A radioactive sample decays by 63% in 10 seconds. Time taken by it to decay by 50% in seconds is [Take ln(2) = 0.693] (Write upto two digits after the decimal point.)

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Solution

It takes a time equal to one time-constant to decay by 63%. Therefore, time-constant, $τ = 10 s$

$N = N_{o} e^{\frac{t}{τ}}$
$\Rightarrow N = N_{o} e^{- \frac{t}{10}}$
For 50% decay,
$∴ \frac{N_{o}}{2} = N_{o} e^{- \frac{t}{10}} \Rightarrow \frac{t}{10} = 0.693 \Rightarrow t = 6.93$

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A radioactive sample decays by 63% in 10 seconds. Time taken by it to decay by 50% in seconds is [Take ln(2) = 0.693] (Write upto two digits after the decimal point.)

It takes a time equal to one time-constant to decay by 63%. Therefore, time-constant, τ=10 s N=Noetτ ⇒ N=No e−t10 For 50% decay, ∴ No2=No e−t10⇒t10=0.693⇒t=6.93

It takes a time equal to one time-constant to decay by 63%. Therefore, time-constant, $τ = 10 s$

$N = N_{o} e^{\frac{t}{τ}}$
$\Rightarrow N = N_{o} e^{- \frac{t}{10}}$
For 50% decay,
$∴ \frac{N_{o}}{2} = N_{o} e^{- \frac{t}{10}} \Rightarrow \frac{t}{10} = 0.693 \Rightarrow t = 6.93$