JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer 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 : [JEE Main 9-4-2019 Morning]

    A) 889.2 nm                    

    B) 642.7 nm

    C) 488.9 nm

    D) 388.9 nm

    Correct Answer: C

    Solution :

    \[\frac{1}{660}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5R}{36}\]             ....(1) \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3R}{16}\]                                ....(2) divide equation (1) with (2) \[\frac{\lambda }{660}=\frac{5\times 16}{36\times 3}\] \[\lambda =\frac{4400}{9}=488.88=488.9nm\] Option [c]                   


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