10, gm of radioactive material of half-life 1 | vedclass

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Question

(c) Remaining material $N = \frac{{{N_0}}}{{{2^{t/T}}}}$

$ \Rightarrow N = \frac{{10}}{{{{(2)}^{20/15}}}} = \frac{{10}}{{2.15}} = 3.96\;gm$

So d

Vedclass · Accepted Answer

$6.04$

Vedclass · Answer

(c) Remaining material $N = \frac{{{N_0}}}{{{2^{t/T}}}}$

$ \Rightarrow N = \frac{{10}}{{{{(2)}^{20/15}}}} = \frac{{10}}{{2.15}} = 3.96\;gm$

So decayed material $ = 10 - 3.96 = 6.04\;gm$

$10\, gm$ of radioactive material of half-life $15$ year is kept in store for $20$ years. The disintegrated material is ............$gm$

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