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The wavelength of the electron emitted, when in a hydrogen atom, electron falls from infinity to stationary state 1, would be (Rydberg constant `= 1.097 xx 10^(7) m^(-1)`)
A. 91 nm
B. 192 nm
C. 406 nm
D. `9.1 xx 10^(-8) nm`

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Best answer
Correct Answer - A
`bar(v) = (1)/(lamda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
`= 1.097 xx 10^(7) m^(-1) ((1)/(1^(2)) - (1)/(oo^(2)))`
`= 1.097 xx 10^(7) m^(-1)`
or `lamda = (1)/(1.097 xx 10^(7)) m = 0.91 xx 10^(-7) m = 91 nm`

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