Verified Solution • 2025–26 Session moderate

QJEEM04Ch8d62: JEE Main 2004 Chemistry - Class 11 JEEM04Ch8d62 Solution: The wavelength of the radiation emitted when in a hydrogen atom …

The wavelength of the radiation emitted when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097 ×\times 107 m-1)

Multiple Choice Options

  • A 406 nm
  • B 192 nm
  • C 91 nm
  • D 9.1 ×\times 10-8 nm

Step-by-Step Verified Solution

We know Rydberg formula,

1λ=R×Z2(1n12−1n22){1 \over \lambda } = R \times {Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)

[ for hydrogen atom Z = 1 ]

     = 1.097×107(112−1∞2)1.097 \times {10^7}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)

∴\therefore λ\lambda = 11.097×107{1 \over {1.097 \times {{10}^7}}}

= 9.11 ×\times 10-8 m

= 91.1 ×\times 10-9 m

= 91.1 nm

[ as 1 nm = 10-9 m ]

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