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CBSE Class 12 Physics 2017 Outside Delhi Set 3 Paper

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Question : 4 of 8
Marks: +1, -0
The ground state energy of hydrogen atom is 13.6 eV. If an electron makes a transition from an energy level −1.51 eV-1.51\,\mathrm{eV} to −3.4 eV-3.4\,\mathrm{eV}, calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.
Solution:  
(a) Calculation of energy difference
(b) Formula
(c) Calculation of wavelength
(d) Name of the series of spectral lines
Energy difference =3.4 eV−1.51 eV=1.89=3.4\,\mathrm{eV}-1.51\,\mathrm{eV}=1.89 eV=3.024×10−19 J\mathrm{eV}=3.024 \times 10^{-19}\,\mathrm{J}
Energy =  hcλ=3.024×10−19 J=\;\frac{h c}{\lambda}=3.024 \times 10^{-19}\,\mathrm{J}
Wavelength =6.57×10−7 m=6.57 \times 10^{-7}\,\mathrm{m}
Series is Balmer series.
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