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Question

Physics Question on Nuclei

If the half life of a radioactive nucleus is 33 days, nearly what fraction of the initial number of nuclei will decay on the 33rd day ? (Given that 0.2530.63)\sqrt[3]{0.25} \approx 0.63))

A

0.63

B

0.5

C

0.37

D

0.13

Answer

0.13

Explanation

Solution

Given, t1/2=3t_{1 / 2}=3 days
Number of active nuclei remaining after time tt,
N=N0(12)tt1/2N=N_{0}\left(\frac{1}{2}\right)^{\frac{t}{t_1/2}}
After t=2t=2 days.
N1=N0(12)2/3=N022/3=N041/3=0.63N0N_{1} =N_{0}\left(\frac{1}{2}\right)^{2 / 3}=\frac{N_{0}}{2^{2 / 3}}=\frac{N_{0}}{4^{1 / 3}}=0.63 N _{ 0 }
N1=0.63N0\therefore N_{1} =0.63\, N _{0}
After t=3t=3 days,
N2=N02=0.5N0N_{2}=\frac{N_{0}}{2}=0.5 \,N_{0}
Number of nuclei that will decay on the 33 rd day,
N3=N2N1=0.63N00.5N0=0.13N0N_{3}=N_{2}-N_{1}=0.63 N_{0}-0.5 N_{0}=0.13 N_{0}
In the term, fraction is 0.130.13