Question

An isotope of californium-252 undergoes a spontaneous fission, producing cesium-135, 3 neutrons, and one other isotope. a) Determine the identity of the other isotope produced in the fission, and write out the reaction equation, using full isotopic notation, for example Pm → C022 + Seag + 2n. b) Calculate the amount of mass (in u) lost during this fission. The masses of the three isotopes are: 252.081 626 u (californium), 134.905 978 u (cesium), 113.935 880 u (unknown). Maintain a precision of no less than 7 SF as you work. c) Calculate the energy (in MeV) released by this fission, accurate to 3 SF.

Answer 1

Answer:

For a: The isotopic symbol of unknown element is $_{43}^{114}\textrm{Tc}$

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 $1.99\times 10^2MeV$

Explanation:

• For a:

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

$^{252}_{98}\textrm{Cf}\rightarrow ^A_Z\textrm{X}+^{135}_{55}\textrm{Cs}+3^1_0\textrm{n}$

• 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 $_{43}^{114}\textrm{Tc}$

• For b:

We are given:

Mass of $_{98}^{252}\textrm{Cf}$ = 252.081626 u

Mass of $_{0}^{1}\textrm{n}$ = 1.008665 u

Mass of $_{55}^{135}\textrm{Cs}$ = 134.905978 u

Mass of $_{43}^{114}\textrm{Tc}$ = 113.935880 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(m_{Cf})-(m_{Cs}+m_{Tc}+3m_{n})\\\\\Delta m=(252.081626)-(134.905978+113.935880+3(1.008665))=0.2137730u$

Hence, the amount of mass lost during this process is 0.2137730 u.

• For c:

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.2137730u)\times c^2$

$E=(0.2137730u)\times (931.5MeV)$ (Conversion factor:  $1u=931.5MeV/c^2$  )

$E=1.99\times 10^2MeV$

Hence, the energy released in the given nuclear reaction is $1.99\times 10^2MeV$