The { }_{92}^{238} U atom disintegrates to { | vedclass

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

Question

(B) The decay process is represented as: ${ }_{92}^{238} U \longrightarrow{ }_{84}^{214} Po + 6({ }_{2}^{4} He) + n({ }_{-1}^{0} e)$.Applying the law

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The decay process is represented as: ${ }_{92}^{238} U \longrightarrow{ }_{84}^{214} Po + 6({ }_{2}^{4} He) + n({ }_{-1}^{0} e)$.Applying the law of conservation of mass number:$238 = 214 + 6(4) + n(0)$$238 = 214 + 24 = 238$ (This is satisfied).Applying the law of conservation of atomic number (charge):$92 = 84 + 6(2) + n(-1)$$92 = 84 + 12 - n$$92 = 96 - n$$n = 96 - 92 = 4$.Therefore,the number of electrons emitted is $4$.

The ${ }_{92}^{238} U$ atom disintegrates to ${ }_{84}^{214} Po$ with a half-life of $4.5 \times 10^9$ years by emitting $6$ $\alpha$-particles and $n$ electrons. Here,$n$ is:

Similar Questions

If $_{92}U^{238} \to _{Z}X^{A} + _{2}He^{4}$,then $Z$ and $A$ will be ..........

$\beta$-decay means emission of electron from

In the following nuclear reaction,$x$ stands for: $n \rightarrow p + e^{-} + x$

$A$ nucleus of mass $218\, amu$ in a free state decays to emit an $\alpha$-particle. The kinetic energy of the emitted $\alpha$-particle is $6.7\, MeV$. The recoil energy (in $MeV$) of the daughter nucleus is:

$A$ radioactive decay forms an isotope of the original nucleus with the emission of which of the following particles?

Vedclass Test Series

Exam Paper Generator

Online Exam Module