A radioactive sample of half life 23. 1days is disintegrating...

Given, half life $\left({T}_{1/2}\right) = 23 . 1$ days

If ${N}_{0}$ be the initial amount, then active nuclei at $t = 15$ days

${N}_{1} = {N}_{0}e^{- 15\lambda} . . . \left(i\right)</p><p>Active nuclei at$t = 16$days</p><p>${N}_{2} = {N}_{0}e^{- 16\lambda} . . . \left(ii\right)$</p><p>Disintegration constant,</p><p>$\lambda = \dfrac{0 . 693}{T1 / 2} = \dfrac{0 . 693}{23 . 1} =0 . 03$</p><p>$∴ %$decay in$15th to 16th$days</p><p>$= \dfrac{{N}_{1} - {N}_{2}}{N1} \times100 = \dfrac{{N}_{0}e^{- 15\lambda} - {N}_{0}e^{- 16\lambda}}{{N}_{0}e^{- 15\lambda}} \times100$</p><p>$= \left(1 - e^{- \lambda}\right) \times100 = \left(1 - e^{- 0 . 03}\right) \times100$</p><p>$= \left(1 - \dfrac{1}{e0 . 03}\right) \times100 = \left(1 - \dfrac{1}{1 . 03}\right) \times100 = 2 . 9 %$