A `12.9 eV` beam of electrons is used to bombard gaseous hydrogen atom at room temperature. Up to which energy level the hydrogen atoms would be excited?
Calculate the wavelength of the first member of Paschen series and first member of Balmer series.

Answer

`n=4; 18750Å; 6547Å`

Explanation

Total energy of gaseous hydrogen
`=(12.9-13.6)eV=-0.7eV`
`:. -13.6/(n^(2))=-0.7, n^(2)=13.6/0.7 =19.4`
`n=sqrt(19.5)lt5`
`:. n=4`
for the first member of Paschen series,
`E=E_(4)-E_(3)=-13.6/(4^(2))-(-13.6/(3^(2)))`
`=-0.85+1.51=0.66eV`
`(hc)/lambda=0.66xx1.6xx10^(-19)J`
`lambda=(hc)/(0.66xx1.6xx10^(-19))m`
`=(6.6xx10^(-34)xx3xx10^(8)xx10^(10)Å)/(0.66xx1.6+10^(-19))`
`=18750Å`
and for first member of Balmer series
`E=E_(3)-E_(2)=-13.6/(3^(2))-(-13.6/(2^(2)))ev`
`=-1.51+3.4=1.89ev`
`(hc)/lambda=1.89xx1.6xx10^(-19)J`
`lambda=(hc)/(1.89xx1.6xx10^(-19))`
`=(6.6xx10^(-34)xx3xx10^(8))/(1.89xx1.6xx10^(-19))m`
`=6.5476xx10^(-7)m=6547Å`