A sample of `._(19)K^(40)` disintegrates into two nuclei Ca & Ar with decay constant `lambda _(Ca)=4.5 xx10^(-10) S^(-1)` and `lambda_(Ar)=0.5xx 10^(-10)S^(-1)` respectively. The time after which 99% of `._(19)K^(40)` gets decayed is

The rate constants for a reaction at 400 K and 500 K are 2 . 60 xx 10^(-5) s^(-1) and 2.60 xx 10^(-3) s^(-1) respectively. The activation energy of the reaction in kJ mol^(-1) is

The rate of constant of reaction at 300 K//320 K are 5 xx 10^(-4) s^(-1) and 2.0 xx 10^(-3)s^(-1) respectively. Calculate the value of activation energy of reaction.

In a radioactive sample. ._(19)^(40)K nuclei either decay into stable ._(20)^(40)Ca nuclei with decay constant 4.5 xx 10^(-10) per year or into stable ._(18)^(40)Ar nuclei with decay constant 0.5 xx 10^(-10) per year. Given that in this sample all the stable ._(20)^(40)Ca and ._(18)^(40)Ar nuclei are produced by the ._(19)^(40)K nuclei only. In time t xx 10^9 years. If the ratio of the sum of stable ._(20)^(40)Ca and ._(18)^(40)Ar nuclei to the radioactive ._(19)^(40)K nuclei is 99. The value of t will be. [Given : In 10 = 2.3]

Identify the reaction order from each of the following rate constants. i) k=2.3 xx10^(5) L mol ^(-1) s ^(-1) ii) K=3 xx10^(-4)s^(-1)

A Bi^(210) radionuclide decays via the chain Bi^(210)(beta^(-)-"decay")/(lambda_(1))Po^(210)(alpha-decay)/(lambda_(2)) Pb^(206) (stable), where the decay constants are lambda_(1) = 1.6 xx 10^(-6)s^(-1), T_(1//2) ~~ 5 days , lambda_(2) = = 5.8 xx 10^(-8)s^(-1), T_(1//2) ~~ 4.6 months. alpha & beta activites of the 1.00 mg a month after its manufacture is (x)/(5) xx 10^(11) Find x.2^((1)/(4.6)) = 0.86

A sample containing same number of two nuclel A and B start decaying. The decay constant of A and B are 10lambda and lambda . The time after which (N_(A))/(N_(B)) becomes (1)/(e) is

Two radioactive nuclei have same number of nuclei initially and decay constants as 5lambda . And 4lambda respectively. The ratio of number of nuclei will be 1/e^(2) after time: