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Chemistry Β· 2026 Β· 3 marks
CBSE 2026 Β· Region 3 Β· Set 1 Β· Q22
The half-life period of a radioactive element is 1.5 Γ 10^10 years. Calculate the time in which the activity of the element is reduced to 75% of itsβ¦
The half-life period of a radioactive element is $\displaystyle 1.5 \times 10^{10}$ years. Calculate the time in which the activity of the element is reduced to $\displaystyle 75$% of its original value. [Given : $\displaystyle \log 2=0 \cdot 30, \log 3=0 \cdot 48, \log 4=0 \cdot 60$ ]
Marking-scheme solution
k = $\displaystyle 0.693$
π‘$\displaystyle 1$
k =
$\displaystyle 1.5$ Γ$\displaystyle 1010$ yearβ$\displaystyle 1$
t = $\displaystyle 2.303$
$\displaystyle 0.693$ Γ $\displaystyle 1.5$ Γ $\displaystyle 1010$ πππ$\displaystyle 100$
t = $\displaystyle 2.303$
$\displaystyle 0.693$ Γ $\displaystyle 1.5$ Γ $\displaystyle 1010$ [πππ$\displaystyle 4$ βπππ$\displaystyle 3$]
t = $\displaystyle 2.303$ Γ $\displaystyle 1.5$ Γ $\displaystyle 1010$
Γ [$\displaystyle 0.60$ β$\displaystyle 0.48$] year
t = $\displaystyle 0.598$ Γ $\displaystyle 1010$ years OR $\displaystyle 6$ Γ 109years
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CBSE Class 12 Chemistry past-paper question from the 2026 board exam, with the answer as CBSEβs own marking scheme gives it. Where our answers come from.