Question

Calculate the ratio of wavelength of first and the ultimate line of balmer series of Li2+ ion.​

Answer 1

To find:

The ratio of wavelength of first and the ultimate line of balmer series of Li2+ ion.

Calculation:

For the first line of Balmer series:

$\therefore \: \dfrac{1}{ \lambda1} = R {z}^{2} \bigg \{ \dfrac{1}{ {(n1)}^{2} } - \dfrac{1}{ {(n2)}^{2} } \bigg \}$

$= > \: \dfrac{1}{ \lambda1} = R {z}^{2} \bigg \{ \dfrac{1}{ {(2)}^{2} } - \dfrac{1} {{(3)}^{2} } \bigg \}$

$= > \: \dfrac{1}{ \lambda1} = R {z}^{2} \bigg \{ \dfrac{1}{4 } - \dfrac{1} {9 } \bigg \}$

$= > \: \dfrac{1}{ \lambda1} = R {z}^{2} \bigg \{ \dfrac{5 } {36} \bigg \}$

For the ultimate line in Balmer series:

$\therefore \: \dfrac{1}{ \lambda2} = R {z}^{2} \bigg \{ \dfrac{1}{ {(n1)}^{2} } - \dfrac{1}{ {(n2)}^{2} } \bigg \}$

$= > \: \dfrac{1}{ \lambda2} = R {z}^{2} \bigg \{ \dfrac{1}{ {(2)}^{2} } - \dfrac{1}{ { \infty}^{2} } \bigg \}$

$= > \: \dfrac{1}{ \lambda2} = R {z}^{2} \bigg \{ \dfrac{1}{4} - 0\bigg \}$

$= > \: \dfrac{1}{ \lambda2} = R {z}^{2} \bigg \{ \dfrac{1}{4} \bigg \}$

Required Ratio:

$\therefore \: \dfrac{ \lambda1}{ \lambda2} = \dfrac{9}{5}$

So, final answer:

$\boxed{ \large{ \bold{ \lambda1 : \lambda2 = 9 : 5}}}$