Question

Group Displacement Law was given by:
a.) Becquerel
b.) Rutherford
c.) Mendeleev
d.) Soddy and Faza

Answer 1

Hint: So to understand the question, the foremost thing is that we should know about the law and what it states, how it works. In short, the concept behind it. Down here it is explained in a proper way and hence by this information we can easily solve this question.

Complete step by step solution: This law gives us information about reduction of weight and atomic number of a radioactive element:
There is a decrease in atomic number and atomic weight of the resulting new nuclei which is formed from earlier radioactive elements. If a α-particle is realized by a radioactive element by its nucleus, the atomic no. of the fresh element which will be formed is deducted by 2 units and its atomic weight is deducted by 4 units. So, by this we can come to a result that the fresh element which is created will relocate itself by two groups on its left side of the periodic table.
$_z^{}{A^M} \to _{z + 1}^{}{C^M} + He\left( {\alpha - \,particle} \right)$

There is a rise in atomic number and atomic weight of the resulting new nuclei will remain the same which is formed from earlier radioactive elements. If a β–particle is released by a radioactive element, the atomic number of the fresh element which will be formed takes a rise by one unit. So, by this we can come to a result that the fresh element which is created will relocate itself by one group on its left side of the periodic table.
$_Z^{}{A^M} \to \,_{Z + 1}^{}{C^M} + \beta \left( {\beta - particle} \right)$
If a α-particle is released by the radioactive nucleus and then 2β-particles are released in the next two alterations, the fresh element will be an isotope of the starting element. The fresh and starting element will have the similar atomic number. So by Group displacement law position of the fresh and daughter element in the periodic table will remain as it was earlier.
                              $_Z^{}{A^M} \to \,_Z^{}{A^M} + 2\beta + He$


The law which is explained above “Group Displacement Law” was specified in 1913 by Fajan and Soddy.

Hence, the correct option is (D) - Soddy and Fazan

Note: In this problem we came across a law specified by Soddy and Fajan in 1913. Both of them told us that whenever an a-particle is removed from a radioactive material there are some changes that occur in the new element. We also studied what happens when β particles are released.