Physics · 2019
JEE Main · 9 April 2019, Shift 1 · Q25
Taking the wavelength of first Balmer line in hydrogen spectrum ( n =3 to n =2 ) as 660 nm, the wavelength of the 2^nd Balmer line ( n =4 to n =2 )…
Taking the wavelength of first Balmer line in hydrogen spectrum ( $\displaystyle \mathrm{n}=3$ to $\displaystyle \mathrm{n}=2$ ) as $\displaystyle 660$ nm, the wavelength of the $\displaystyle 2^{\text {nd }}$ Balmer line ( $\displaystyle \mathrm{n}=4$ to $\displaystyle \mathrm{n}=2$ ) will be :
Official answer
(1)
488.$\displaystyle 9$ nm
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.