Radioactive material decays by simultaneous emission of two particles with half-lives of 1620 and 810 years. What is the time, in years, after which one-fourth of the material remains?
Correct answer: B. 1080 years
- A. 2430 years
- B. 1080 years
- C. 3240 years
- D. 4260 years
Explanation
According to the law of Rutherford - Soddy the number of atoms left after n number of Half-life is N=N0(½)nWhere N0 is the original number of atoms. The number of half-livesn=time of decay/effective half−lifeThe relation between effective disintegration constant(lambda) and half-life(T) is; λ=ln 2÷TNow, λ1+λ2=ln 2/T1+ln 2/T2and the effective half-life is ; 1/T=1/T1+1/T21/T=1/1620+1/8101/T=1+2/1620T=540Therefore, the number of half-lives is, n=time of decay/ effect half-life = t/540Now according to the question, N=N0(½)nTherefore, N/N0=(1/2)2 =(1/2)t/540 (1/2)2=(1/2)t/5402 = t/540t=1080 years Therefore, the time in years, after which one-fourth of the material remains is 1080 years.
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