24.109 CP - Chemistry: Molecular Nature Of Matter And Change

Question

The isotopic mass of ${}_{86}{}^{210}Rn$  is 209.989669 . When this nuclide decays by electron capture, it emits  2.368MeV. What is the isotopic mass of the resulting nuclide?

Step-by-Step Solution

Answer

The isotopic mass of the resulting nuclide is $Mas{s}_{85}^{210}At=209.9927 amu$ .

Step 1: Electron capture

When an unstable nucleus accepts an electron, the phenomena is known as electron capture. The electron combines with a proton to change into a neutron. The result of electron capture is that the daughter nucleus formed has the same mass number but the atomic number decreases by one unit.

  ${}_1{}^{1}p+{}_{-1}{}^{0}e\to {}_0{}^{1}n$

Step 2: Reaction taking place during decay.

The  ${}_{86}{}^{210}Rn$ decays by capturing an electron to form ${}_{85}{}^{210}At$  . The equation representing the decay is given below.

 ${}_{86}{}^{210}Rn{+}_{-1}^{0}e{\to }_{85}^{210}At$

Step 3: Find the isotopic mass of the resulting nuclide

In the problem, it state that ${}_{86}{}^{210}Rnemits2.368MeV$  . The energy can be converted into atomic mass units with the help of the following conversion relations.

 $$\begin{gathered} 1amu=931.5x{10}^{6}eV=931.5MeV \\ 2.368MeV=2.368 MeV\left(\frac{1amu}{931.5MeV}\right)=0.0025amu \end{gathered}$$

We will use to calculate the isotopic mass,

 

 $$\begin{gathered} Mas{s}_{85}^{210}At=Mas{s}_{86}^{210}Rn+massofelectron-equivalent \\ Mas{s}_{85}^{210}At=209.9897amu+0.000549amu-0.0025amu \\ Mas{s}_{85}^{210}At=209.9927 amu \end{gathered}$$

 

Therefore, the required value of  $Mas{s}_{85}^{210}At=209.9927 amu$.