If the half-life of any sample of radioactive substance is 4 days, then the fraction of sample will remain undecayed after 2 days, will be?

Question

If the half-life of any sample of radioactive substance is 4 days, then the fraction of sample will remain undecayed after 2 days, will be? (A)  √2  (B)  (1/√2)  (C)  (√2-1/√2)  (D)  (1/2)

Tardigrade Admin · Accepted Answer

From Rutherford-Soddy law, the fraction left after

n

half-lives is

N = N_{0} (\frac{1}{2})^{n}

where

n

is number of half-lives.

n = \frac{time ( t )}{half-life ( T _{1/2} )} = \frac{2}{4} = \frac{1}{2}

∴ \frac{N}{N _{0}} = (\frac{1}{2})^{1/2}

\Rightarrow \frac{N}{N _{0}} = \frac{1}{2}

Q. If the half-life of any sample of radioactive substance is $4$ days, then the fraction of sample will remain undecayed after $2$ days, will be?

$2$

$\frac{1}{2}$

$\frac{2 - 1}{2}$

$\frac{1}{2}$

From Rutherford-Soddy law, the fraction left after $n$ half-lives is
$N = N_{0} (\frac{1}{2})^{n}$
where $n$ is number of half-lives.
$n = \frac{time ( t )}{half-life ( T _{1/2} )} = \frac{2}{4} = \frac{1}{2}$
$∴ \frac{N}{N _{0}} = (\frac{1}{2})^{1/2}$
$\Rightarrow \frac{N}{N _{0}} = \frac{1}{2}$

Q. If the half-life of any sample of radioactive substance is 4 days, then the fraction of sample will remain undecayed after 2 days, will be?

2​

2​1​

2​2​−1​

21​

From Rutherford-Soddy law, the fraction left after n half-lives is N=N0​(21​)n where n is number of half-lives. n= half-life (T1/2​) time (t)​=42​=21​ ∴N0​N​=(21​)1/2 ⇒N0​N​=2​1​

Q. If the half-life of any sample of radioactive substance is $4$ days, then the fraction of sample will remain undecayed after $2$ days, will be?

$2$

$\frac{1}{2}$

$\frac{2 - 1}{2}$

$\frac{1}{2}$

From Rutherford-Soddy law, the fraction left after $n$ half-lives is
$N = N_{0} (\frac{1}{2})^{n}$
where $n$ is number of half-lives.
$n = \frac{time ( t )}{half-life ( T _{1/2} )} = \frac{2}{4} = \frac{1}{2}$
$∴ \frac{N}{N _{0}} = (\frac{1}{2})^{1/2}$
$\Rightarrow \frac{N}{N _{0}} = \frac{1}{2}$