Question 279: Atoms & Nuclei

Question

Taking the wavelength of first Balmer line in
hydrogen spectrum (n = 3 to n = 2) as 660 nm,
the wavelength of the 2nd Balmer line (n = 4 to
n = 2) will be :

Accepted Answer

$${1 \over {{\lambda _1}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} ight]$$
$${1 \over {{\lambda _2}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{4^2}}}} ight]$$
$${{5{\lambda _1}} \over {36}} = {{12{\lambda _2}} \over {4 	imes 16}}$$
$${\lambda _2} = {{5 	imes 660 	imes 64} \over {36 	imes 12}} = 489\,nm$$

Answer

642.7 nm

Answer

488.9 nm

Answer

889.2 nm

Answer

388.9 nm

Atoms & Nuclei

Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be :

Choose the correct answer: