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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» Correct Answer - (a) `(9)/(8R)=103nm` (b) `(16)/(3R)=487 nm` 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`. |
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