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For emission line of atomic hydrogen from n_{i}=8\, to\, n_{f}=n, the plot of wave number\overline{\left ( v \right )} against \left ( \frac{1}{n^{2}} \right ) will be  (The R_{Y} dberg constant ,is in wave nimber unit)

  • Option 1)

    Linear with intercept--R_{H}

  • Option 2)

    Non linear

  • Option 3)

     

    Linear with slopeR_{H}

  • Option 4)

    Linear with slope---R_{H}

Answers (1)

best_answer

 

Balmer Series Spectrum -

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

Where n_{1}=2\ and\ n_{2}=3, 4, 5, 6 .....

It lies in the visible region

As we know the formula \frac{1}{\lambda }= v= R_{H}Z^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )

n_{1}\rightarrow 8\, \, to\, \, n_{2}\rightarrow n\, \, and\, \, Z=1

v=R_{H}\left ( 1 \right )^{2}\left ( \frac{1}{8^{2}}-\frac{1}{n^{2}} \right )\frac{R_{H}}{64}-\frac{R_{H}}{n^{2}}

comparing y=mx+c, \, \, so,v=\left ( -RH \right )\left ( \frac{1}{n} \right )\left ( \frac{1}{n^{2}} \right )+\left ( \frac{R_{H}}{64} \right )

slope\rightarrow \left ( -R_{H} \right )-

 

 

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 ……

 

 

 

 

-

 

  


Option 1)

Linear with intercept--R_{H}

Option 2)

Non linear

Option 3)

 

Linear with slopeR_{H}

Option 4)

Linear with slope---R_{H}

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