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  • Let  be the frequency of the series limit of the Lyman series,

Let \mathrm{v}_1 be the frequency of the series limit of the Lyman series,\mathrm{v}_2 be the frequency of the first line of the Lyman series, and \mathrm{v}_3 be the frequency of the series limit of the Balmer series.

Option: 1

\mathrm{v}_1-\mathrm{v}_2=\mathrm{v}_3


Option: 2

v_2-v_1=v_3


Option: 3

v_3=\frac{1}{2}\left(v_1+v_2\right)


Option: 4

\mathrm{v}_1+\mathrm{v}_2=\mathrm{v}_3


Answers (1)

best_answer

Series limit means the shortest possible wavelength (maximum photon energy), and
first means the longest possible wavelength (minimum photon energy) in the series.

v=c\left[\frac{1}{n^2}-\frac{1}{m^2}\right], \text { where } c=\text { constant }

For series limit of the Lyman series,    \mathrm{n}=1, \mathrm{~m}=\infty, \mathrm{v} _{1}=\mathrm{c}

For first line of the Lyman series,         \mathrm{n}=1, \mathrm{~m}=2, \mathrm{v}_2=\frac{3 \mathrm{c}}{4}

For series limit of the Balmer series,    \mathrm{n}=2, \mathrm{~m}=\infty, \mathrm{v}_3=\frac{\mathrm{c}}{4}

\therefore \quad \mathrm{v}_1-\mathrm{v}_2=\mathrm{v} _{3} .

Posted by

Suraj Bhandari

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