Question

if 15/16 of portion of radioactive nuclie decays in 16/15 min
then determin the half
life​

Answer 1

Answer:

Let

$( \frac{1}{2} ) {}^{n} = \frac{15}{16} \\ \\ - n \: log(2) = log( \frac{15}{16} ) \\ \\ n = \frac{ log(16) - log(15) }{ log(2) }$

Now time taken to decay n half portion will be equal to n half lifes

So

$n \times t_{ \frac{1}{2} } = \frac{16}{15} \\ \\ t_{ \frac{1}{2} } = \frac{1}{n} \times \frac{16}{15} \\ \\ t_{ \frac{1}{2} } = \frac{16}{15} . \: \frac{ log(2) }{ log(16) - log(15) } \\ \\ t_{ \frac{1}{2} } = \frac{16}{15} . \: \frac{ log(2) }{4 log(2) - log(3) - log(5) } \\ \\ t_{ \frac{1}{2} } \approx \frac{16}{15} . \: \frac{0.3}{4 \times 0.3 - 0.48 - 0.7} \\ \\ t_{ \frac{1}{2} } \approx \frac{16}{15} \: . \frac{0.3}{0.02} = \frac{16}{15} \times \frac{30}{2} \\ \\ \huge \red{t_{ \frac{1}{2} } \approx16 \: min}$

Answer 2

Explanation:

Let

\begin{gathered}( \frac{1}{2} ) {}^{n} = \frac{15}{16} \\ \\ - n \: log(2) = log( \frac{15}{16} ) \\ \\ n = \frac{ log(16) - log(15) }{ log(2) } \end{gathered}(21)n=1615−nlog(2)=log(1615)n=log(2)log(16)−log(15)

Now time taken to decay n half portion will be equal to n half lifes

So

\begin{gathered}n \times t_{ \frac{1}{2} } = \frac{16}{15} \\ \\ t_{ \frac{1}{2} } = \frac{1}{n} \times \frac{16}{15} \\ \\ t_{ \frac{1}{2} } = \frac{16}{15} . \: \frac{ log(2) }{ log(16) - log(15) } \\ \\ t_{ \frac{1}{2} } = \frac{16}{15} . \: \frac{ log(2) }{4 log(2) - log(3) - log(5) } \\ \\ t_{ \frac{1}{2} } \approx \frac{16}{15} . \: \frac{0.3}{4 \times 0.3 - 0.48 - 0.7} \\ \\ t_{ \frac{1}{2} } \approx \frac{16}{15} \: . \frac{0.3}{0.02} = \frac{16}{15} \times \frac{30}{2} \\ \\ \huge \red{t_{ \frac{1}{2} } \approx16 \: min}\end{gathered}n×t21=1516t21=n1×1516t21=1516.log(16)−log(15)log(2)t21=1516.4log(2)−log(3)−log(5)log(2)t21≈1516.4×0.3−0.48−0.70.3t21≈1516.0.020.3=1516×230t21≈16min