Q17 PE
Question
\({{\rm{\beta }}^{\rm{ - }}}\) decay of \(^{\rm{3}}{\rm{H}}\) (tritium), a manufactured isotope of hydrogen used in some digital watch displays, and manufactured primarily for use in hydrogen bombs
Step-by-Step Solution
VerifiedThe \({\beta ^ - }\) Decay equation of \(_1^3{H_2} is {\rm{}}_1^3{H_2} \to _2^3H{e_1} + {\beta ^ - } + \overline {{\nu _e}} \).
The amount of matter contained in an atom of an element is called its atomic mass.
\(A = N + Z\)
Where A is atomic mass number
Z= Number of protons in a nucleus
X= symbol for the element
In the expression below:
\(_Z^A{X_N}\)
Z = Number of protons in a nucleus
X= symbol for the element
We know that
\(A = N + Z\)
Where A is atomic mass number
The atomic mass of \(_1^3{{\rm{H}}_2}\) is 3 and
\(\begin{align}A &= 3\\Z &= 1\;\\N &= 2\end{align}\)
An electron and an antineutrino are released as a result of Beta decay.
\(\begin{align}A &= 3\\N &= 1 - ( - 1) = 2\\Z &= A - N\\ & = 3 - 2\\ &= 1\end{align}\)
As charge on beta particle is negative, so \({\beta ^ - }\) Decay equation of \(_1^3{H_2}{\rm{ is }}_1^3{H_2} \to _2^3H{e_1} + {\beta ^ - } + \overline {{\nu _e}} \)
Therefore, \({\beta ^ - }\) Decay equation of \(_1^3{H_2}{\rm{ is }}_1^3{H_2} \to _2^3H{e_1} + {\beta ^ - } + \overline {{\nu _e}} \).