Question

Samarium – 151 is a beta emitter. Write an equation for this nuclear reaction and identify the product nucleus.

Answer 1

A "beta emitter" (or "beta emission" generally) is a bit ambiguous since, insofar as beta emission refers to beta decay, there are two types of beta decay: beta-positive and beta-negative.

The general nuclear reaction for isobaric beta decay of both types is given by

$\[_{Z}^{A}\textrm{X}\rightarrow _{Z\mp 1}^{A}\textrm{Y} + \beta ^\pm + \nu + \mathrm{Q}\]$

where A is the mass number, Z is the atomic number, X is the element symbol of the parent nucleus, Y is the element symbol of the daughter nucleus, β is the beta particle, ν is a neutrino, and Q is energy change.

Samarium-151 happens to undergo beta-negative decay, which takes the following form:

$\[_{Z}^{A}\textrm{X}\rightarrow _{Z+1}^{A}\textrm{Y} + \beta ^- + \bar{\nu} + \mathrm{Q}\]$

where

$\beta^- = \ce{^{0}_{-1}e^-}$ and $\bar{\nu}$ is an antineutrino.

So, the mass number remains the same, but the atomic number increases by one, which means a neutron is converted into a proton. Samarium has an atomic number (Z) of 62; the element with an atomic number of 62 + 1 = 63 is europium. You probably do not have to worry about the Q value in this case (if you do, please feel free to follow up), so we can omit it from the final equation.

Stitching everything together, our nuclear equation for the beta-negative decay of samarium-151 would be

$\mathrm{_{62}^{151}Sm \rightarrow \mathrm{_{63}^{151}Eu}} + {}_{-1}^{0}e^-} +\bar{\nu}.$

You could simplify it further by using just e⁻ to denote the electron:

$\mathrm{_{62}^{151}Sm \rightarrow \mathrm{_{63}^{151}Eu}} + e^- +\bar{\nu}.$