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  • One of the lines in the emission spectrum of  has the same wavelength as that of the <img alt="

One of the lines in the emission spectrum of Li^{2+} has the same wavelength as that of the 2^{nd} line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is :

Option: 1

n=4\rightarrow n=2

 

 


Option: 2

n=8\rightarrow n=2


Option: 3

n=8\rightarrow n=4


Option: 4

n=12\rightarrow n=6


Answers (1)

best_answer

For 2nd line of Balmer series in the hydrogen spectrum

\frac{1}{\lambda }=R\left ( 1 \right )\left ( \frac{1}{2^{2}} -\frac{1}{4^{2}}\right )=\frac{3}{16}R

For Li ^{2+} \frac{1}{\lambda }=R\left ( 3 \right )^{2}\left ( \frac{1}{6^{2}}-\frac{1}{12^{2}} \right )=\frac{3}{16}R

Which is satisfied by only (D).

Posted by

Ajit Kumar Dubey

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