Question

Calculate the number of {\text{\alpha }} and {\text{\beta }} particles in the following change:
${\text{U}}_{92}^{235} \to {}_{82}^{207}{\text{Pb}}$
A) {\text{\alpha }} - 4;{\text{\beta }} - 3
B) {\text{\alpha }} - 2;{\text{\beta }} - 3
C) {\text{\alpha }} - 5;{\text{\beta }} - 3
D) {\text{\alpha }} - 7;{\text{\beta }} - 4

Answer 1

Alpha particle \left( {\text{\alpha }} \right) is a nucleus of helium atom. The {\text{\alpha }}-decay is {}_{\text{Z}}^{\text{A}}{\text{X}}\xrightarrow{{\text{\alpha }}}{}_{{\text{Z}} - 2}^{{\text{A}} - 4}{\text{Y}}. The {\text{\beta }}-decay is {}_{\text{Z}}^{\text{A}}{\text{X}}\xrightarrow{{\text{\beta }}}{}_{{\text{Z}} + 1}^{\text{A}}{\text{Y}}. Here, A is the atomic mass and Z is the atomic number. Use these reactions to calculate the number of {\text{\alpha }} and {\text{\beta }} particles.

Complete step by step answer:
In {\text{\alpha }}-decay, a nucleus transforms itself into a different nucleus by emitting and {\text{\alpha }}-particle. {\text{\alpha }}-particle is a nucleus of helium. Thus, ${}_2^4{\text{He}}$ is the {\text{\alpha }}-particle.
The helium nucleus contains two neutrons and two protons. Thus, when {\text{\alpha }}-particle is emitted the atomic mass decreases by four and the atomic number decreases by two.
Thus, ${}_{\text{Z}}^{\text{A}}{\text{X}}$ changes to ${}_{{\text{Z}} - 2}^{{\text{A}} - 4}{\text{Y}}$ by emitting ${}_2^4{\text{He}}$. The general reaction is as follows:
${}_{\text{Z}}^{\text{A}}{\text{X}} \to {}_{{\text{Z}} - 2}^{{\text{A}} - 4}{\text{Y}} + {}_2^4{\text{He}}$
In {\text{\beta }}-decay, electrons or positrons are emitted. {\text{\beta }}-decay is of two types: beta plus decay and beta minus decay.
In {{\text{\beta }}^ + }-decay, positron is emitted. The general reaction is as follows:
${}_{\text{Z}}^{\text{A}}{\text{X}} \to {}_{{\text{Z}} - 1}^{\text{A}}{\text{Y}} + {e^ + } + \upsilon$
In {{\text{\beta }}^ - }-decay, electrons are emitted. The general reaction is as follows:
${}_{\text{Z}}^{\text{A}}{\text{X}} \to {}_{{\text{Z}} + 1}^{\text{A}}{\text{Y}} + {e^ - } + \upsilon$
We have to calculate the number of {\text{\alpha }} and {\text{\beta }} particles emitted.
${\text{U}}_{92}^{235}\xrightarrow{{\alpha ,\beta }}{}_{82}^{207}{\text{Pb}} + {}_2^4{\text{He}} + {}_{ - 1}^0e$
Calculate the number of {\text{\alpha }}-particles as follows:
{\text{Atomic mass of Pb}} = {\text{Atomic mass of U}} - 4{\text{\alpha }}
{\text{207}} = {\text{235}} - 4{\text{\alpha }}
{\text{\alpha }} = \dfrac{{235 - 207}}{4}
{\text{\alpha }} = 7
Thus, the number of {\text{\alpha }}-particles is 7.
Calculate the number of {\text{\beta }}-particles as follows:
{\text{Atomic number of U}} = {\text{Atomic number of Pb}} + 2{\text{\alpha }} - {\text{\beta }}
{\text{92}} = {\text{82}} + 2 \times {\text{7}} - {\text{\beta }}
{\text{\beta }} = - 92 + 82 + 14
{\text{\beta }} = 4
Thus, the number of {\text{\beta }}-particles is 4.
Thus, the number of {\text{\alpha }} and {\text{\beta }} particles in the change ${\text{U}}_{92}^{235} \to {}_{82}^{207}{\text{Pb}}$ are {\text{\alpha }} - 7;{\text{\beta }} - 4.

Thus, the correct option is (D) {\text{\alpha }} - 7;{\text{\beta }} - 4.

Note: The given reaction ${\text{U}}_{92}^{235} \to {}_{82}^{207}{\text{Pb}}$, is not a direct decay. This is because many alpha and beta decays occur. Remember that positrons and protons are different. Positrons are positively charged electrons and it has no mass. In alpha decay, no external energy is required. The alpha decay decreases the atomic number while the beta decay increases or decreases the atomic number.