Home
Class 12
CHEMISTRY
Assuming the age of the earth to be 10^(...

Assuming the age of the earth to be `10^(th)` years , if the percentage of original amount of `U^(238)` still in existance on earth is x% (nearly) (`t_(1//2)` of `U^(238)` is `4.5 xx 10^(9)` years) . Then 'x/10' is ____

Text Solution

Verified by Experts

The correct Answer is:
2
`lambda t = 2.303 xx "log" ((N_(0))/(N_(t))) , (0.693)/(4.5 xx 10^(9)) xx 10^(10) = 2.303 xx log ((N_0)/(N_t)) , (3.010)/(4.5) = log ((N_0)/(N_t))`
`(10)/(15) = log ((N_0)/(N_t)) , log ((N_0)/(N_t)) = (2)/(3) , log ((N_0)/(N_t)) = 0.667 , ((N_(0))/(N_(t))) `= anti log `(0.667) , ((N_0)/(N_t)) = (4.65)/(1)`
`((N_t)/(N_0)) = (1)/(4.65%) , % ((N_t)/(N_(0))) = (1)/(4.65) xx 100 = 21.5%` , Here `((x)/(10)) = (21.5%)/(10) = 2.15 -=2`
Promotional Banner

Topper's Solved these Questions

  • NUCLEAR CHEMISTRY

    AAKASH SERIES|Exercise PRACTICE SHEET - 2 (Match the following questions)|2 Videos

  • NOBLE GASES

    AAKASH SERIES|Exercise PRACTICE SHEET - 2 (Single more than one option questions)|5 Videos

  • PERIODIC TABLE

    AAKASH SERIES|Exercise QUESTIONS FOR DESCRIPTIVE ANSWERS|37 Videos

Similar Questions

Explore conceptually related problems

In a sample of rock, the ratio of number of ""^(206)Pb to ""^(238)U nuclei is found to be 0.5. The age of the rock is (18//n)xx10^(9)(ln((3)/(2)))/(ln2) year (Assume that all the Pb nuclides in he rock was produced due to the decay of Uranium nuclides and T_(1//2)(""^(238)U)=4.5xx10^(9) year). Find n.

If the Earth's magnetic field at the equator is about 4 xx 10^(-5)T , What is its approximate magnetic dipole moment ?

Calculate the solubility of CaF_(2) in water at 298 K which is 70% dissociated. K_(sp) of CaF_(2) is 1.7 xx 10^(-10) . If answer is x xx 10^(-4) mol/ltr then x = ____?

Radioactive disintegration is a first order reaction and it's rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure . The rate of radioactive disintegration (Activity) is represented as - (dN)/(dt) = lambda N , Where lambda = decay constant , N number of nuclei at time t , N_(0) = initial no. of nuclei. The above equation after integration can be represented as lambda = (2.303)/(t) "log" ((N_(0))/(N)) Half-life period of U^(232) is 2.5 xx 10^(5) years . In how much time will the amount of U^(237) remaining be only 25% of the original amount ?