Question 16: Atoms & Nuclei

Question

After two hours one-eight of the starting amount of a certain radioactive isotope remained undecayed. The half-life of the isotope is

Accepted Answer

To determine the half-life of the radioactive isotope, we need to use the information given and the formula for radioactive decay. The formula for radioactive decay is:
$$N(t) = N_0 \left( \frac{1}{2} ight)^{\frac{t}{T_{\frac{1}{2}}}}$$
where:
$N(t)$ is the remaining quantity of the isotope at time $t$.
$N_0$ is the initial quantity of the isotope.
$T_{\frac{1}{2}}$ is the half-life of the isotope.
$t$ is the elapsed time.
According to the given information, after 2 hours (which is 120 minutes), one-eighth of the starting amount of the isotope remains undecayed. Therefore, if $\frac{N(t)}{N_0} = \frac{1}{8}$, we can write:
$$\frac{N(t)}{N_0} = \left( \frac{1}{2} ight)^{\frac{t}{T_{\frac{1}{2}}}}$$
Substituting the values we have:
$$\frac{1}{8} = \left( \frac{1}{2} ight)^{\frac{120}{T_{\frac{1}{2}}}}$$
We know that $\frac{1}{8}$ can be written as $\left( \frac{1}{2} ight)^3$, therefore:
$$\left( \frac{1}{2} ight)^3 = \left( \frac{1}{2} ight)^{\frac{120}{T_{\frac{1}{2}}}}$$
By equating the exponents, we get:
$$3 = \frac{120}{T_{\frac{1}{2}}}$$
Solving for $T_{\frac{1}{2}}$:
$$T_{\frac{1}{2}} = \frac{120}{3} = 40 \ 	ext{minutes}$$
Thus, the half-life of the isotope is 40 minutes. Therefore, the correct option is:
Option B: 40 min

Answer

15 min

Answer

40 min

Answer

45 min

Answer

4 h

Atoms & Nuclei

After two hours one-eight of the starting amount of a certain radioactive isotope remained undecayed. The half-life of the isotope is

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