Question

The radioactivity of an old sample of whisky due to tritium ( half life $25$yr) was found to be only about $5$% of that measured in a recently purchased bottle marked $7$ yr old. The sample must have been prepared about

• A. $70yr$
• B. $95yr$
• C. $115yr$
• D. $125yr$

Answer 1

The correct option is C $115yr$

Radioactivity initially $=N_0$

After $7$ years$=N_o{e}^{-\lambda 7}\cdots \cdots \cdots \cdots \cdots \left[1\right]$

After $x$ years $=N_o{e}^{-\lambda x}\cdots \cdots \cdots \cdots \cdots \left[2\right]$

Given, [2] is 5% of [1]

$\frac{5}{100}\times N_o{e}^{-7\lambda }=N_o{e}^{-\lambda x}and\lambda =\frac{ln2}{25}[\therefore given]$

$\Rightarrow {\frac{e}{20}}^{\frac{-7ln2}{25}}=e^{\frac{-ln2x}{25}}$

$\Rightarrow ln\left(\frac{1}{20}\right)-\frac{7ln2}{25}=\frac{-ln2x}{25}$

$\Rightarrow x=\frac{(3+00.19408)\times 25)}{ln2}$

$\Rightarrow x=115.226years$