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  • The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like on is: <

The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like on is:

 

Option: 1

5


Option: 2

6


Option: 3

3


Option: 4

2


Answers (1)

best_answer

The wavelength of the first line of Lyman series for hydrogen atom is \frac{1}{\lambda}=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]

The wavelength of the second line of Balmer series for hydrogen like ion is \frac{1}{\lambda^{\prime}}=\mathrm{Z}^2 \mathrm{R}\left[\frac{1}{2^2}-\frac{1}{4^2}\right]

According to question, \lambda=\lambda^{\prime}

\Rightarrow \mathrm{R}\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\mathrm{Z}^2 \mathrm{R}\left[\frac{1}{2^2}-\frac{1}{4^2}\right]

Or \frac{3}{4}=\frac{3 z^2}{16}

or \mathrm{Z}^2=4$ or $\mathrm{Z}=2

Posted by

Gautam harsolia

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