Question

Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-lifeof ²⁴¹Am is 433 years, and it decays by emitting alpha particles. How many alpha particles are emitted each second by a 5.00-g sample of ²⁴¹Am?

Short answer

$6.35\times {10}^{-11}{\text{decay.sec}}^{-1}$

Step-by-step solution

Half-life.

Half-life refers to the time required for a substance to decrease to its half value in terms of concentration. Rate constant $\left(\lambda \right)$and half-life$\mathbf{\left(}{\mathit{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{\right)}$ of a reaction are inversely correlated.

Determining the decay constant .

The half-life of americium 241 is 433 years, and the decay mode is alpha particle emission.

The formula is:

${\mathit{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{693}}{\mathbf{\lambda }}\to \mathbf{\lambda }\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{693}}{{\mathit{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}}$

The known value is substituted.

$\begin{array}{c}\mathrm{\lambda }=\frac{0.693}{433}\times \frac{1\mathrm{year}}{365\mathrm{days}}\times \frac{1\mathrm{day}}{24\mathrm{hour}}\times \frac{1\mathrm{hour}}{3600\sec }\\ =5.075\times {10}^{-11}{\sec }^{-1}\end{array}$

Determining the decay rate.

Let N be the number of nuclides. Here, the decay rate (A) is evaluated as:

$\begin{array}{c}Rate=\lambda \times N\\ =\left(5.075\times {10}^{-11}{\sec }^{-1}\right)\times \left(5.00g\times \frac{1mol}{241}\times \frac{6.022\times {10}^{23}nuclei}{mol}\right)\\ =6.35\times {10}^{-11}\mathrm{decays}.{\sec }^{-1}\end{array}$

Hence, the decay rate is role="math" localid="1663793465033" $\mathbf{6}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{}\mathit{d}\mathit{e}\mathit{c}\mathit{a}\mathit{y}\mathit{s}\mathbf{.}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{-}\mathbf{1}}$.