6.5-bo'lim : Ilovalar
4. Bizda dastlab 100 gramm radioaktiv element bor va 1250 yildan keyin 80 gramm qoladi.
- Ushbu element uchun eksponensial yemirilish tenglamasini aniqlang.
- Elementning yarmi parchalanishi uchun qancha vaqt kerak bo'ladi?
- Elementdan atigi 1 gramm qolguncha qancha vaqt ketadi?
a Yechim
Biz radioaktiv elementning dastlabki miqdori 100 va shuning uchun \({Q_0} = 100\) ekanligini bilishimizni tan olishimiz bilan boshlashimiz mumkin. Shunday qilib, tenglama shunday bo'ladi:
\[Q\left( t \right) = 100{{\bf{e}}^{k\,t}}\]Endi biz \(Q\left( {1250} \right) = 80\) va uni yuqoridagi tenglamaga kiritish shuni ham bilamizki,
\[80 = Q\left( {1250} \right) = 100{{\bf{e}}^{1250k}}\]dan texnikalardan foydalanishimiz mumkin Logarifm tenglamalarini yechish \(k\) qiymatini aniqlash uchun bo'lim.
\[\begin{align*}80 & = 100{{\bf{e}}^{1250k}}\\ \frac{{80}}{{100}} & = {{\bf{e}}^{1250k}}\\ \ln \left( {\frac{4}{5}} \right) & = 1250k\\ k = \frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right) & = - 0.000178515\end{align*}\]Sizning afzalliklaringizga qarab, biz aniq qiymat yoki kasr qiymatidan foydalanishimiz mumkin. Shuni yodda tutingki, \(k\) eksponensial funktsiyaning ko'rsatkichida bo'lganligi sababli, agar biz haddan tashqari yaxlitlashimiz mumkin bo'lgan qiymatdagi potentsial katta farqlarga yo'l qo'ymaslik uchun bir nechta o'nli kasrlardan foydalanishimiz kerak.
Bularning barchasini jamlagan holda, ushbu populyatsiya uchun eksponensial yemirilish tenglamasi,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{Q = 100{{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t}}}}\]b Yechim
Bu erda bizdan haqiqatan ham so'ralayotgan narsa tenglamani yechishdir,
\[50 = Q\left( t \right) = 100{{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t}}\]va biz dan bilamiz Logarifm tenglamalarini yechish buni qanday qilish kerak bo'limi. Mana bu qism uchun yechim ishi.
\[\begin{align*}\frac{{50}}{{100}} & = {{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t}}\\ \ln \left( {\frac{1}{2}} \right) & = \frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{1250\ln \left( {\frac{1}{2}} \right)}}{{\ln \left( {\frac{4}{5}} \right)}} = 3882.8546}}\end{align*}\]Elementning yarmi parchalanishi uchun 3882,8546 yil kerak bo'ladi. Yon eslatmada bu vaqt deyiladi yarim yemirilish davri elementdan.
c Yechim
Ushbu qismda bizdan tenglamani yechish so'raladi,
\[1 = Q\left( t \right) = 100{{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t}}\]Ushbu qismni hal qilish jarayoni avvalgi qism bilan bir xil. Mana bu qism uchun yechim ishi.
\[\begin{align*}\frac{1}{{100}} & = {{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t}}\\ \ln \left( {\frac{1}{{100}}} \right) & = \frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{1250\ln \left( {\frac{1}{{100}}} \right)}}{{\ln \left( {\frac{4}{5}} \right)}} = 25797.1279}}\end{align*}\]25 797,1279 yildan keyin faqat 1 gramm element qoladi.