Question 578: Atoms & Nuclei

Question

Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is

Accepted Answer

We see that wavelength of Lyman series,
n
f
= 3, n
i
= 4
$${1 \over {{\lambda _L}}} = R\left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} ight]$$ = $${{7R} \over {144}}$$
We see that wavelength of Balmer series :
n
f
= 2, n
i
= 3
$${1 \over {{\lambda _B}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} ight]$$ = $${{5R} \over {36}}$$
Now ratio of longest wavelengths corresponds to
Lyman and Balmer series:
$${{{\lambda _L}} \over {{\lambda _B}}} = {5 \over {36}} 	imes {{144} \over 7}$$ = $${5 \over {27}}$$

Answer

$${7 \over {29}}$$

Answer

$${9 \over {31}}$$

Answer

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

Answer

$${3 \over {23}}$$

Atoms & Nuclei

Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is

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