Question

Q13B.12 A sample of a certain radioactive isotope has an activity that we observe to decrease by a factor of 47 dur- ing a year. What is the isotope's half-life?

Answer 1

Solution:

Using the radio active decay realtion,

R = R_o e^{- $\lambda$ t}

Given, R = R_o/47 and t = 1yr

R_o/47 = R_o e^{- $\lambda$ *1yr}

Solve for $\lambda$ decay constant.

1/47 = e^{- $\lambda$}

- $\lambda$ = ln [1/47]

- $\lambda$ = -3.85

$\lambda$ = 3.85/yr

Now Using the realtion of T_1/2,

T_1/2 = ln[2]/ $\lambda$

= 0.693/3.85

= 0.18 yr.

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