Inside a vessel, there are 4 × 10^ 23 tritium atoms. The half-life...

The number of remaining atoms after time $t$ is given by the radioactive decay law:

$N = N_0 \left(\dfrac{1}{2}\right)^{\dfrac{t}{T_{1/2}}}$

Given $N_0 = 4 \times 10^{23}$, $T_{1/2} = 12.3 \text{ y}$, and $t = 6.15 \text{ y}$.

The number of half-lives elapsed is $n = \dfrac{6.15}{12.3} = \dfrac{1}{2}$.

Substituting the values:

$N = 4 \times 10^{23} \left(\dfrac{1}{2}\right)^{\dfrac{1}{2}} = \dfrac{4 \times 10^{23}}{\sqrt{2}}$

The number of decays that occur is the difference between the initial number of atoms and the remaining atoms:

$\Delta N = N_0 - N = 4 \times 10^{23} - \dfrac{4 \times 10^{23}}{\sqrt{2}}$

\Delta N = 4 \times 10^{23} \left(1 -\dfrac{1}{\sqrt{2}}\right)

Using \sqrt{2} \approx$1.414$:

\Delta N = 4 \times 10^{23} \left(1 -\dfrac{1}{1.414}\right) \approx$4 \times 10^{23} (1 - 0.707)$

$\Delta N = 4 \times 10^{23} \times 0.293 = 1.172 \times 10^{23}$

The number of decays is approximately $1.17 \times 10^{23}$.