Answer :
Answer:
For a: The isotopic symbol of unknown element is [tex]_{43}^{114}\textrm{Tc}[/tex]
For b: The amount of mass lost during this process is 0.2137730 u.
For c: The energy released in the given nuclear reaction is [tex]1.99\times 10^2MeV[/tex]
Explanation:
- For a:
In a nuclear reaction, the total mass and total atomic number remains the same.
For the given fission reaction:
[tex]^{252}_{98}\textrm{Cf}\rightarrow ^A_Z\textrm{X}+^{135}_{55}\textrm{Cs}+3^1_0\textrm{n}[/tex]
- To calculate A:
Total mass on reactant side = total mass on product side
252 = A + 135 + 3
A = 114
- To calculate Z:
Total atomic number on reactant side = total atomic number on product side
98 = Z + 55 + 0
Z = 43
Hence, the isotopic symbol of unknown element is [tex]_{43}^{114}\textrm{Tc}[/tex]
- For b:
We are given:
Mass of [tex]_{98}^{252}\textrm{Cf}[/tex] = 252.081626 u
Mass of [tex]_{0}^{1}\textrm{n}[/tex] = 1.008665 u
Mass of [tex]_{55}^{135}\textrm{Cs}[/tex] = 134.905978 u
Mass of [tex]_{43}^{114}\textrm{Tc}[/tex] = 113.935880 u
To calculate the mass defect, we use the equation:
[tex]\Delta m=\text{Mass of reactants}-\text{Mass of products}[/tex]
Putting values in above equation, we get:
[tex]\Delta m=(m_{Cf})-(m_{Cs}+m_{Tc}+3m_{n})\\\\\Delta m=(252.081626)-(134.905978+113.935880+3(1.008665))=0.2137730u[/tex]
Hence, the amount of mass lost during this process is 0.2137730 u.
- For c:
To calculate the energy released, we use the equation:
[tex]E=\Delta mc^2\\E=(0.2137730u)\times c^2[/tex]
[tex]E=(0.2137730u)\times (931.5MeV)[/tex] (Conversion factor: [tex]1u=931.5MeV/c^2[/tex] )
[tex]E=1.99\times 10^2MeV[/tex]
Hence, the energy released in the given nuclear reaction is [tex]1.99\times 10^2MeV[/tex]