Suppose in certine condition only those transition are allowed to hydrogen atoms in which the principal quantum number a changes by`2` (a) Find the smaller wavelength emitted by hydrogen (b) list the wavelength emitted by hydrogen in the visible range `(380 nm to 780 nm)`

Answer

(a) `(9)/(8R)=103nm` (b) `(16)/(3R)=487 nm`

Explanation

Allowed energy of hyrogen atom are
`{:(E_(1)=-13.6 eV",",E_(2)= -3.4 eV",",E_(3)=-1.5eV",",E_(4)= -0.85 eV",",),(E_(5)=-0.54eV",",E_(6)=37 eV",",E_(7)=-27eV",",E_(8)=0.2eV",",):}`
For emission of the smallest wavelength we must wave `DeltaE_(max)`
`DeltaE_(max)=E_(3)-E_(1)`
`13.6-1.5=12.1eV`
corresponding `lambda=(hc)/(DeltaE)=(1242eV-nm)/(12.1)=103 nm`.
Any transition from `n=4` to `n=2` will be falling in the visible range
`Delta E-E_(4)-E_(2)=2.55eV`
corresponding `lambda=(hc)/(DeltaE)=487 nm`.