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 )

Question

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 ) (A)  91 nm  (B)  192 nm  (C)  406 nm  (D)  9.1 × 10-8 nm

Tardigrade Admin · Accepted Answer

(1/λ)= barvH= barRH[(1/n21)-(1/n22)] =1.097×107[(1/12)-(1/∞2)] ∴ λ=(1/1.097×107) m =9.11×10-8 m =91.1×10-9 m =91.1 nm (1 nm=10-9 m)

Q. The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state $1$ , would be (Rydberg constant $= 1.097 \times 1 0^{7} m^{- 1}$ )

$91 nm$

$192 nm$

$406 nm$

$9.1 \times 1 0^{- 8} nm$

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

91nm

192nm

406nm

9.1×10−8nm

λ1​=vˉH​=RˉH​[n12​1​−n22​1​] =1.097×107[121​−∞21​] ∴λ=1.097×1071​m =9.11×10−8m =91.1×10−9m =91.1nm(1nm=10−9m)

Q. The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state $1$ , would be (Rydberg constant $= 1.097 \times 1 0^{7} m^{- 1}$ )

$91 nm$

$192 nm$

$406 nm$

$9.1 \times 1 0^{- 8} nm$