Get Answers to all your Questions

header-bg qa

A mixture consists of two radioactive materials A_1 and A_2 with half lives of 20 \mathrm{~s} and 10 \mathrm{~s} respectively. Initially the mixture has 40 \mathrm{~g} of \mathrm{A}_1 and 160 \mathrm{~g} of \mathrm{A}_2. The amount of the two in the mixture will become equal after:

Option: 1

60 s 


Option: 2

80 s


Option: 3

20 s


Option: 4

40 s


Answers (1)

best_answer

Here, \left(\mathrm{T}_{1 / 2}\right)_1=20 \mathrm{~s},\left(\mathrm{~T}_{1 / 2}\right)_2=10 \mathrm{~s}, \mathrm{~N}_{01}=40 \mathrm{~g}, \mathrm{~N}_{02}=160 \mathrm{~g}

Let the amount of the two radioactive materials become equal after t s.

From, \frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}} ,we get

\mathrm{N}_1=\mathrm{N}_{01}\left(\frac{1}{2}\right)^{\mathrm{t} / 20}=40\left(\frac{1}{2}\right)^{\mathrm{t} / 20}

\mathrm{N}_2=\mathrm{N}_{02}\left(\frac{1}{2}\right)^{\mathrm{t} / 10}=160\left(\frac{1}{2}\right)^{\mathrm{t} / 10}

\text { As } \mathrm{N}_1=\mathrm{N}_2

\therefore \quad 40\left(\frac{1}{2}\right)^{t / 20}=160\left(\frac{1}{2}\right)^{t / 10}

\text { or } \quad\left(\frac{1}{2}\right)^{t / 20}=4\left(\frac{1}{2}\right)^{t / 10}=\left(\frac{1}{2}\right)^{\frac{t}{10}-2}

\begin{aligned} & \therefore \quad \frac{t}{20}=\frac{t}{10}-2 \\ & \therefore \quad t=40 \mathrm{~s} \end{aligned}

Posted by

Pankaj Sanodiya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE