Question

calculate wavenumber of 1st line of balmer series

Answer 1

 μ =1/λ = R_{n} [1/ n_{1} ^{2} + 1/ n_{2^{2} }
 
= 1.097 x 10^{7} [1/ 2^{2} - 1/ 3^{2}
 
= 0.152 x 10^{7} m^{-1}
   
wave number = 1/0.152 x 10^{7}
 
=6.57 x 10^{-7} meter 

Answer 2

Answer:

656 nm

656 nmThe wavelength of the first line in the Balmer series of hydrogen spectrum is 656 nm. The wavelength of the last line in the Balmer series of hydrogen spectrum. As wavelength is cannot be negative. The wavelength of the last line in the Balmer series of hydrogen spectrum is 364 nm

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