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Taking the wavelength of first Balmer line in hydrogen spectrum \left ( n=3\; to\; n=2 \right ) as 660\; nm, the wavelength of the 2^{nd} Balmer line \left ( n=4\; to\; n=2 \right ) will be

  • Option 1)

     889.2 \: nm

  • Option 2)

    488.9\: nm

  • Option 3)

     642.7\: nm

  • Option 4)

     388.9\: nm

Answers (1)

best_answer

\frac{1}{\lambda_{1}}=R\left ( \frac{1}{2^{2}}-\frac{1}{3^{2}} \right )=\frac{5R}{36}

\frac{1}{\lambda_{2}}=R\left [ \frac{1}{2^{2}}-\frac{1}{4^{2}} \right ]=\frac{3R}{36}

\frac{\lambda _{1}}{\lambda _{2}}=\frac{27}{20}

\lambda _{2}=\frac{20}{27}\times 660

\lambda _{2}=488.8


Option 1)

 889.2 \: nm

Option 2)

488.9\: nm

Option 3)

 642.7\: nm

Option 4)

 388.9\: nm

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