If the electron jumps to the ground state from the third excited state in hydrogen atom, calculate the wavelength and energy of corresponding emitted photon in eV. Rydberg's constant `=1.097xx210^(7)m^(-1)`,
`c=3xx10^(8)m s^(-1)=6.62xx10^(-34)Js`

Explanation

`R=1.097xx10^(7)m^(-1)`
`c=3xx10^(8)m s^(-1)`
`h=6.62xx10^(-34)Js`
`n_(i)=4` [ Third excited state =3+1]
`n_(f)=1`
`(1)/(lambda)=R[(1)/(n_(f)^(2))-(1)/(n_(i)^(2))]`
`=1.097xx10^(7)[(1)/((1)^(2))-(1)/((4)^(2))]`
`=1.097xx10^(7)[(15)/(16)]`
`lambda=(16xx10^(-7))/(1.097xx15)`
`lambda=9.72xx10^(-8)m`
`E=(hc)/(lambda)=(6.62xx10^(-34)xxx3xx10^(8))/(9.72xx10^(-8))`
`=2.04xx10^(-18)J`
`=(2.04xx10^(-18))/(1.6xx10^(-19))eV`
`:.E=1.277eV`s