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# Radioactive substance has 10 ^8 nuclei, it`s half life is 30s the number of nuclei left after 15s is

4 years ago
Dear Student,
0.693/t1/2 = k
t1/2= half life
k=disintegration or rot consistent
=>0.693/30 = k
=>k=0.0231
k=2.303log(a/a-x)/t
a=initial sum
a-x= sum left after deterioration
t=time of rot
=>0.0231=2.303log(108/108-x)/15
=>108/108-x=1.41
=>Amount left=108-x=7,09,21,986 nuclei.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
one year ago

A radioactive for the substance has 10^8 nuclei

k=0.693/t1/2

t1/2= half life (Its half life is 30s)

k=disintegration or rot consistent  0.693/30 = k

k=0.0231

k=2.303log(a/a-x)/t

a=initial sum

a-x= sum left after deterioration

t=time of rot

0.0231=2.303log(108/108-x)/15

108/108-x=1.41

Amount left=108-x=7,09,21,986 nuclei.

Therefore The no, of nuclei left after 15s is nearly = 7,09,21,986 nuclei