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Home / Doubts and Solutions/JEE/Physics

THE WAVELENGTH OF Ka X-RAY OF TWO METAL A AND B ARE 4/1875R AND 1/675R RESPECTIVELY. WHERE R IS RYDBERG CONSTANT THE NO ELEMENTS LYING BETWEEN A AND B ACC. TO THE ATOMIC NO. ARE
 
 
 
 
PLS ANSWER THIS QUESTION.

Asked by sudheerkapoor67 10th October 2017, 9:51 AM
Answered by Expert
Answer:
According to Moseley's equation for Kα radiation 1 over lambda space equals space R left parenthesis z minus 1 right parenthesis squared space open parentheses 1 over 1 squared space minus space 1 over 2 squared close parentheses
 
Hence lambda subscript A over lambda subscript B space equals space fraction numerator begin display style open parentheses Z subscript B space minus space 1 close parentheses squared end style over denominator begin display style open parentheses Z subscript A minus 1 close parentheses squared end style end fraction space equals space fraction numerator 4 space x space 675 space over denominator 1875 end fraction space equals space fraction numerator 4 space cross times space 3 space cross times 225 over denominator 625 space cross times space 3 end fraction space equals space fraction numerator 4 space cross times space 225 over denominator 625 end fraction space equals space fraction numerator 2 squared space space cross times space 15 squared over denominator 25 squared end fraction space equals space 30 squared over 25 squared
 
So ZB = 31 and ZA = 26. As per atomic numbers, metal-A is Fe and metal-B is Ga.
 
In between Fe and Ga, there are 4 elements Co(z=27), Ni(z=28), Cu(z=29) and Zn(z=30)   


Answered by Expert 11th October 2017, 1:37 PM
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