Complete the equation for the following fission process
$$_{92}{U^{235}}{ + _0}{n^1}{ \to _{38}}{n^{90}} + .........$$
A.
$$_{54}X{e^{143}} + {3_0}{n^1}$$
B.
$$_{54}X{e^{145}}$$
C.
$$_{57}X{e^{142}}$$
D.
$$_{57}X{e^{142}}{ + _0}{n^1}$$
Answer :
$$_{54}X{e^{143}} + {3_0}{n^1}$$
Solution :
$$_{92}{U^{235}}{ + _0}{n^1}{ \to _{38}}S{r^{90}}{ + _{54}}X{e^{143}} + {3_0}{n^1}$$
If total atomic number on $$LHS = 92 + 0 = 92$$
Total atomic number on $$RHS = 38 + 54 + 0 = 92$$
Total mass number on $$LHS = 235 + 1 = 236$$
Total mass number on $$RHS = 90 + 143 + 3 \times 1 = 236$$
So, option (A) is correct. NOTE
For a nuclear reaction to be completed, the mass number and charge number on both sides should be same.
Releted MCQ Question on Modern Physics >> Atoms or Nuclear Fission and Fusion
In the nuclear fusion reaction
$$_1^2H + _1^3H \to _2^4He + n$$
given that the repulsive potential energy between the two
nuclei is $$ \sim 7.7 \times {10^{ - 14}}J,$$ the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s Constant $$k = 1.38 \times {10^{ - 23}}J/K$$ ]
The binding energy per nucleon of deuteron $$\left( {_1^2H} \right)$$ and helium nucleus $$\left( {_2^4He} \right)$$ is $$1.1\,MeV$$ and $$7\,MeV$$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is