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  4. The first line of Balmer series has wavelength 6563 mathringA. What will be the wavelength of the first member of Lyman series

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Q. The first line of Balmer series has wavelength $6563\, \mathring{A}$. What will be the wavelength of the first member of Lyman series

JIPMERJIPMER 2018Atoms

Solution:

$\frac{1}{Balmer}\quad R\, \frac{1}{2^{2}}\, \frac{1}{3^{2}}\,=\frac{5R}{36},$

$\frac{1}{Lyman}\quad R\, \frac{1}{1^{2}}\, \frac{1}{2^{2}}\,=\frac{3R}{4}$

$\therefore \,\,\lambda_{Lyman}\,=\,\lambda_{Balmer}\times\frac{5}{27}= 1215.4\mathring{A}$