Q. If the series limit of Lymen series for Hydrogen atom is equal to the series limit of Balmer series for a hydrogen like atom, then atomic number of this hydrogen like atom will be

A) 1
B) 2
C) 3
D) 4
Explanation

(b) By using $\frac{1}{\lambda} = RZ^{2}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
For Hydrogen atom $\frac{1}{\left( \lambda_{\min} \right)_{H}} = R\left\lbrack \frac{1}{1^{2}} - \frac{1}{\infty} \right\rbrack = R$
$\Rightarrow \left( \lambda_{\min} \right)_{H} = \frac{1}{R}$ …(i)
For hydrogen like atom $\left( \frac{1}{\lambda_{\min}} \right)_{atom} = RZ^{2}\left( \frac{1}{2^{2}} - \frac{1}{\infty} \right)$
$\Rightarrow \left( \lambda_{\min} \right)_{atom} = \frac{4}{RZ^{2}}$ …(i)
From equation (i) and (ii), $\frac{1}{R} = \frac{4}{RZ^{2}} \Rightarrow Z = 2$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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