A freshly prepared sample of a radioisotope of half - life `1386 s ` has activity `10^(3) ` disintegrations per second Given that `ln 2 = 0.693` the fraction of the initial number of nuclei (expressed in nearest integer percentage ) that will decay in the first `80 s ` after preparation of the sample is

A freshly prepared sample of a radioisotope of half - life `1386 s ` h

1.	A freshly prepared sample of a radioisotope of half - life `1386 s ` has activity `10^(3) ` disintegrations per second Given that `ln 2 = 0.693` the fraction of the initial number of nuclei (expressed in nearest integer percentage ) that will decay in the first `80 s ` after preparation of the sample is
Answer» Correct Answer - 4 We know `(0.693)/(t_(1//2)) = (2.303)/(t) log ((N_(0))/(N))` Given `t_(1//2) = 1386` sec t = 80 sec, `N_(0) = 1` N = remaining fraction after 80 sec `(0.693)/(1386) = (2.303)/(80) log ((1)/(N))` On solving N = 0.96 Fration decayed = 1 - 0.96 = 0.04 precentage decay `= 0.04 xx 100 = 4`

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A freshly prepared sample of a radioisotope of half - life `1386 s ` has activity `10^(3) ` disintegrations per second Given that `ln 2 = 0.693` the fraction of the initial number of nuclei (expressed in nearest integer percentage ) that will decay in the first `80 s ` after preparation of the sample is
Uranium (`""_(92)U""^(238)`) decayed to `""_(82)Pb""^(206)`. The process is `""_(92)U""^(238)underset(xalpha,ybeta)to.""_(82)Pb""^(206)` `t""_(1//2)of U""^(238)=4.5xx10""^(9)` years A sample of rock from South America contains equal number of atoms of `U""^(238)` and `Pb""^(""^(206)`. The age of the rock will beA. `4.5 xx 10^(9)` yearsB. `9 xx 10^(9)` yearsC. `13.5 xx 10^(9)` yearsD. `2.25 xx 10^(9)` years
In the atmosphere, carbon dioxide is found in two forms, i.e., `.^(12)CO_(2)` and `.^(14)CO_(2)`. Plants absorb `CO_(2)` during photosynthesis. In presence of chlorophyll, plants synthesise glucose. `6 CO_(2) + 6 H_(2) O to overset(hv)(to) C_(6)H_(12)O_(6) + 6O_(2) uarr` Half life of `.^(14)C` is 5760 years. The analysis of wooden artifacts for `.^(14)C` and `.^(12)C` gives useful information for deermination of its age. all living organisms, because of their constant exchange of `CO_(2)` with the surrounding have the same ratio of `.^(14)C` to `.^(12)C`, i.e., `1.3 xx 10^(-12)`. When an organism dies, the `.^(14)C` in it keeps on decaying as follows: `._(6)^(14)C to ._(7)^(14)N + ._(-1)^(0)e` + Energy Thus, the ratio `.^(14)C//^(12)C` decrease with the passage of time. we can be used to date anything made of organic matter, e.g., bone, skeleton, wood, etc. Using carbon dating material have been dated to about 50,000 years with accuracy. `.^(14)C` exists in atmosphere due toA. conversion of `.^(12)C` to `.^(14)C`B. Combustion of fossil fuelC. bombardement of atmosphere nitrogen by cosmic ray neutronsD. none of the above
Uranium `._(92)U^(238)` decayed to `._(82)Pb^(206)`. They decay process is `._(92)U^(238) underset((x alpha, y beta))(rarr ._(82)Pb^(206))` `t_(1//2)` of `U^(238) = 4.5 xx 10^(9)` years The analysis of a rock shows the relative number of `U^(238)` and `Pb^(206)` atoms `(Pb//U = 0.25)` The age of rock will beA. `(2.303)/(0.693) xx 4.5 xx 10^(9) log 1.25`B. `(2.303)/(0.693) xx 4.5 xx 10^(9) log 0.25`C. `(2.303)/(0.693) xx 4.5 xx 10^(9) log 4`D. `(2.303)/(4.5 xx 10^(9)) xx 0.693 log 4`
The total number of `alphaandbeta` particles emitted in the nuclear reaction `""_(92)^(238)Uto""_(82)^(214)Pb` is

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