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Let `N_(beta)` be the number of `beta` particles emitted by `1` gram of `Na^(24)` radioactive nuclei (half life `= 15` hrs) in `7.5` hours, `N_(beta)` is close to (Avogadro number `= 6.023 xx 10^(23) // "g. mole"`) :-A. `7.5 xx 10^(21)`B. `1.75 xx 10^(22)`C. `6.2 xx 10^(21)`D. `1.25 xx 10^(22)` |
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Answer» Correct Answer - A No. of nuclei remains at any instant `N= N_(0)/(2^(t//T_(1//2)))=N_(0)/sqrt(2)` `N_("decayed")=N_(0)-N=N_(0) [1-1/sqrt(2)]` `=1/24xx6.023xx10^(23)xx0.3` `~~7.5xx10^(21)` |
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