Question

Electron in a hydrogen atoms first jumps from fifth excited state to third excited and then from third state to second excited state the ratio of wavelengths emitted in two cases is

Answer 1

Given:

Electron in a hydrogen atoms first jumps from fifth excited state to third excited and then from third state to second excited state.

To find:

Ratio of wavelength in the two cases.

Calculation:

In the 1st case:

$\dfrac{1}{ \lambda_{1} } = R \bigg \{ \dfrac{1}{ {(n2)}^{2} } - \dfrac{1}{ {(n1)}^{2} } \bigg \}$

$= > \dfrac{1}{ \lambda_{1} } = R \bigg \{ \dfrac{1}{ {(3)}^{2} } - \dfrac{1}{ {(5)}^{2} } \bigg \}$

$= > \dfrac{1}{ \lambda_{1} } = R \bigg \{ \dfrac{1}{ 9} - \dfrac{1}{ 25 } \bigg \}$

$= > \dfrac{1}{ \lambda_{1} } = R \bigg \{ \dfrac{25 - 9}{ 225} \bigg \}$

$= > \dfrac{1}{ \lambda_{1} } = R \bigg \{ \dfrac{16}{ 225} \bigg \}$

In the 2nd case:

$\dfrac{1}{ \lambda_{2} } = R \bigg \{ \dfrac{1}{ {(n2)}^{2} } - \dfrac{1}{ {(n1)}^{2} } \bigg \}$

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

$= > \dfrac{1}{ \lambda_{2} } = R \bigg \{ \dfrac{1}{ 4 } - \dfrac{1}{ 9} \bigg \}$

$= > \dfrac{1}{ \lambda_{2} } = R \bigg \{ \dfrac{9 - 4}{ 36 } \bigg \}$

$= > \dfrac{1}{ \lambda_{2} } = R \bigg \{ \dfrac{5}{ 36 } \bigg \}$

So, required ratio:

$= > \dfrac{ \lambda_{1} }{ \lambda_{2} } = \dfrac{( \frac{5}{36} )}{ (\frac{16}{225}) }$

$= > \dfrac{ \lambda_{1} }{ \lambda_{2} } = \dfrac{1125}{576}$

$= > \lambda_{1} : \lambda_{2} = 1125 : 576$

So, final answer is:

$\boxed{ \bf{\lambda_{1} : \lambda_{2} = 1125 : 576}}$