An atomic nucleus $A$ is composed of $84$ protons and $128$ neutrons. The nucleus $A$ emits an alpha particle and is transformed into nucleus $B$ . What is the composition of nucleus B.

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Solution

Step1:- Given data
Number of protons of nucleus $A$ is equals to $84$
Number of neutrons of nucleus $A$ is equals to $128$
Step2:- Finding mass number of nucleus $A$
$= m a s s n u m b e r o f n u c l e u s A = n u m b e r o f p r o t o n s + n u m b e r o f n e u t r o n s = 84 + 128 = 212$
Step:- Write the nuclear equation
We known that emission of $α$ -particle's mass number decreased by $4 u n i t s$ and Its atomic number decreased by $2 u n i t s$
$\begin{array}{rcl} N o w, m a s s n u m b e r o f n u c l e u s B & = & 212 - 4 \\ = & 208 \\ A t o m i c n u m b e r o f n u c l e u s B & = & 84 - 2 \\ = & 82 \end{array}$
$α$ -particles symbol ${}_{2}^{4}H e$
So, The Nuclear equation can be represented as
${}_{84}A^{212} \to {}_{82}B^{208} + {}_{2}^{4}H e$
Hence, The composition of nucleus B is $82$ protons and $126$ neutrons.

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