Group Displacement Law
Category : JEE Main & Advanced
Soddy, Fajans and Russell (1911-1913) observed that when an a-particle is lost, a new element with atomic number less by 2 and mass number less by 4 is formed. Similarly, when b-particle is lost, new element with atomic number greater by 1 is obtained. The element emitting then a or b-particle is called parent element and the new element formed is called daughter element. The above results have been summarized as,
(1) When an a-particle is emitted, the new element formed is displaced two positions to the left in the periodic table than that of the parent element (because the atomic number decreases by 2).
(2) When a b-particle is emitted, the new element formed is displaced one position to the right in the periodic table than that of the parent element (because atomic number increased by 1).
(3) When a positron is emitted, the daughter element occupies its position one group to the left of the parent element in periodic table.
Group displacement law should be applied with great care especially in the case of elements of lanthanide series (57 to 71), actinide series (89 to 103), VIII group (26 to 28; 44 to 46; 76 to 78), IA and IIA groups.
To determine the number of a- and b- particles emitted during the nuclear transformation. It can be done in following manner, \[_{c}^{a}X\to \,_{d}^{b}Y+x\,_{2}^{4}He+y{{\,}_{-1}}{{e}^{0}}\]
\[a=b+4x\] or \[\text{ }x=\frac{a-b}{4}\] .....(i)
\[c=d+2x-y\] .....(ii)
where x = no. of a-emitted, y = no. of b-emitted
substituting the value of x from eq. (i) in eq. (ii) we get \[c=d+\left( \frac{a-b}{4} \right)\,2-y\]; \[y=d+\left[ \frac{a-b}{2} \right]-c\]
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