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  • A radioactive material decays by simultaneous emission of two particles with halflives 810 yr and 405 yr respectively. The time in years after which one-fourth of material remains, is :<b

A radioactive material decays by simultaneous emission of two particles with halflives 810 yr and 405 yr respectively. The time in years after which one-fourth of material remains, is :
 

Option: 1

405 yr
 


Option: 2

810 yr
 


Option: 3

540 yr
 


Option: 4

205 yr


Answers (1)

best_answer

Since, from Rutherford-Soddy law, the number of atoms left after half lives is given by

\mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}}

where, \mathrm{N}_0 is the original number of atoms.
Relation between effective disintegration constant (\lambda) and half life (\mathrm{T})

\begin{aligned} & \lambda=\frac{\ln 2}{\mathrm{~T}} \\ \\& \therefore \lambda_1+\lambda_2=\frac{\ln 2}{\mathrm{~T}_1}+\frac{\ln 2}{\mathrm{~T}_2} \end{aligned}

Effective half life, 

\begin{aligned} & \frac{1}{\mathrm{~T}}=\frac{1}{\mathrm{~T}_1}+\frac{1}{\mathrm{~T}_2}=\frac{1}{810}+\frac{1}{405} \\ \\& \frac{1}{\mathrm{~T}}=\frac{1+2}{810} \Rightarrow \mathrm{T}=270 \mathrm{yr} \\ \\& \therefore \mathrm{n}=\frac{\mathrm{T}}{270} \\ \\& \therefore \mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{t} / 270} \Rightarrow \frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^{\mathrm{t} / 270} \\ \\& \Rightarrow \frac{\mathrm{t}}{270}=2 \Rightarrow \mathrm{t}=2 \times 270=540 \mathrm{yr} \end{aligned}

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Rakesh

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