Question 520: Atoms & Nuclei

Question

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is

Accepted Answer

In case of Balmer series :
n
1
= $$\infty $$, n
2
= 2
$$ 	herefore $$ $${1 \over {{\lambda _B}}} = Rc\left( {{1 \over {{2^2}}} - {1 \over {{\infty ^2}}}} ight)$$ = $${{Rc} \over 4}$$
In case of Lyman series :
n
1
= $$\infty $$, n
2
= 1
$$ 	herefore $$ $${1 \over {{\lambda _L}}} = Rc\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} ight)$$ = Rc
$$ 	herefore $$ $${{{\lambda _B}} \over {{\lambda _L}}}$$ = $${4 \over 1}$$ = 4

Answer

1

Answer

4

Answer

0.5

Answer

2

Atoms & Nuclei

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is

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