1. 1/2 2. 1/2√(2) 3. 2/3 4. 2/3√(2)

Answer: 2. 1/2√(2) Key idea: activity decays as A=A₀(1/2)^t/T_(1/2) . dfrac AA₀=(1/2)^(150/100)=(1/2)^(3/2)=dfrac12sqrt2 . Option 2.

The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be :

1. 1/2

2. \(\frac{1}{2\sqrt{2}}\)

3. \(\frac{2}{3}\)

4. \(\frac{2}{3\sqrt{2}}\)

Doubtly Asked 30 Oct 2026 NEET UG 2021 neet Save WhatsApp Report

Best answer

Answer: 2. \(\frac{1}{2\sqrt{2}}\)

Key idea: activity decays as \(A=A_0\left(\tfrac12\right)^{t/T_{1/2}}\).

\(\dfrac A{A_0}=\left(\dfrac12\right)^{150/100}=\left(\dfrac12\right)^{3/2}=\dfrac1{2\sqrt2}\). Option 2.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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