A radioactive nucleus $ A $ has a single decay mode with half life $ \tau_{\mathrm{A}} $. Another radioactive nucleus $ B $ has two decay modes, $ 1 $ and $ 2 $. If decay mode $ 2 $ is absent, the half life of $ B $ would be $ \frac{\tau_{\mathrm{A}}}{2} $. If decay mode $ 1 $ is absent, the half life of $ B $ would be $ 3 \tau_{\mathrm{A}} $ . If the actual half life of $ B $ is $ \tau_{\mathrm{B}} $, then the ratio $ \frac{\tau_{\mathrm{B}}}{\tau_{\mathrm{A}}} $ is,

Question

A radioactive nucleus $ A $ has a single decay mode with half life $ \tau_{\mathrm{A}} $. Another radioactive nucleus $ B $ has two decay modes, $ 1 $ and $ 2 $. If decay mode $ 2 $ is absent, the half life of $ B $ would be $ \frac{\tau_{\mathrm{A}}}{2} $. If decay mode $ 1 $ is absent, the half life of $ B $ would be $ 3 \tau_{\mathrm{A}} $ . If the actual half life of $ B $ is $ \tau_{\mathrm{B}} $, then the ratio $ \frac{\tau_{\mathrm{B}}}{\tau_{\mathrm{A}}} $ is,, $ \frac{3}{7} $, $ \frac{7}{2} $, $ \frac{7}{3} $, $ 1 $

MARKS App · Accepted Answer

This is an example of parallel decay of a nuclide. For this decay, we can write, - d N B d t = d N C d t + d N D d t ⇒ N 0 λ B = N 0 λ 1 + N 0 λ 2 λ = deecay constant ⇒ λ B = λ 1 + λ 2 Decay constant is related to half life as, λ = ln 2 τ . ⇒ ln 2 τ B = ln 2 τ 1 + ln 2 τ 2 , where τ 1 and τ 2 are half life for paths 1 and 2 , respectively. ⇒ τ B = τ 1 × τ 2 τ 1 + τ 2 = τ A 2 × 3 τ A τ A 2 + 3 τ A = 3 7 τ A ⇒ τ B τ A = 3 7

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