Question 569: Atoms & Nuclei

Question

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

Accepted Answer

For the longest wavelength in the Lyman series,
n
1
= 1 and n
2
= 2
$${1 \over {{\lambda _L}}} = R\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} ight)$$ = $${{3R} \over 4}$$
For the longest wavelength in the Balmer series,
n
1
= 2 and n
2
= 3
$${1 \over {{\lambda _B}}} = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} ight)$$ = $${{5R} \over {36}}$$
$$ 	herefore $$ $${{{\lambda _L}} \over {{\lambda _B}}} = {{{4 \over {3R}}} \over {{{36} \over {5R}}}}$$ = $${5 \over {27}}$$

Answer

$${{27} \over 5}$$

Answer

$${5 \over {27}}$$

Answer

$${4 \over 9}$$

Answer

$${9 \over 4}$$

Atoms & Nuclei

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

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