Nuclear Physics: Question 1

Worked solution

Step 1: Recall the relationship between decay constant and half-life

For exponential radioactive decay, the number of undecayed nuclei falls to half its value after one half-life, so: $t_{1/2}=\frac{\ln2}{\lambda}$

Step 2: Substitute the values

$t_{1/2}=\frac{\ln2}{4.62\times10^{-2}}=\frac{0.6931}{4.62\times10^{-2}}$

$t_{1/2}=15.0\text{ s}\ (3\text{ s.f.})$

Recompute as a check, working the other way: if $t_{1/2}=15.0\text{ s}$, then $\lambda=\dfrac{\ln2}{t_{1/2}}=\dfrac{0.6931}{15.0}=4.62\times10^{-2}\text{ s}^{-1}$, which matches the given decay constant exactly.

Why the other options are wrong

• A ($7.50\text{ s}$): half of the correct value. Arises from a slip such as using $\lambda t_{1/2}=0.5$ (as if $\lambda$ were “the fraction decayed per second”) instead of the correct relation $\lambda t_{1/2}=\ln2\approx0.693$.
• C ($21.6\text{ s}$): comes from forgetting the $\ln2$ factor entirely and computing $1/\lambda=1/(4.62\times10^{-2})=21.6\text{ s}$. This is the mean lifetime, not the half-life. The two are different by a factor of $\ln2$.
• D ($43.3\text{ s}$): comes from doubling the mean lifetime, $2/\lambda=2\times21.6=43.3\text{ s}$, compounding the missing-$\ln2$ error with an extra factor of 2.

Final answer

• The half-life of the tracer is $\boxed{15.0}\text{ s}$, option B.