The missing particle in the reaction,_{92}^{2 | vedclass

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

Question

(C) In a nuclear fission reaction,both the mass number and the atomic number must be conserved on both sides of the equation.Let the missing particle

Vedclass · Accepted Answer

$_{36}^{87}Kr$

Vedclass · Answer

(C) In a nuclear fission reaction,both the mass number and the atomic number must be conserved on both sides of the equation.Let the missing particle be $_{Z}^{A}X$.The equation is: $_{92}^{235}U + _{0}^{1}n 	o _{56}^{146}Ba + _{Z}^{A}X + 3_{0}^{1}n$.Conservation of mass number: $235 + 1 = 146 + A + 3(1) \implies 236 = 149 + A \implies A = 87$.Conservation of atomic number: $92 + 0 = 56 + Z + 3(0) \implies 92 = 56 + Z \implies Z = 36$.The element with atomic number $36$ is Krypton $(Kr)$.Thus,the missing particle is $_{36}^{87}Kr$.

The missing particle in the reaction,$_{92}^{235}U + _{0}^{1}n \to _{56}^{146}Ba + ... + 3_{0}^{1}n$ is

Similar Questions

The isotope $_{92}U^{235}$ decays in a number of steps to an isotope of lead $_{82}Pb^{207}$. The groups of particles emitted in this process will be

The number of $\alpha$ and $\beta$-particles emitted during the transformation of $_{90}Th^{232}$ to $_{82}Pb^{208}$ are respectively

$_{5}B^{10} + _{2}He^{4} \rightarrow X + _{0}n^{1}$
In the above nuclear reaction '$X$' will be

In the sequence of the following nuclear reactions $_{92}X^{238}$ $\xrightarrow{-\alpha} Y$ $\xrightarrow{-\beta} Z$ $\xrightarrow{-\beta} L$ $\xrightarrow{-n\alpha} _{84}M^{218}$. The value of $n$ will be:

Which of the following can be used to convert $_7^{14}N$ into $_8^{17}O$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module