Questions
From the graph versus t, the mean life of the radioactive sample ( in sec )is
a
1
b
2
c
3
d
0.5
detailed solution
Correct option is A
Decay Law , N=N0e−λt ⇒dNdt=−λN0e−λt ⇒ dNdt=λN0e−λt ⇒lndNdt=lnλN0−λt Comparing with y = mx +c , Slope of graph is −λ ∴−λ=8−42−6=−1 ⇒λ=1 ⇒1Tmean=1 ⇒Tmean = 1 secTalk to our academic expert!
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In a radioactive material, fraction of active material remaining after time t is 9/16. The fraction that was remaining after t/2 is :
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