Experiments show that Radium disintegrates at a rate proportional to the amount of Radium present at the moment.Its half-life is 159 yrs.What % will disappear in 1 year?

Asked by aksharaprachi | 15th May, 2015, 04:22: PM

Expert Answer:

fraction numerator d R over denominator d t end fraction proportional to R rightwards double arrow fraction numerator d R over denominator d t end fraction equals K R comma space w h e r e space K space i s space a space c o n s tan t. rightwards double arrow d R equals K R d t G i v e n space t h a t space h a l f minus l i f e space left parenthesis t i m e space t a k e n space f o r space t h e space a m o u n t space o f space R a d i u m space t o space b e c o m e space h a l f space o f space i n i t i a l right parenthesis space i s space 159 space y e a r s. A s s u m i n g space i n i t i a l space a m o u n t space o f space r a d i u m space t o space b e space R subscript 0 comma space i n t e g r a t e space t h e space a b o v e space e q u a t i o n space w i t h space l i m i t s. integral subscript R subscript 0 end subscript superscript R subscript 0 divided by 2 end superscript fraction numerator d R over denominator R end fraction equals integral subscript 0 superscript t subscript 1 divided by 2 end subscript end superscript K d t rightwards double arrow ln open parentheses 1 half close parentheses equals K t subscript 1 divided by 2 end subscript rightwards double arrow K equals negative fraction numerator ln 2 over denominator t subscript 1 divided by 2 end subscript end fraction equals negative fraction numerator ln 2 over denominator 159 end fraction N o w space f o r space 1 space y e a r integral subscript R subscript 0 end subscript superscript R fraction numerator d R over denominator R end fraction equals K integral subscript 0 superscript 1 d t rightwards double arrow ln open parentheses R over R subscript 0 close parentheses equals K open parentheses 1 minus 0 close parentheses rightwards double arrow R over R subscript 0 equals e to the power of K equals e to the power of ln open parentheses 1 half close parentheses to the power of 1 divided by 159 end exponent end exponent  rightwards double arrow R over R subscript 0 equals open parentheses 1 half close parentheses to the power of 1 divided by 159 end exponent rightwards double arrow 1 minus R over R subscript 0 equals 1 minus open parentheses 1 half close parentheses to the power of 1 divided by 159 end exponent P e r c e n t a g e space d i s a p p e a r e d space equals space open parentheses 1 minus R over R subscript 0 close parentheses cross times 100 equals open parentheses 1 minus open parentheses 1 half close parentheses to the power of 1 divided by 159 end exponent close parentheses cross times 100 almost equal to 0.43 %

Answered by satyajit samal | 17th May, 2015, 12:05: AM