Skip to content
Exercise 5 · Q8

Q.Half-life of radioactive carbon-14 is 5700 years. A certain bone was observed to contain 75% of carbon-14 as compared to what is present in the leaving creatures. Determine its antiquity.

Delhi CbseNCERTSubjective· 3mImportance★★★★★est
94% · 51/54 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Carbon-14 decays with k=log⁡25700k=\frac{\log 2}{5700}; a sample at 75% of the living level gives an age t=5700log⁡(4/3)log⁡2≈2366t=\frac{5700\log(4/3)}{\log 2}\approx 2366 years.

N(t)=N0e−ktN(t)=N_0e^{-kt} — radioactive decay, where N0N_0 = C-14 in a living creature, N(t)N(t) = C-14 in the bone now, k=log⁡2t1/2k=\dfrac{\log 2}{t_{1/2}} = decay constant, tt = age in years, t1/2=5700t_{1/2}=5700 yr.

  1. Decay constant. k=log⁡25700=0.69315700=1.2161×10−4 yr−1k=\dfrac{\log 2}{5700}=\dfrac{0.6931}{5700}=1.2161\times10^{-4}\ \text{yr}^{-1}.
  2. Set up condition. Bone contains 75% of the living level: 0.75 N0=N0e−kt⇒e−kt=0.750.75\,N_0=N_0e^{-kt}\Rightarrow e^{-kt}=0.75. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.