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The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×107 m–1) [2004]
  • a)
    406 nm
  • b)
    192 nm
  • c)
    91 nm
  • d)
    9.1×10–8 nm
Correct answer is option 'C'. Can you explain this answer?
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The wavelength of the radiation emitted, when in a hydrogen atom elect...
TIPS/Formulae :
λ = 91.15 x 10 -9 m ≈ 91nm
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The wavelength of the radiation emitted when an electron in a hydrogen atom falls from infinity to stationary state 1 can be calculated using the Rydberg formula. The Rydberg formula is given by:

1/λ = R*(1/n1^2 - 1/n2^2)

Where λ is the wavelength of the emitted radiation, R is the Rydberg constant (1.097 x 10^7 m^-1), n1 is the initial state (infinity), and n2 is the final state (1).

Plugging in the values:

1/λ = 1.097 x 10^7 * (1/infinity^2 - 1/1^2)

Since the initial state is infinity, the term 1/infinity^2 is effectively 0.

1/λ = 1.097 x 10^7 * (0 - 1/1^2)

1/λ = 1.097 x 10^7 * (-1)

1/λ = -1.097 x 10^7

Taking the reciprocal of both sides:

λ = -1/1.097 x 10^7

λ = -0.912 nm

Therefore, the wavelength of the radiation emitted when the electron falls from infinity to stationary state 1 in a hydrogen atom is approximately 0.912 nm.
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The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×107 m–1) [2004]a)406 nmb)192 nmc)91 nmd)9.1×10–8 nmCorrect answer is option 'C'. Can you explain this answer?
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