$^{131} I$ has half life period $133$ hour. After $79.8$ hour, what fraction of $^{131} I$ will remain?

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Solution

$k = \frac{0.693}{133}$
$N = N_{o} e^{(- \frac{0.693}{133} \times 79.8)}$
$\frac{N}{N_{o}} = 0.66$

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131I has half life period 133 hour. After 79.8 hour, what fraction of 131I will remain?

k=0.693133N=Noe(−0.693133×79.8)NNo=0.66

$^{131} I$ has half life period $133$ hour. After $79.8$ hour, what fraction of $^{131} I$ will remain?

$k = \frac{0.693}{133}$
$N = N_{o} e^{(- \frac{0.693}{133} \times 79.8)}$
$\frac{N}{N_{o}} = 0.66$