^{226}Ra disintegrates at such a rate that af | vedclass

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(C) The radioactive decay formula is given by $N = N_o (1/2)^n$,where $n$ is the number of half-lives.Given that after $3160 \ years$,the remaining am

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$1580$

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(C) The radioactive decay formula is given by $N = N_o (1/2)^n$,where $n$ is the number of half-lives.Given that after $3160 \ years$,the remaining amount $N = (1/4)N_o$,so $N/N_o = 1/4$.Substituting this into the formula: $1/4 = (1/2)^n$.Since $1/4 = (1/2)^2$,we get $n = 2$.The number of half-lives $n$ is related to total time $t$ and half-life $t_{1/2}$ by $n = t / t_{1/2}$.Therefore,$2 = 3160 / t_{1/2}$.$t_{1/2} = 3160 / 2 = 1580 \ years$.

$^{226}Ra$ disintegrates at such a rate that after $3160 \ years$ only one-fourth of its original amount remains. The half-life of $^{226}Ra$ will be .......... $years$.

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