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VITEEE 2014 Paper

Section: Physics

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Question : 28 of 120

Marks: +1, -0

The activity of a radioactive sample is measured as

N_0

counts per minute at t = 0 and

N_0

/C counts per minute at t = 5 min. The time, (in minute) at which the activity reduces to half its value, is

$\frac{\log_e 2}{5}$
$\frac{5}{\log_e 2}$
5 $\log_{10}$ 2
5 $\log_e$ 2

Solution:

After n half-lives

\frac{N}{N_0}

=

\left(\frac{1}{2}\right)^n

=

\left(\frac{1}{2}\right)^{\frac{t}{7}}

N =

N+\frac{0}{e}

=

\frac{cN_0}{cN_0}

=

\left(\frac{1}{2}\right)^{\frac{5}{7}}

⇒

\frac{1}{e}

=

\left(\frac{1}{2}\right)^{\frac{5}{7}}

Taking log on both sides, we get

log 1 - log e =

\frac{5}{7}

log

\frac{1}{2}

- 1 =

\frac{5}{7}

(- log 2)

T =

5 \log_e 2

Now, let t’ be the time after which activity reduces to half

\left(\frac{1}{2}\right)

=

\left(\frac{1}{2}\right)^{\frac{1'}{5 \log_e 2}}

t' = 5

\log_e

2

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