A { }_{92} U^{235} nucleus absorbs a neutron | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The nuclear reaction is given by: ${ }_{92} U^{235} + { }_{0} n^{1} 	o { }_{54} X e^{139} + { }_{38} S r^{94} + X$.To find $X$,we balance the mas

Vedclass · Accepted Answer

$3$ neutrons

Vedclass · Answer

(A) The nuclear reaction is given by: ${ }_{92} U^{235} + { }_{0} n^{1} 	o { }_{54} X e^{139} + { }_{38} S r^{94} + X$.To find $X$,we balance the mass numbers and atomic numbers on both sides.Sum of mass numbers on the left side: $235 + 1 = 236$.Sum of mass numbers on the right side: $139 + 94 + A = 233 + A$,where $A$ is the mass number of $X$.Equating them: $236 = 233 + A \implies A = 3$.Sum of atomic numbers on the left side: $92 + 0 = 92$.Sum of atomic numbers on the right side: $54 + 38 + Z = 92 + Z$,where $Z$ is the atomic number of $X$.Equating them: $92 = 92 + Z \implies Z = 0$.Since $A=3$ and $Z=0$,the product $X$ consists of $3$ neutrons $(3 { }_{0} n^{1})$.

$A$ ${ }_{92} U^{235}$ nucleus absorbs a neutron and disintegrates into ${ }_{54} X e^{139}$,${ }_{38} S r^{94}$,and $X$. What will be the product $X$?

Similar Questions

The binding energy of deuteron $_1^2H$ is $1.112 \, MeV$ per nucleon and an $\alpha$-particle $_2^4He$ has a binding energy of $7.047 \, MeV$ per nucleon. Then in the fusion reaction $_1^2H + _1^2H \to _2^4He + Q$,the energy $Q$ released is ........ $MeV$.

The Sun radiates energy continuously and maintains its brightness because:

Let $E_1$ and $E_2$ be the binding energies of two nuclei $A$ and $B.$ It is observed that two nuclei of $A$ combine together to form a $B$ nucleus. This observation is correct only if

Which nuclear process occurs in a hydrogen bomb? Or,state the principle of a hydrogen bomb.

Explain the phenomenon of nuclear fission.

Vedclass Test Series

Exam Paper Generator

Online Exam Module