Question

A sample of 1.0x10¹⁰atoms that decay by alpha emission

has a half-life of 100 min. How many alpha particles are emitted

between t = 50 min and t = 200 min?

Short answer

Therefore $4.57\times {10}^9\text{atoms}$are emitted between t=50 min and t = 200 min.

Step-by-step solution

Given information

$$\begin{gathered} t_{1/2}=100\min \\ {\mathrm{N}}_{\mathrm{o}}=\left(1.0\times {10}^{10}\right)\text{atoms} \end{gathered}$$

Explanation

To begin, we must determine the number of radioactive atoms at time t = 50 minutes:

$$\begin{gathered} N_{50}=N_o{\left(\frac{1}{2}\right)}^{t/t_{1/2}} \\ {N}_{50}=\left(1.0\times {10}^{10}\right)\text{atoms}{\left(\frac{1}{2}\right)}^{50\min /100\min } \\ {N}_{50}=7.07\times {10}^9\text{atoms} \end{gathered}$$

We determine the number of radioactive atoms at time t = 200 minutes:

$$\begin{gathered} N_{200}=N_o{\left(\frac{1}{2}\right)}^{t/t_{1/2}} \\ {N}_{200}=\left(1.0\times {10}^{10}\right)\text{atoms}{\left(\frac{1}{2}\right)}^{200\min /100\min } \\ {N}_{200}=2.50\times {10}^9\text{atoms} \end{gathered}$$

Explanation

We determine the number of radioactive atoms between time t = 50 minutes and 200 minutes:

$$\begin{gathered} N_{\text{decay}}=N_{50}-N_{200} \\ {N}_{\text{decay}}=\left(7.07\times {10}^9\text{atoms}\right)-\left(2.50\times {10}^9\text{atoms}\right) \\ {N}_{\text{decay}}=4.57\times {10}^9\text{atoms} \end{gathered}$$