{l n_{2}}{t_{\frac{1}{2}}} ≃ 3 \cdot 78 \times 10^{9} y e a r s$

Suggest Corrections

Similar questions

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

A sample of uranium rock from moon contains equimolar amounts of $U^{238}$ and $P b^{206}$ . The age of the rock will be $(t_{\frac{1}{2}} f o r U^{238} = 4.5 \times 10^{9} y r s)$

Q. A sample of

U^{238}

(half life

= 4.5 \times 10^{9}

year), ore is found to contain

23.8 g

U^{238}

and

20.6 g

P b^{206}

The age of the ore is

X \times 10_{9}

year. Calculate the value of

2 X

Q. In a sample of rock, the ratio of

^{206} P b

^{238} U

nuclei is found to be 0.5. The age of the rock is (given half-life of

U^{238}

4.5 \times 10^{9}

years)

A radioactive sample of $U^{238}$ decay to pb through a process for which half-life is $4.5 \times 10^{9}$ years. The ratio of number of nuclei of pb to $U^{238}$ after a time of $4.5 \times 10^{9}$ years $(g i v e n 2^{\frac{1}{3}} = 1.26)$

Join BYJU'S Learning Program

The weight based ratio of U238 and Pb226 in a sample of rock is 4 : 3. If the half life of U238 is 4.5×109 years, then the age of rock is:

The correct option is C 3.78×109yearst=ln(A0At)k=ln(4238+32264238)ln2t12k=ln2t12≃3⋅78×109years

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:

The correct option is C $3.78 \times 10^{9} y e a r s$
$t = \frac{l n (\frac{A_{0}}{A_{t}})}{k} = \frac{l n (\frac{\frac{4}{238} + \frac{3}{226}}{\frac{4}{238}})}{l n 2} t_{\frac{1}{2}}$
$k = \frac{l n_{2}}{t_{\frac{1}{2}}} ≃ 3 \cdot 78 \times 10^{9} y e a r s$