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For the emission line of atomic hydrogen, from n_{i}=8\, to\, n_{f}=n, what is the plot equation of wave number{\left ( \bar{v} \right )} against \left ( \frac{1}{n^{2}} \right ) energy, the Rydberg constant is in wave number units?

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Given - $n_i = 8$ and $n_f = n$

Solution - For the emission line of atomic hydrogen, the wave number is given by $\bar{\nu} = R\left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)$. 
So,  $\bar{\nu} = R\left( \dfrac{1}{n^2} - \dfrac{1}{64} \right) = R \cdot \dfrac{1}{n^2} - \dfrac{R}{64}$. 
This equation is of the form $y = mx + c$ , here, 

  • $y = \bar{\nu}$, $x = \dfrac{1}{n^2}$
  • Slope $m = R$, 
  • Intercept $c = -\dfrac{R}{64}$.

Hence, the plot of wave number $\bar{\nu}$ against $\dfrac{1}{n^2}$ is a straight line, and the Rydberg constant $R$ is the slope of this line. 

Posted by

Saniya Khatri

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