Question

The radionuclide ${}^{\mathbf{56}}\mathbf{Mn}$has a half-life of 2.58 hand is produced in a cyclotron by bombarding a manganese target with deuterons. The target contains only the stable manganese isotope${}^{\mathbf{55}}\mathbf{Mn}$, and the manganese–deuteron reaction that produces${}^{\mathbf{56}}\mathbf{Mn}$is

${}^{\mathbf{55}}\mathbf{Mn}\mathbf{+}\mathbf{d}\mathbf{\to }{}^{\mathbf{55}}\mathbf{Mn}\mathbf{+}\mathbf{p}$.

If the bombardment lasts much longer than the half-life of ${}^{\mathbf{55}}\mathbf{Mn}$, the activity of the ${}^{\mathbf{55}}\mathbf{Mn}$produced in the target reaches a final value of $\mathbf{8}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathbf{Bq}$. (a) At what rate is${}^{\mathbf{56}}\mathbf{Mn}$being produced? (b) How many${}^{\mathbf{56}}\mathbf{Mn}$nuclei are then in the target? (c) What is their total mass?

Short answer

a) The rate at which ${}^{55}\mathrm{Mn}$is being produced is $8.88\times {10}^{10}{s}^{-1}$.

b) The number of ${}^{55}\mathrm{Mn}$nuclei present in the target is $1.19\times {10}^{15}$.

c) The total mass of the ${}^{55}\mathrm{Mn}$nuclei is $0.11\mathrm{\mu g}$.

Step-by-step solution

Given data

Half-life of radionuclide${}^{55}\mathrm{Mn},{\mathrm{T}}_{1/2}=2.58\mathrm{h}$,

The reaction is${}^{55}\mathrm{Mn}+\mathrm{d}\to {}^{55}\mathrm{Mn}+\mathrm{p}$.

The activity of the produced in the target, $\mathrm{\lambda N}=8.88\times {10}^{10}\mathrm{Bq}$

Molar mass of manganese isotope ${}^{55}\mathrm{Mn},\mathrm{A}=56\mathrm{g}/\mathrm{mol}$,

Understanding the concept of decay  

In the given problem, the radionuclide is produced due to the bombardment of the deuteron isotope with the stable manganese isotope with a proton. In a secular equilibrium condition, the decay rate of the substance is equal to the production rate. Thus, using the given concept, we can get the disintegration constant of the isotope. Now, the number of nuclei is found using the stochiometric concept of the number of atoms calculation. Thus, the total mass of the nuclei is calculated using the mass of each manganese isotope.

Formulae:

The rate of decay,$R=\lambda N\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

where, $\lambda$is the disintegration constant.

Nis the number of undecayed nuclei.

The disintegration constant, $\mathrm{\lambda }=\frac{\mathrm{In}2}{{\mathrm{T}}_{1/2}}-\dots \left(2\right)$

where, $T_{1/2}$is the half-life of the substance.

The number of nuclei in a given mass of an atom,

$N=\frac{m}{A}{N}_A.......\left(3\right)\mathrm{Where}{\mathrm{N}}_{\mathrm{A}}=6.022\times {10}^{23}\mathrm{nuclei}/\mathrm{mol}$

a) Calculation of the decay rate

The sample is in secular equilibrium with the source, and the decay rate equals the production rate.

Let be the rate of production of ${}^{56}\mathrm{Mn}$and let $\lambda$be the disintegration constant.

Thus, from the given data and above value, we have the decay rate of ${}^{56}\mathrm{Mn}$using equation (1) as:$R=8.88\times {10}^{10}Bq$

Hence, the decay rate is $8.88\times {10}^{10}{\mathrm{s}}^{-1}$.

Calculation of the number of manganese nuclei

Substituting equation (2) in equation (1) with the given data, we can get the number of${}^{56}\mathrm{Mn}$ nuclei present in the target as follows:

$$\begin{gathered} N=\frac{R{T}_{1/2}}{In2} \\ =\frac{\left(8.88\times {10}^{10}/\mathrm{s}\right)\left(2.58\mathrm{h}\right)}{\mathrm{In}2}\left(\because 1\mathrm{Bq}=1/\mathrm{s}={\mathrm{s}}^{-1}\right) \\ =1.19\times {10}^{15}\mathrm{nuclei} \end{gathered}$$

Hence, the number of nuclei in the target is $1.19\times {10}^{15}$.

c) Calculation of the total mass of the manganese nuclei

The total mass of the manganese nuclei can be calculated as follows:

$$\begin{gathered} M_{total}=Nm \\ =N\left(\frac{NA}{N_A}\right)\left(fromequation\left(3\right)\right) \\ =\frac{\left(1.19\times {10}^{15}\mathrm{nuclei}\right)\left(56\mathrm{g}/\mathrm{mol}\right)}{\left(6.022\times {10}^{23}\mathrm{nuclei}/\mathrm{mol}\right)} \\ =1.11\times {10}^{-7}\mathrm{g} \\ =0.11\mathrm{\mu g} \end{gathered}$$

Hence, the value of total mass is $0.11\mathrm{\mu g}$.