Chapter 19, Problem 19.84SP

Interpretation Introduction

Interpretation:

The product formed in the reaction.

Conceptual Introduction:

Artificial radioactivity or induced radioactivity is a phenomenon in which a stable element is converted in to a radioactive isotope by an artificial disintegration.
Example: On Boron bombardment of ??particle into radioactive isotope of nitrogen.

  ${}_{\text{5}}{\text{B}}^{\text{10 }}{\text{+ }}_{\text{2}}{\text{He}}^{\text{4}}{\to }_{\text{7}}{\text{N}}^{\text{13 }}{\text{+ }}_{\text{0}}{\text{n}}^{\text{1}}$

The radioactive nitrogen decay into stable isotope of carbon by emitting a positron.

  ${}_{\text{7}}{\text{N}}^{\text{13 }}{\to }_{\text{6}}{\text{C}}^{\text{13 }}{\text{ + }}_{\text{1}}{\text{β}}^{\text{0}}\text{ + v}$

??decay : The emission of alpha particle from the radioactive nucleus is called alpha decay. a-particle is helium nucleus of radioactive atom having atomic no. 2 and mass no. $4$. Therefore, when an alpha-particle is emitted from the nucleus of an radioactive atom ,the atomic no. Z is increased by two and mass no. A is increased by $4$.

  ${}_{\text{Z}}{\text{X}}^{\text{A}}{\to }_{\text{Z-2}}{\text{Y}}^{\text{A-4 }}{\text{+ }}_{\text{2}}{\text{He}}^{\text{4 }}\text{+ Q}$

  $\text{parent particle }{\to }_{\text{Z-2}}{\text{Y}}^{\text{A-4 }}\text{+ helium particle + energy released}$

b-decay : The emission of b-particle from radioactive nucleus is called ??decay. On emission of beta particle atomic number is increased by one while mass no. remain unchanged.
In Beta emmision a neutron is converted into proton.

  ${}_{\text{0}}{\text{n}}^{\text{1 }}{\to }_{\text{1}}{\text{H}}^{\text{1 }}{\text{+ }}_{\text{-1}}{\text{β}}^{\text{0 }}\text{+ v}$

  ${}_{\text{Z}}{\text{X}}^{\text{A}}{\to }_{\text{Z+1}}{\text{Y}}^{\text{A }}{\text{+ }}_{\text{-1}}{\text{β}}^{\text{0}}\left(\text{electron}\right)\text{ + v }\left(\text{antineutrino}\right)\text{ + Q}$

Gamma decay : The emission of gamma-rays along with a particle and beta particle from the radioactive nucleus is called gamma decay.

There is no change in atomic no. and mass no. by emission of ?-rays.

Solutions:

${}_{\text{92}}{\text{U}}^{\text{238 }}{\text{ + }}_{\text{1}}{\text{H}}^{\text{2 }}{\to }_{\text{93}}{\text{Np}}^{\text{238}}{\text{ + 2}}_{\text{0}}{\text{n}}^{\text{1}}$

To determine :

balanceequation

  ${}_{\text{92}}{\text{U}}^{\text{238}}{\text{+ }}_{\text{1}}{\text{H}}^{\text{2 }}\to {\text{ ? + 2}}_{\text{0}}{\text{n}}^{\text{1}}$