Question

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

• A. 406 nm
• B. 192 nm
• C. 91 nm
• D. $9.1\times {10}^{-8}$ nm

Answer 1

The correct option is C

91 nm

$\frac{1}{\lambda }$=$R_n$$Z^2$($\frac{1}{n_1^2}$-$\frac{1}{n_2^2}$)

$\frac{1}{\lambda }$=$1.097\times {10}^7{m}^{-1}$($\frac{1}{1^2}$-$\frac{1}{\infty }$)

Therefore, $\lambda$=$91\times {10}^{-9}$m

We know ${10}^{-}9$=1nm so $\lambda$=91nm