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.

Explanation

Energy of `n^(th)` energy level in hydrogen atom is given by
`E_n=(-13.6)/n^2` eV
Energy in ground state (n=1) is :
`E_1=(-13.6)/1^2`=-13.6eV
Energy in third excited state (n=4) is :
`E_4=(-13.6)/4^2`=-0.85 eV
Energy required to excite the hydrogen atoms from ground state (n=1) to the excited state (n=4) is :
`E=E_4-E_1=-0.85-(-13.6)`=12.75 eV
Hence the hydrogen atoms will be excited upto 4th energy level (n=4)
From Balmer.s formula , we have
`1/lambda=R[1/n_1^2-1/n_1^2]`
For the first line of Paschen series : `n_i`=4 and `n_f`=3
`1/lambda=R[1/3^2-1/4^2]=(7R)/144`
`rArr lambda=144/(7R)=144/(7xx1.09xx10^7)`=1,887 nm
For the first line of Balmer series , `n_i`=3 and `n_f` =2
`therefore 1/lambda=R[1/2^2-1/3^2]=(5R)/36`
`rArr lambda=36/(5R) =36/(5xx1.097xx10^7)`=656 nm