Question

The periodic table has 18 groups and 7 columns. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. The group number of element X is:

${}_{29}^{63}Cu{+}_1^{1}H$ $\stackrel{}{\to }$ $6_0^{1}n+\alpha +2_1^{1}H+X$

• A. $8$
• B. $7$
• C. $6$
• D. $5$

Answer 1

Answer: (A)

$8$

${}_{29}^{63}Cu{+}_1^{1}H\to 6_0^{1}n{+}_2^{4}He+2_1^{1}H{+}_Z^{A}X$

The sum of mass numbers on LHS of the equation is equal to the sum of mass numbers on RHS of the equation.

$63+1=1\times 6+4+1\times 2+A$

$A=52$

The sum of atomic numbers on LHS of the equation is equal to the sum of atomic numbers on RHS of the equation.

$29+1=6\times 0+2+2\times 1+Z$

$Z=26$

The element X is iron.

${}_Z^{A}X={}_{26}^{52}Fe$

Hence, X is in group ${}^{\prime }{8}^{\prime }$ in the periodic table.