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  • The ground state energy of hydrogen atom is – 13·6 eV. If an electron makes a transition from an energy level – 1·51 eV to – 3·4 eV, calculate the wavelength of the spectral line emitted and name the series of hyd

The ground state energy of hydrogen atom is – 13·6 eV. If an electron makes a transition from an energy level – 1·51 eV to – 3·4 eV, calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.

 

Answers (1)

E_{2}-E_{1}=-1.51-(-3.4)

                    =1.89\; eV

\lambda =\frac{hc}{E_{2}-E_{1}}

       =\frac{6.6\times 10^{-34}\times 3\times 10^{8}}{1.89\times 1.6\times 10^{-19}}

        =6.548\times 10^{-7}m

        =6548\; A^{\circ}

E_{1}=\frac{-13.6}{h^{2}}

   h=\sqrt{\frac{13.6}{1.51}}=3

similarly for

E_{2},\; \; n=\sqrt{\frac{13.6}{3.4}}=2

The transition is from n=3 to n=2,

which belongs to Balmer series.

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Safeer PP

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