For emission line of atomic hydrogen from
Question:

For emission line of atomic hydrogen from $n_{i}=8$ to $n_{f}=n$, the plot of wave number $(\bar{v})$ against $\left(\frac{1}{\mathrm{n}^{2}}\right)$ will be (The Rydberg constant, $R_{H}$ is in wave number uint)

  1. Linear with intercept $-\mathrm{R}_{\mathrm{H}}$

  2. Non linear

  3. Linear with slope $\mathrm{R}_{\mathrm{H}}$

  4. Linear with slope $-\mathrm{R}_{\mathrm{H}}$


Correct Option:

Solution:

As we know,

$\bar{v}=-R_{H}\left(\frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}}\right) Z^{2}($ where,$Z=1)$

After putting the values, we get

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

$\Rightarrow \bar{v}=\frac{R_{H}}{64}-\frac{R_{H}}{n^{2}}$

Comparing to $y=m x+c$, we get

$x=\frac{1}{n^{2}}$ and $m=-R_{H}$ (slope)

Administrator

Leave a comment

Please enter comment.
Please enter your name.