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The wave number of the spectral line corresponding to the transition from n_{1}=2\:to\:n_{2}=4 is

  • Option 1)

    20565 per m

  • Option 2)

    205.65 per cm

  • Option 3)

    20565 per cm

  • Option 4)

    205.65 per m

 

Answers (1)

best_answer

As learnt in

Line Spectrum of Hydrogen like atoms -

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

 

- wherein

Where R is called Rhydberg constant, R = 1.097 X 107 , Z is atomic number

n1= 1,2 ,3….

n2= n1+1, n1+2 ……

 

 

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

     = 10973731.6 \ m^{-1} \times \left ( 1 \right )^{2}\left ( \frac{1}{4}- \frac{1}{16} \right )

     = 10973731.6 \times \frac{3}{16}

     = 20575.74 cm-1

     \sim 20565 cm 


Option 1)

20565 per m

This option is incorrect

Option 2)

205.65 per cm

This option is incorrect

Option 3)

20565 per cm

This option is correct

Option 4)

205.65 per m

This option is incorrect

Posted by

divya.saini

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