Verified Solution • 2025–26 Session moderate

QJEEM19Ph010c: JEE Main 2019 Physics - Class 12 JEEM19Ph010c Solution: Taking the wavelength of first Balmer line in hydrogen spectrum …

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 :

Multiple Choice Options

  • A 642.7 nm
  • B 488.9 nm
  • C 889.2 nm
  • D 388.9 nm

Step-by-Step Verified Solution

1λ1=R[122−132]{1 \over {{\lambda _1}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right]

1λ2=R[122−142]{1 \over {{\lambda _2}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right]

5λ136=12λ24×16{{5{\lambda _1}} \over {36}} = {{12{\lambda _2}} \over {4 \times 16}}

λ2=5×660×6436×12=489 nm{\lambda _2} = {{5 \times 660 \times 64} \over {36 \times 12}} = 489\,nm

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