Atomic StructureHard
Question
When an electron jumps from the second orbit to fourth orbit, its distance from nucleus increases by 2.116 Å. The atom or ion should be
Options
A.H atom
B.He+ ion
C.Li2+ ion
D.Be3+ ion
Solution
$r_{4} - r_{2} = 2.116\overset{o}{A} \Rightarrow 0.529 \times \frac{4^{2}}{z} - 0.529 \times \frac{2^{2}}{z} = 2.116$
$\therefore z = 3 \Rightarrow Li^{2 +}\text{ ion}$
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