A sample of a radioactive substance undergoes 80% decomposition in 345 minutes its half life is
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Answer:
Let the initial concentration of radioactive substance be 100.
Then, the concentration left is =100-80
= 20
log 5=0.7
NOTE- DECOMPOSITION OF RADIOACTIVE SUBSTANCE IS THE FIRST ORDER REACTION.
SOLUTION-
t=2.303/k*logAi/Af
345=2.303/k*log100/20
k=2.303/345*log5
k= 2.303/345*0.7
k=0.004672
t half= 0.693/k
= 0.693/0.004672
=148.33 minutes.
I HOPE THAT IT HELPED YOU.
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