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Mathematics 13 Online
OpenStudy (anonymous):

An isotope has a half-life of 15 days. If one is to make a table showing the half-life decay of a sample of this isotope from 32 grams to 1 gram; list the time (in days, starting with t = 0) in the first column and the mass remaining (in grams) in the second column, which type of sequence is used in the first column and which type of sequence is used in the second column?

rishavraj (rishavraj):

@allydiaz wht do u mean by sequence here?? u must be knowing tht decaying process is 1 order reaction....

OpenStudy (anonymous):

I don't fully understand what the question is asking...

OpenStudy (anonymous):

I think once sequence would be geometric and another arithmetic. Maybe

rishavraj (rishavraj):

ok |dw:1427585569054:dw|

rishavraj (rishavraj):

\[k = \lambda \]

rishavraj (rishavraj):

on right side its remaining conc. of radioactive substance.......

rishavraj (rishavraj):

11 days - 16 grms??????

OpenStudy (anonymous):

oh sorry I would change it for every 15 days instead ( read the numbers wrong)

rishavraj (rishavraj):

yup now the table is correct....

rishavraj (rishavraj):

on left side its A.P and right side its G.P

OpenStudy (anonymous):

What does that mean?

rishavraj (rishavraj):

i just told about the nature of sequence on each side of equality

OpenStudy (anonymous):

ok so the right side is arithmetic and the left side is geometric?

rishavraj (rishavraj):

no no.....the days column is arithmetic and mass column is geometric..... left side - arithmetic right side - geometric

OpenStudy (anonymous):

Ah ok. So how do I apply these numbers to the table you previously made?

rishavraj (rishavraj):

hey man , thts ur answer.... u got those values when u plug data in equations i provided in tht table

OpenStudy (anonymous):

The remaining conc. part has me confused b/c idk what k= (weird symbol) means

rishavraj (rishavraj):

see for radioactivity half life i.e \[t_{\frac{ 1 }{ 2 }} = \frac{ \ln 2 }{ \lambda } \] \[\lambda ~~~ is ~~~ decay~~~~~constant \]

OpenStudy (anonymous):

Ok. I don't think I need that for this problem though.

OpenStudy (anonymous):

Thanks for your help @rishavraj

rishavraj (rishavraj):

u r most welcome....

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