1.0 g of radioactive sodium decays to 0.25 | vedclass

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(B) The radioactive decay follows the formula $N_t = N_0 	imes (1/2)^n$,where $n = t/T_{1/2}$.First,find the half-life $(T_{1/2})$: $0.25 \ g = 1.0 \

Vedclass · Accepted Answer

$32$

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(B) The radioactive decay follows the formula $N_t = N_0 	imes (1/2)^n$,where $n = t/T_{1/2}$.First,find the half-life $(T_{1/2})$: $0.25 \ g = 1.0 \ g 	imes (1/2)^n$,which gives $(1/2)^2 = (1/2)^n$,so $n = 2$.Since $n = t/T_{1/2}$,$2 = 16 \ hours / T_{1/2}$,thus $T_{1/2} = 8 \ hours$.Now,for $48 \ g$ to become $3.0 \ g$: $3.0 = 48 	imes (1/2)^n$.$(1/2)^n = 3.0 / 48 = 1/16 = (1/2)^4$,so $n = 4$.Total time $t = n 	imes T_{1/2} = 4 	imes 8 \ hours = 32 \ hours$.

$1.0 \ g$ of radioactive sodium decays to $0.25 \ g$ in $16 \ hours$. Calculate the time required for $48 \ g$ of the same radioactive sodium to decay to $3.0 \ g$. (in $hours$)

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