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2.33 What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of $\textup{He}^+$ spectrum?

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For the transition of H-like particles,

$\vartheta = \frac{1}{\lambda} = R_{H}Z^2\left [ \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right ]$

For $\textup{He}^+$ transition spectrum,

$Z=2$ , $n_{2}=4$ , and $n_{1}=2$

Therefore,

$\vartheta = \frac{1}{\lambda} = R_{H}2^2\left [ \frac{1}{2^2}-\frac{1}{4^2} \right ] = \frac{3R_{H}}{4}$

Then for the hydrogen spectrum,

$\vartheta = \frac{3R_{H}}{4}$ and $Z=1$

Therefore,

$\vartheta = \frac{1}{\lambda} = R_{H}\times1\left [ \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right ]$

$\Rightarrow R_{H}\times1\left [ \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right ] = \frac{3R_{H}}{4}$

$\Rightarrow \left [ \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right ] = \frac{3}{4}$

The values of $n_{1}\ and\ n_{2}$ can be found by the hit and trial method in the above equation.

So, we get $n_{1} = 1$ and $n_{2} = 2$ , i.e., the transition is from $n =2$ to $n =1$ .

Posted by

Divya Prakash Singh

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