Question

A radioisotope ${}_{z}^{m}A$ (${{t}_{{}^{1}/{}_{2}}}$ = 10 days) decays to give ${}_{z-6}^{m-12}B$ stable atom along with $\alpha \text{-particles}$. If m g of A are taken and kept in a sealed tube, the volume of He in litre that will accumulate in 20 days at STP is _____.
(If answer is X, then write 10X)

Answer 1

Hint: Write a balanced radioactive decay equation for the reaction. Then calculate the number of half-life periods (n) in total time of decay. Calculate the amount decayed and then accordingly equate with the number of moles of He using the reaction. Multiply the amount of He with the volume of a gas at STP to get the answer.

Complete step by step solution:
- Let’s start by writing the equation of radioactive decay.
\[{}_{z}^{m}A\to {}_{z-6}^{m-12}B+3\,{}_{2}^{4}He\]
- A radioactive isotope is decaying and giving rise to stable atoms having atomic number 6 less than the radioactive isotope and atomic mass number 12 less than the mass of radioactive isotope.
- Half-life is given as 10 days for radioactive isotope, A.
- The total time of decay is 20days.
- Number of half-life periods in total time of decay is $\dfrac{T}{{{t}_{{}^{1}/{}_{2}}}}=\dfrac{20}{10}=2$
- The amount of radioactive isotope is left behind is $\dfrac{1}{{{2}^{2}}}=\dfrac{1}{4}mol$
- Therefore, the amount of radioactive isotope which got decayed is $1-\dfrac{1}{4}=\dfrac{3}{4}mol$
- Now, one mole of isotope is getting decayed and forming three moles of He atoms.
- Therefore, amount of He atoms formed is $\dfrac{3}{4}\times 3=\dfrac{9}{4}mol$
- Therefore, volume of He atoms at STP is $22.4\times \dfrac{9}{4}=50.4L$

Therefore, the volume of He that will accumulate in 20 days at STP is 50.4L.

Note: Remember, at STP, temperature is $298K$ and pressure is $1atm$ and volume of gas is $22.4L$ for one mole of a gas. Always write the radioactive decay equation to understand the equivalence of one mole of radioisotope with the products formed. Remember, $\alpha \text{-particles}$ are helium atoms.