Calculate
(a) The wavelength and the frequency of the `H_(beta)` line of the Balmer series for hydrogen.
(b) Find the longest and shortest wavelengths in the Lyman series for hydrogen. In what region of the electromagnetic spectrum does this series lie ?
(c) Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in LYman series. What wavelength does this latter photon correspond to ?
(d) The wavelength of the first of Lyman series for hydrogen is identical to that of the second line of Balmer series for some hydrogen-like ion X. Find Z and energy for first four levels.

Explanation

`H_(beta)` line i.e. second line i.e. `n_(1)=2, n_(2)=2+2=4`
`(1)/(lambda)=R((1)/(2^(2))-(1)/(4^(2)))=R((1)/(4)-(1)/(16))=(3R)/(16)`
`lambda=(16)/(3R) =(16)/(3xx1.097xx10^(7))=4.86xx10^(-7) m=486 nm`
`v=(c )/(lambda)=(3xx10^(8))/(4.86xx10^(-7))=0.62xx10^(15) Hz`
(b) Longest wavelength i.e. energy difference should be minimum.
`n_(1), n_(2)=2`
` (1)/(lambda_(1))=R((1)/(1^(2))-(1)/(2^(2)))=(3R)/(4)`
`lambda _(1)=(4)/(3R)=(4)/(3xx1.097xx10^(7))`
`1.215xx10^(-7) m=121.5 nm`
Shortest wavelength i.e. energy difference should be maximum
`n_(1), n_(2)=oo`
`(1)/(lambda_(s))=R ((1)/(1^(2))-(1)/(oo))=R implies lambda_(s)=(1)/( R) =(1)/(1.097xx10^(7))`
`=0.91xx10^(-7) m=91 nm`
These wavelength correspond to ultraviolet region.
Balmer series ends at n=2. One more transition corresponds to n=2 to n=1
`(1)/(lambda)=R((1)/(1^(2))-(1)/(2^(2)))`
`lambda =121.5 nm`
(d) For second line of Balmer series, `n_(1) =2, n_(2)=2+2=4`
`(1)/(lambda)=R((1)/(1^(2))-(1)/(2^(2)))=Z^(2)R ((1)/(2^(2))-(1)/(4^(2)))`
`(3)/(4)=Z^(2).(3)/(16) implies Z^(2)=4 implies Z=2`
` E_(n)=-13.6(Z^(2))/(n^(2))=-(13.6(2)^(2))/(n^(2))=-(54.4)/(n^(2)) eV`
`n=1, E_(1)=-(54.4)/(1^(2))=-54.4 eV`
`n=2,E_(2)=-(54.4)/(2^(2))=-13.6 eV`
`n=3, K_(3)=-(54.4)/(3^(2))=-6.04 eV`
`n=4, E_(4)=-(54.4)/(4^(2))=-=3.4 eV`