The wavelength of the radiations emitted when in a hydrogen atom electron falls from infinity to stationary state is ` : (R_H = 1. 097 xx10^7 m^(-1))` .

Options

A. `91 nm`
B. `192 nm`
C. `406 nm`
D. `9.1xx10^(-8)nm`

Answer

A

Explanation

According to Rydberg's formula
`barv_(H) - (1)/(lambda_(H)) = R_(H) ((1)/(n^(2))-(1)/(m^(2)))`
Thus,
`(1)/(lambda_(H)) = 1.097 xx 10^(7)m^(-1) ((1)/(1^(2))-(1)/(oo^(2)))`
`(1)/(lambda_(H)) - 1.097 xx 10^(7)m^(-1)`
`lambda_(H) = (1)/(1.097)xx10^(7)m^(-1)`
`= 0.9116xx10^(-7)m = 91.16 xx 10^(-9)m = 91.16 nm`