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CBSE Class 12 Physics 2017 Outside Delhi Set 2 Paper

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Question : 4 of 9
Marks: +1, -0
The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 Å. Calculate the short wavelength limit for Balmer series of the hydrogen spectrum.
Solution:  
  1λ=R(  1n12−  1n22)\; \frac{1}{\lambda} = R \left( \; \frac{1}{n_1^2} - \; \frac{1}{n_2^2} \right)
∴\therefore For Balmer Series: (λB)  short   =4R\left( \lambda_B \right)_{\; \text{short } \;} = \frac{4}{R}
and For Lyman Series: (λL)  short   =1R\left( \lambda_L \right)_{\; \text{short } \;} = \frac{1}{R}
∴λB=913.4×4A˚=3653.6A˚\therefore \lambda_B = 913.4 \times 4 \text{\AA} = 3653.6 \text{\AA}
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