JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 5)
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 :
642.7 nm
488.9 nm
889.2 nm
388.9 nm
Explanation
$${1 \over {{\lambda _1}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right]$$
$${1 \over {{\lambda _2}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right]$$
$${{5{\lambda _1}} \over {36}} = {{12{\lambda _2}} \over {4 \times 16}}$$
$${\lambda _2} = {{5 \times 660 \times 64} \over {36 \times 12}} = 489\,nm$$
$${1 \over {{\lambda _2}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right]$$
$${{5{\lambda _1}} \over {36}} = {{12{\lambda _2}} \over {4 \times 16}}$$
$${\lambda _2} = {{5 \times 660 \times 64} \over {36 \times 12}} = 489\,nm$$
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