Q37P - Fundamentals Of Physics

Question

The radionuclide ${}^{\mathbf{64}}\mathbf{Cu}$ has a half-life of 12.7h. If a sample contains 5.50g of initially pure ${}^{\mathbf{64}}\mathbf{Cu}$at,  t = 0how much of it will decay between t = 14h and t = 16h?

Step-by-Step Solution

Answer

The radionuclide ${}^{64}\mathrm{Cu}$will decay 0.265g between t = 14h and t = 16h.

Step 1: The given data

a)   Mass of the sample ${}^{64}\mathrm{Cu} , m=5.50g$, 

b)   Half-life of the radionuclide, ${}^{64}\mathrm{Cu} , T_{1/2}=12.7 \mathrm{h}$ 

c)   Time of decay, $t_1=14 \mathrm{h} \mathrm{and} {\mathrm{t}}_2=16 \mathrm{h}$

d)   Molar mass of the radionuclide, ${}^{64}\mathrm{Cu} , A=64\frac{\mathrm{g}}{mol}$

Step 2: Understanding the concept of decay

The radioactive decay constant or the disintegration constant represents the fraction of radioactive atoms that disintegrates in a unit of time. The given relation using the number of nuclei formula will give the undecayed nuclei at the two-time states. Thus, their difference will be the required answer for the undecayed nuclei within the time gap.

 

Formulae:

The undecayed sample remaining after a given time is as follows:

 $N=N_0{e}^{\frac{\mathrm{In} 2}{T_{1/2}}}$                                        ….. (i)

The number of nuclei in a given mass of an atom is as follows:

$N=\frac{m}{A}{N}_A$                                                    …… (ii)

 

Here, $N_A=6.022\times {10}^{23}\frac{\mathrm{nuclei}}{\mathrm{mol}}$

Step 3: Calculate the amount of decay of the radioactive nuclide

From equation (ii), we can see that the number of undecayed nuclei is directly proportional to the mass of the sample. Thus, using this condition in equation (i), get the amount of decayed nuclei between t = 14 h and t = 16 h as follows:

$$\begin{gathered} \left|∆m\right|=m{\left|{}_{t_f=14 \mathrm{h}}-m\right|}_{t_i==16h} \\ =m_0\left(1-e^{-t_f\mathrm{In} 2/{\mathrm{T}}_{1/2}}\right)-m_0\left(1-e^{-t_i\mathrm{In} 2/{\mathrm{T}}_{1/2}}\right) \\ =m_0\left(e^{-t_i\mathrm{In} 2/{\mathrm{T}}_{1/2}}-e^{-t_f\mathrm{In} 2/{\mathrm{T}}_{1/2}}\right) \end{gathered}$$

 

Substitute the values and solve as:

$$\begin{gathered} \left|∆\mathrm{m}\right|=\left(5.5 \mathrm{g}\right)\left[e^{-16 \mathrm{hIn} 2/\left(12.7 \mathrm{h}\right)} -e^{-14 \mathrm{hIn} 2/\left(12.7 \mathrm{h}\right)}\right] \\ =0.265 \mathrm{g} \end{gathered}$$

Hence, the value of the decayed nuclei is 0.265g.