If 3/4 of a radioactive element disintegrates | vedclass

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(A) The amount of radioactive element remaining after $2$ hours is $N = N_o - \frac{3}{4} N_o = \frac{1}{4} N_o$.We know that the amount remaining aft

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$1$ hour

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(A) The amount of radioactive element remaining after $2$ hours is $N = N_o - \frac{3}{4} N_o = \frac{1}{4} N_o$.We know that the amount remaining after $n$ half-lives is given by $N = N_o (1/2)^n$.Substituting the values,$\frac{1}{4} N_o = N_o (1/2)^n$,which implies $(1/2)^2 = (1/2)^n$.Therefore,the number of half-lives $n = 2$.Since $n = \frac{t}{t_{1/2}}$,we have $2 = \frac{2 \ hours}{t_{1/2}}$.Thus,$t_{1/2} = 1 \ hour$.

If $3/4$ of a radioactive element disintegrates in $2$ hours,its half-life would be

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