Question

A radioactive decay chain starts from $_{93}N{p^{237}}$ and produces $_{90}T{h^{229}}$ by successive emissions. The emitted particles can be
A. Two $\alpha $ particles and one $\beta $ particle
B. Three ${\beta ^ + }$ particles
C. One $\alpha $ particle and two ${\beta ^ + }$ particles
D. One $\alpha $particle and two ${\beta ^ - }$particles

Answer 1

Hint: In radioactive reactions, when $\alpha $ decay happens, the parent nuclei loses two protons and two neutrons, which is represented as $_2H{e^4}$ nucleus, while, in $\beta $ decay, it loses either an electron (${e^ - }$) or a positron (${e^ + }$).

Complete step by step solution:
Radioactive decay can be described as a nuclear phenomenon where an unstable nucleus emits $\alpha ,\beta $ or $\gamma $ emissions and undergoes decay. When an unstable nucleus emits a $\alpha $ particle, which is $_2H{e^4}$ nucleus, it loses two protons and two neutrons. This can also be said as the parent nucleus has its atomic number (Z) decreased by two and the atomic mass (A) decreased by four and forms a daughter nucleus. It can be represented by,
\[_z{X^A}{ \to _{Z - 2}}{Y^{A - 4}}{ + _2}H{e^4}\]

Similarly, when an unstable nucleus emits a $\beta $particle, which is either an electron (${e^ - }$) or positron (${e^ + }$). Positron is a similar particle like electron with negligible mass and oppositely positive charge. If an electron (${e^ - }$) is emitted in a radioactive decay, then the atomic mass (A) remains the same but the atomic number (Z) increases by one.
${\beta ^ - }decay{ \Rightarrow _Z}{X^A}{ \to _{Z + 1}}{Y^A} + {e^ - } + \overline \nu $

If a positron (${e^ + }$) is emitted, then the atomic mass (A) remains the same but the atomic number (Z) decreases by one. It goes as follows,
${\beta ^ + }decay{ \Rightarrow _Z}{X^A}{ \to _{Z - 1}}{Y^A} + {e^ + } + \nu $
$\nu $ and $\overline \nu $ are neutrinos and antineutrinos respectively. They are neutral and have negligible mass. They are hard to detect because they have weak interactions with matter.

To find the number of $\alpha $and $\beta $ particles emitted, let us consider the following,
$_{93}N{p^{237}}{ \to _{90}}T{h^{229}} + {x_2}H{e^4} + y{e^ \pm }$
Let us equate the atomic masses (A) on the reaction,
\[237 = 229 + 4x + 0\]
\[\Rightarrow 237 - 229 = 4x\]
\[\Rightarrow 8 = 4x\]
\[\Rightarrow x = \dfrac{8}{4}\]
\[\Rightarrow x = 2\]
So the number of $\alpha $particles emitted are 2.

Now, equate the atomic number (Z),
\[93 = 90 + 2x + y\]
\[\Rightarrow 93 = 94 + y\]
\[\Rightarrow y = 93 - 94\]
\[\therefore y = - 1\]
So the number of $\beta $particle emitted is 1.

The correct answer is option (A).

Note:In this reaction $\beta $ is negative which means an electron (${e^ - }$) is emitted, if an electron is emitted then the atomic number (Z) is increased by one. So in this reaction two $\alpha $ particles were emitted, then Z will be decreased by four, (i.e., 93 - 4 = 89) and the electron adds one to Z (i.e., 89+1=90) which results in $_{90}T{h^{229}}$ from$_{93}N{p^{237}}$.