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If the shortest wavelength of the spectral line of He+ in Layman series is x then the longest wavelength of the line in Balmer series of Li2+ is: (in terms of x)

Option: 1

1.25


Option: 2

0.8


Option: 3

3.2


Option: 4

9


Answers (1)

best_answer

We know,

 \frac{1}{\lambda }= RZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )

And the data with given with wavelength and energy.

 

For the shortest wavelength for He+ in Lyman Series. So n2 = infinity and n1 = 1

\mathrm{\frac{1}{x}=R \times 4 \times(\frac{1}{1}-\frac{1}{\infty})}

\Rightarrow R = \frac{1}{4x}

For the longest wavelength of Li2+ in Balmer series, So  n2 = 3 and n1 = 2

\mathrm{\frac{1}{\lambda}=R \times9 \times(\frac{1}{4}-\frac{1}{9})=\frac{5R}{4}}

\therefore\frac{1}{\lambda} =\frac{5}{16x}

\therefore \lambda= \frac{16x}{5}

 

 

Posted by

Divya Prakash Singh

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