Question 458: Atoms & Nuclei

Question

Half-lives of two radioactive elements A and B are 30 minute and 60 minute respectively. Initially the samples have equal number of nuclei. After 120 minute the ratio of decayed numbers of nuclei of $B$ to that of $A$ will be

Accepted Answer

For radioactive decay, amount of sample remaining is given by,
$$\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}ight)^\lambda\quad	ext{.... (i)}$$
$\mathrm{N}_{\mathrm{o}}$ is initial sample and $\lambda$ is decay constant.
Decay constant, $\lambda=\left(\frac{t}{t_{1 / 2}}ight)=\frac{120}{t_{1 / 2}}\quad	ext{.... (given)}$
For element A,
$$\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}ight)^{\frac{120}{30}}=\frac{\mathrm{N}_{\mathrm{o}}}{2^4}$$
...[From(i)]
Amount of sample A decayed, $\mathrm{N}_{\mathrm{A}}^{\prime}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}-\frac{\mathrm{N}_{\mathrm{o}}}{2^4}=\frac{15}{16} \mathrm{~N}_{\mathrm{O}}\quad	ext{.... (ii)}$
For element B,
$$N_B=N_0\left(\frac{1}{2}ight)^{\frac{120}{60}}=\frac{N_0}{2^2}$$ 
...[From(i)]
Amount of sample B decayed,
$$\mathrm{N}_{\mathrm{B}}{ }^{\prime}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}}-\frac{\mathrm{N}_{\mathrm{o}}}{2^2}=\frac{3}{4} \mathrm{~N}_{\mathrm{o}}$$
From (ii) and (iii),
$$\frac{\mathrm{N}_{\mathrm{A}^{\prime}}}{\mathrm{N}_{\mathrm{B}^{\prime}}}=\frac{4}{5}$$

Answer

$1: 15$

Answer

$1: 4$

Answer

$4: 5$

Answer

$5: 4$

Atoms & Nuclei

Half-lives of two radioactive elements A and B are 30 minute and 60 minute respectively. Initially the samples have equal number of nuclei. After 120 minute the ratio of decayed numbers of nuclei of B to that of A will be

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