A `12.5 eV` electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelengths and the corresponding series of the lines emitted.

Explanation

The energy absorbed is given by `Delta E = Rhc ((1)/(1^(2)) - (1)/(n_(2)^(n)))`
Given, `Delta E = 12.5 eV Rhc = 13.6 e V `
`12.5 eV = 13.6 eV (1 - (1)/(n_(2)^(2)))`
Following transitions are permitted
`n_(3) overset(lambda_(1))to n_(1)`
`n_(2) overset(lambda_(2))to n_(1) and n_(3) overset(lambda_(3))to n_(2)`
= `(6.63xx10^(-34) xx3 xx 10^(8))/(1.6 xx10^(-19)) = 12.43 xx10^(-7) Delta E` in joule
For `n_(3) overset (lambda_(1))(to) n_(1) DeltaE_(1) = 12.5 eV`
`lambda_(1) = (12.43)/(12.5) xx10^(-7) m = 9.944 xx10^(-8) m`
` = 994A^(@) rArr ` Lyman series
Lymen Series
`For n_(2)overset(lambda_(2))(to) n_(1) DeltaE_(2) = 3.1 + 13.6 = 10.2 e V`
`lambda_(2) = (12.43)/(10.2) xx10^(-7) m = 1.219 xx 10^(-7) m = 1219 A^(2)`
Balmer Series
For ` n_(3)overset(lambda_(3))(to) n_(2) DeltaE_(2) = (-1.5 + 3.4) = 1.9 eV`
` lambda_(3)= (12.43 xx10^(-7) )/(1.9) = 6.542xx10^(-7) m=6542A^(@)`,
Alternate Solution:
Energy of `e^(-)`beam = 12 5 eV
Energy of hydrogen atom =` ((-13.6) + 12.5) eV = - 1.1 eV`
Now, ` E_(1)= - 13.6 e V`
`E_(2)= (- 13.6)/(2^(2)) = (-13.6)/(4) = - 3.4 e V`
`E_(3)= (- 13.6)/(3^(2)) = - 1.5eV`
`E_(4)= (- 13.6)/(4^(2)) = - 8.5eV`
`rArr ` H will be excited up to `3^(nd)` energy level.
Also, `(1)/(lambda) = R[(1)/(n_(1)^(2) )- (1)/(n_(2)^2)]`
where, ` R = 1 .097 xx10^(-7) m^(-1)`
For Lyman series : ` n_(1) = 1 , n_(2) = 2` (for I member
`(1)/(lambda) = 1.097xx[(1)/(1) - (1)/(4)] = 1.097 xx10^(7) xx(3)/(4)`
`rArr lambda = (4)/(3xx1.097 xx10^(7)) = 1.215 xx10^(-7) = 121.5nm therefore ` UV region
For Balmer series : `2, n_(2) = 3`
`(1)/(lambda) = 1.097 xx10^(7) [(1)/(4) - (1)/(9) ] rArr (1)/(lambda )= 1.097xx10^(7) xx((1)/(4)- (1)/(9))`
`lambda = (36)/(5xx(1.097 xx10^(7)) )rArr lambda = 656 nm rArr ` Visible region.