A nucleus of _{84}^{210}Po originally at rest | vedclass

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

Question

(A) The decay process is given by: $_{84}^{210}Po \rightarrow _{82}^{206}X + _{2}^{4}He$.According to the law of conservation of linear momentum,the i

Vedclass · Accepted Answer

$\frac{4v}{206}$

Vedclass · Answer

(A) The decay process is given by: $_{84}^{210}Po \rightarrow _{82}^{206}X + _{2}^{4}He$.According to the law of conservation of linear momentum,the initial momentum of the system is zero because the nucleus is at rest.Let $M_d$ be the mass of the daughter nucleus ($206$ amu) and $M_{\alpha}$ be the mass of the $\alpha$ particle ($4$ amu).Let $v'$ be the recoil speed of the daughter nucleus and $v$ be the speed of the $\alpha$ particle.$M_d v' + M_{\alpha} v = 0$$206 v' + 4v = 0$$v' = -\frac{4v}{206}$The magnitude of the recoil speed is $|v'| = \frac{4v}{206}$.

$A$ nucleus of $_{84}^{210}Po$ originally at rest emits an $\alpha$ particle with speed $v$. What will be the recoil speed of the daughter nucleus?

Similar Questions

When a radioactive substance emits an $\alpha$-particle,its position in the periodic table is lowered by

The nucleus ${ }_{88}^{226} Ra$ is converted into ${ }_{82}^{206} Pb$ by the emission of alpha $(\alpha)$ and beta $(\beta)$ particles. The number of alpha and beta particles emitted are respectively:

In which radioactive disintegration does a neutron dissociate into a proton and an electron?

In a radioactive reaction $_{92}X^{232} \to _{82}Y^{204}$,the number of $\alpha$-particles emitted is:

$A$ 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module