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  4. If the wavelength of first line of the Balmer series of hydrogen atom is 656.1 nm, the wavelength of second line of this series would be

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Q. If the wavelength of first line of the Balmer series of hydrogen atom is $656.1$ nm, the wavelength of second line of this series would be

Structure of Atom

Solution:

$\frac{1}{\lambda_{1}}=R_{H}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R_{H}\left(\frac{5}{36}\right)$
$R_{H}=\frac{36}{5\times656.1}$
$\frac{1}{\lambda_{2}} =R_{H} \left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)$
$\frac{1}{\lambda_{2}}=\frac{36}{5\times626.1}\left(\frac{3}{16}\right) $
$\lambda=\frac{5\times656.1 \times16}{3\times36}=486.0\, nm$