Taking the wavelength of first Balmer line in hydrogen spectrum ( n = 3 to n = 2 ) as 660 n m , the…

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 :
  1. 889.2 nm
  2. 488.9 nm
  3. 388.9 nm
  4. 642.7 nm

Solution

From Rydberg’s equation,
1λ=R1n2-1n22
1660×10-9=R122-132
1λ=R122-142
λ600×10-9=59×4×16×412
λ=488.9 nm

Asked in: JEE Main 2019 (09 Apr Shift 1)

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