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The number of `alpha`-particles emitted per second by 1g fo `.^(226)Ra` is `.7 xx 10^(10)`. The decay constant is:A. `1.39 xx 10^(-11) sec^(-1)`B. `13.9 xx 10^(-11) sec^(-1)`C. `139 xx 10^(-10) sec^(-1)`D. `13.9 xx 10^(-10) sec^(-1)` |
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Answer» Correct Answer - a `("No. of atoms disintegration per second")/("Total number of atoms present")` or `(3.7 xx 10^(10))/((6.02 xx 10^(23))/(226)) = (226 xx 3.7 xx 10^(10))/(6.02 xx 10^(23)) = lambda` |
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