Question

\[_{86}{A^{222}}{ \to _{84}}{B^{210}}.\] In this reaction, how many \[\alpha \] and\[\beta \] particles are emitted?
a) \[6\alpha ,3\beta \]
b) \[3\alpha ,4\beta \]
c) \[4\alpha ,3\beta \]
d) \[3\alpha ,6\beta \]

Answer 1

Hint: \[\alpha \]- decay happens when an unstable nucleus emits an \[\alpha \]- particle (\[_2H{e^4}\] nucleus), where it loses two protons and two neutrons. \[\beta \]- decay happens when an unstable nuclei emits a \[\beta \]- particle (\[{e^ - }\] - electron or \[{e^ + }\]- positron).

Complete step by step solution:
In \[\alpha \]- decay, the unstable nuclei loses two protons and two neutrons, which means the atomic number (Z) decreases by 2 and the atomic mass (A) decreases by 4. It can be represented as follows:
\[_z{X^A}{ \to _{Z - 2}}{Y^{A - 4}}{ + _2}H{e^4}\]

In \[\beta \]- decay, if the unstable nuclei lose an electron (\[{e^ - }\]), then the atomic number (Z) increases by 1 and the atomic mass remains same. If it loses a positron (\[{e^ + }\]), then the atomic number (Z) decreases by 1. It can be shown as
\[{\beta ^ - }{\rm{ decay}}{ \Rightarrow _Z}{X^A}{ \to _{Z + 1}}{Y^A} + {e^ - } + \overline \nu \]
\[{\beta ^ + }{\rm{ decay}}{ \Rightarrow _Z}{X^A}{ \to _{Z - 1}}{Y^A} + {e^ + } + \nu \]
Where,\[\nu {\rm{ and }}\overline \nu \] are neutrino and antineutrino respectively.

To find the number of \[\alpha \]and\[\beta \] particles emitted in the above reaction, we need to equate the atomic number (Z) and atomic mass (A).
Let,
\[_{86}{A^{222}}{ \to _{84}}{B^{210}} + {x_2}H{e^4} + {y_{ - 1}}{e^0}\]

Equating the atomic numbers, we get
\[86 = 84 + 2x - y\] (1)

Equating the atomic masses, we get
\[222 = 210 + 4x\]
\[12 = 4x\]
\[x = \dfrac{{12}}{4}\]
\[x = 3\]

Substituting x=3 in (1)
\[86 = 84 + 2(3) - y\]
\[86 = 84 + 6 - y\]
\[86 = 90 - y\]
\[y = 90 - 86\]
\[y = 4\]

So the above reaction emits 3\[\alpha \] and 4\[\beta \] particles.

Hence, option (b) is correct.

Note:Most reactions with \[\beta \]- decay would lose an electron (\[{e^ - }\]), unless it is mentioned as positron(\[{e^ + }\]). So if \[\beta \]- decay is given without positron specification, then consider it as electron decay.