Question

A nuclear transformation is denoted by $X\left( {n,\alpha } \right) \to 3L{i^7}$. The nucleus of the element X is
A. $6{C^{12}}$
B. $5{B^{10}}$
C. $5{B^9}$
D. $4B{e^{11}}$

Answer 1

Hint:To solve this problem we have to use the concept of nuclear bombardment where a larger atom splits into a smaller atom with some alpha particle. Now using neutron bombardment we can compare the RHS and LHS terms so as to find the element X. Some other particles are also released.

Complete step by step answer:
When an element is bombarded with a neutron having mass number A equals to \[1\] and atomic number Z as $0$ then the element breaks into pieces one is an element of smaller atomic number and mass number with respect to that of the bombarding elements and also an $\alpha $ particle which is know as helium atom whose mass number is \[4\] and atomic number is \[2\].

Representing the above statement mathematically we will get,
$Z{X^A} + 0{n^1} \to Z{Y^A} + 2H{e^4}$
Where, the bombarding element = $X$, the breaking element = $Y$, Mass Number = $A$, Atomic Mass = $Z$, Neutron = $n$ and Helium atom = $He$.
Now from the problem we know the value of Y that is $3L{i^7}$
Writing the equation $\left( 1 \right)$ we will get,
$Z{X^A} + 0{n^1} \to 3L{i^7} + 2H{e^4}$
Comparing both sides we can say that both sides must be balanced.

So according to the conservation of mass number we can write,
$A + 1 = 7 + 4$
$ \Rightarrow A + 1 = 11$
Rearranging the above equation we will get,
$A = 10$
Now according to conservation of charge number we will get,
$Z + 0 = 2 + 3$
$ \therefore Z = 5$
Hence the element X has atomic number equals to $5$ and mass number equals to $10$.Therefore Boron is the element whose atomic number is $5$ and ,ass number is $10$.

Hence, the correct option is $\left( B \right)$.

Note:Generally students get confused between beryllium and boron as both of them start from B but their notations are different so before answering make sure that boron is denoted as B. Always keep in mind while comparing the RHS and LHS terms only the common terms get compared with each other similarly while adding the RHS side or LHS side atomic mass one atom gets added with atomic mass of another atom smiliary in case of mass number.