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(A) The effective decay constant $\lambda$ is the sum of individual decay constants: $\lambda = \lambda_1 + \lambda_2$.Since $\lambda = \frac{\ln 2}{T

Vedclass · Accepted Answer

$1080$

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(A) The effective decay constant $\lambda$ is the sum of individual decay constants: $\lambda = \lambda_1 + \lambda_2$.Since $\lambda = \frac{\ln 2}{T}$,we have $\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2}$,where $T$ is the effective half-life.Substituting the given values: $\frac{1}{T} = \frac{1}{1620} + \frac{1}{810} = \frac{1 + 2}{1620} = \frac{3}{1620} = \frac{1}{540}$.Thus,the effective half-life $T = 540 \, years$.The amount of radioactive material remaining is given by $N = N_0 \left( \frac{1}{2} ight)^{t/T}$.We want $N = \frac{1}{4} N_0$,so $\frac{1}{4} = \left( \frac{1}{2} ight)^{t/540}$.Since $\frac{1}{4} = \left( \frac{1}{2} ight)^2$,we have $\frac{t}{540} = 2$.Therefore,$t = 2 	imes 540 = 1080 \, years$.

List-$I$	List-$II$
$A$. Butane $\to$ Isobutane	$a$. Cracking
$B$. Butane $\to$ Lower hydrocarbons	$b$. Isomerisation
$C$. $n$-Heptane $\to$ Toluene	$c$. Reed reaction
$D$. Propane $\to$ $CH_3CH_2CH_2SO_2Cl$	$d$. Aromatisation

$A$ radioactive material decays by simultaneous emission of two particles with respective half-lives $1620 \, years$ and $810 \, years$. The time (in years) after which one-fourth of the material remains is:

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