The β ^ - activity of a sample of C O _ 2 prepared from a contemporary wood gave a count rate of 25.5 counts per minute ( cpm ) . The same mass of C O _ 2 from an ancient wooden statue gave a count rate of 20.5 c pm

Correct answer: A. 1822years r = 20.5cpm, r _ 0 = 25.5cpm ∵ r _ 0 ∝ N _ 0 andr ∝ N ∴ (r0)/(r) = (N0)/(N) Also, t = (2.303)/(λ)log ((N0)/(N)) = (2.303)/(λ) ((r0)/(r)) t = (2.303 × 5770)/(0.693) ((25.5)/(20.5)) =1822years

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Nuclear Physics

NTA Abhyas 2020 · Physics · Nuclear Physics

The β ^ - activity of a sample of C O _ 2 prepared from a...

The

{β}^{-}

activity of a sample of

C{O}_{2}

prepared from a contemporary wood gave a count rate of

\text{25.5}

counts per minute

( cpm )

. The same mass of

C{O}_{2}

from an ancient wooden statue gave a count rate of

\text{20.5} c

pm under& the same conditions. If the half life of

{}^{14}C

5770years

, then the age of the statue is close to [Take

\log_{10}\left(\dfrac{255}{205}\right) \approx 0. 095

]

Options

A $1822years$
B $182years$
C $822years$
D $18220years$

Correct answer

A. $1822years$

Step-by-step solution

r = 20.5cpm, {r}_{0} = 25.5cpm $<br />$ ∵ {r}_{0} ∝ {N}_{0}andr ∝ N $<br />$ ∴ \dfrac{r0}{r} = \dfrac{N0}{N} $<br />Also,$ t = \dfrac{2.303}{λ}log $\left(\dfrac{N0}{N}\right) = \dfrac{2.303}{λ}$ \log \left(\dfrac{r0}{r}\right)< $br />$ t = \dfrac{2.303 \times 5770}{0.693}\text\log $\left(\dfrac{25.5}{20.5}\right) =1822years$

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