Physics, asked by Puspakio, 1 year ago

The age of 40 years of a radioactive substance is decayed and 16 or 1/16 of it will survive? What is the 6 determinants of this substance? (log_e2 = 0.693)

Answers

Answered by Swarnimkumar22
16
\bold{\huge{\underline{Question}}}

The age of 40 years of a radioactive substance is decayed and 16 or 1/16 of it will survive? What is the 6 determinants of this substance? (log_e2 = 0.693) ?



\bold{\huge{\underline{Answer-}}}



\bold{\underline{Solution-}}



\bold{\underline{Given\:that}}

Half-life T = 40 years

N = N0/16



Now, Using the Formula



  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \boxed  { \boxed { \bf{N = N _{0}( \frac{1}{2} ) {}^{n}}}} \:




 \bf \: now \: putting \: the \: above \: values \\  \bf \: in \: formula \:  \\  \\  \\  \\  \bf\frac{N_{0}}{16}\: = N_{0}(\frac{1}{2} ) {}^{n}  \\   \\ \\  \bf \:  \frac{ \cancel{N_{0}}}{16 \times  \cancel{N_{0}} }  =  (\frac{1}{2}  ){}^{n}  \\  \\  \\  \bf \:  \frac{1}{16}  = ( \frac{1}{2} ) {}^{n}  \\  \\  \\  \bf \tiny \: you \: know \: now \: we \: can \: write \: it \: like \\  \\  \bf \: ( \frac{1}{2} ) {}^{4}  =  {( \frac{1}{2} })^{n}  \\  \\  \\  \\ so \bf \:  \:  \:  \:  \:  \:  \:  \:  \:  \: n \:  = 4





Therefore , We need to find Total years you know what's the formula for finding Total years



\bold{\underline{Formula\:for\:Finding\:Total\:years}}



 \boxed { \boxed{ \bf{t = nT}}}



Now , putting the above values in formula



So, 4 × 40 = 160 Years,




Now, decayed radioactivity substance is,




 \lambda \:  =  \frac{0.693}{T}  \\  \\  \implies \frac{0.693}{40}  \\  \\  \implies0.0173 \:  \:  \: per \: year
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