The time elapsed between 25% and 50% decay of a radioactive sample having half life of 1 hour is
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a
ln3−ln2ln2hr
b
ln3ln2hr
c
2ln2ln3hr
d
ln4−ln3hr
answer is A.
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Detailed Solution
N1=N0−25% of N0=3N04and N2=N0−50% of N0=N02∴ 3N04=N0e−λt1⇒eλt1=43⇒t1=ln43λ=T.ln43ln2Again, N02=N0e−λt2⇒eλt2=2⇒t2=ln2λ=T∴ t2−t1=T−T.ln43ln2=Tln2−2ln2+ln3ln2⇒ t2−t1=1.ln3−ln2ln2hr