The ground state energy of hydrogen atom is -13.6eV. If an electron makes a transition from an energy level -0.85 eV to -1.51 eV, calculate the wavelength of spectral line emitted. To which series of hydrogen spectrum does this wavelength belongs?

Explanation

Since energy of hydrogen atom is given as `E_(n)=(-13.6)/(n^(2))eV`
Ground state energy `E_(1)=-13.6 eV`
Now electron makes a transition from an energy level
`n_(p)=-0.85 eV" to "n_(p)=-1.51 eV`
`"Now, "E_(np)=(-13.6)/(n_(p)^(2))eV`
`n_(p)^(2)=(-13.6)/(-0.85)=16`
n=4
`"Again, "E_(np)=(-13.6)/(n_(p)^(2))eV`
`n_(p)^(2)=(-13.6)/(-1.51)=9`
`n_(q)=3`
`"Thus, we have transition from "n=4" to "n=3.`
Since transition corresponds to the transition of an electron from some higher energy state to an orbit having n=3. It is Paschen series of the hydrogen spectrum. Wavelength is given as
`(1)/(lamda)=R((1)/(3^(2))-(1)/(x^(2)))`
Where R is Rydberg's constant.
`R=1.097xx10^(7)m^(-1)(1)/(lamda)=1.097xx10^(7)[(1)/(9)-(1)/(16)]=1.097xx10^(7)[(7)/(9xx16)]`
`lamda=(9xx16)/(1.097xx10^(7)xx7)`
`=18.752xx10^(-7)m=1875 nm.`