If the wavelengths of the first line of the Lyman series for the hydrogen atom is $$1210$$ Å then the wavelength of the first line of the Balmer series of the hydrogen spectrum is

Question

If the wavelengths of the first line of the Lyman series for the hydrogen atom is $$1210$$ Å  then the wavelength of the first line of the Balmer series of the hydrogen spectrum is

Infinity Learn · Accepted Answer

we know that $$1$$λ$$=R1n12$$−$$1n22$$  ForfirstlineofLymanseries, $$n1=1$$ and $$n2=2$$∣ $$1$$λ$$1=R11$$−$$14=3R4$$ or $$R=43$$λ$$1$$  ForfirstlienofBalmerseries $$n1=2$$ and $$n2=3$$ $$1$$λ$$2=R14$$−$$19=5R36=53643$$λ$$1$$ or $$1$$λ$$2=536$$×$$43$$×$$11210$$ or λ$$2=36$$×$$3$$×$$12105$$×$$4=6534$$Å

If the wavelengths of the first line of the Lyman series for the hydrogen atom is 1210 $Å$ then the wavelength of the first line of the Balmer series of the hydrogen spectrum is

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If the wavelengths of the first line of the Lyman series for the hydrogen atom is 1210 Å then the wavelength of the first line of the Balmer series of the hydrogen spectrum is

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If the wavelengths of the first line of the Lyman series for the hydrogen atom is 1210 $Å$ then the wavelength of the first line of the Balmer series of the hydrogen spectrum is