The activity of a radioactive substance decreases in its 32 years, its initial value goes down to 1/16, find the half-age of the substance?
Answers
Heya ____
Ques ___ The activity of a radioactive substance decreases in its 32 years, its initial value goes down to 1/16, find the half-age of the substance ??
Ans _______
The half age of a substance in radioactive substance is meany by the time taken to decay an isotope it's half....
It is 50.0 gram in 3 years....
So after the 32 years it would be remain only 25 gram...
Thank you
Answer:
\begin{lgathered}\large{\sf\underline{ \red{\underline{Question }}}}:- \: \\\end{lgathered}
Question
:−
→ The activity of a radioactive elements drops to \dfrac{1}{16}th
16
1
th of its initial value in 32 years .find the mean life of the sample .
\begin{lgathered}\large\sf\underline{\underline{Answer }}:- \: \\ \implies \sf mean \: life (\tau) = 11.544 \: \\ \\ \large\sf\underline{ \green{\underline{To \: Find }}}:- \\ \to \sf mean \: life \: of \: the \: sample \\ \\ \large\sf\underline{ \pink{\underline{Explanation }}}:- \:\end{lgathered}
Answer
:−
⟹meanlife(τ)=11.544
ToFind
:−
→meanlifeofthesample
Explanation
:−
Given that :
Activity of element \dfrac{1}{16}
16
1
time (t) = 32 years
We know that ,
→ Activity of a radioactive substances at any time "t" is -
\begin{lgathered}\implies \sf \red{ N = N_0 \: }{e}^{ - \lambda \:t} \: \: \: \\ \\ \implies \sf \frac{N}{N_0} = {e}^{ - \lambda \: t} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \because \boxed{ \sf\frac{N}{N_0} = \frac{1}{16} } \\ \\ \implies \sf \: \frac{1}{16} = {e}^{ - \lambda \: t} \\ \\ \implies \sf \: {e}^{ 32\lambda \: } = 16 \\ \\ \sf \: taking \: \log() \: on \: both \: sides \: \\ \\ \implies \sf \log( {e}^{32 \lambda} ) = \log(16) \\ \\ \: \: \: \: \: \because \boxed{ \sf \: \log( {m}^{n} ) = n \log m} \\ \\ \implies \sf \: 32 \lambda \log e = \log( {2}^{4} ) \\ \\ \because \boxed{ \sf \log e = 1} \\ \\ \implies \: \: 32 \lambda = 4 \log 2 \\ \\ \implies \boxed{ \sf \: \lambda = \frac{ \log 2}{8} \: }\end{lgathered}
⟹N=N
0
e
−λt
⟹
N
0
N
=e
−λt
∵
N
0
N
=
16
1
⟹
16
1
=e
−λt
⟹e
32λ
=16
takinglog()onbothsides
⟹log(e
32λ
)=log(16)
∵
log(m
n
)=nlogm
⟹32λloge=log(2
4
)
∵
loge=1
⟹32λ=4log2
⟹
λ=
8
log2
We know that ,
\begin{lgathered}\to \sf \boxed{ \sf mean \: life \:(\tau) = \frac{ 1}{ \lambda} } \\ \\ \to \sf \tau= \frac{ 1}{ \frac{ \log 2}{8} } \\ \\ \to { \sf \tau = \frac{8}{ \log 2} \: } \: \: \because \boxed{ \sf \log 2 = 0.693} \\ \\ \to \sf \: \tau= \frac{8}{0.693} \\ \\ \to \boxed{ \sf \tau= 11.544 \: }\end{lgathered}
→
meanlife(τ)=
λ
1
→τ=
8
log2
1
→τ=
log2
8
∵
log2=0.693
→τ=
0.693
8
→
τ=11.544