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CBSE Class 12 Physics 2016 Outside Delhi Set 1 Paper

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Question : 8 of 26
Marks: +1, -0
Calculate the de-Broglie wavelength of the electron orbiting in the n=2n=2 state of hydrogen atom.
Solution:  
Formulae of kinetic energy and de-Broglie a wavelength
Calculation and Result
Kinetic Energy for the second state-
Ek=13.6 eVn2=13.6 eV4E_k = \frac{13.6\ \mathrm{eV}}{n^2} = \frac{13.6\ \mathrm{eV}}{4} =3.4×1.6×10−19 J= 3.4 \times 1.6 \times 10^{-19}\ \mathrm{J}
     de-Broglies wavelength   λ=  h2mEk\;\;\text{ de-Broglies wavelength }\; λ = \; \frac{h}{\sqrt{2 m E_k}}
  =6.63×10−342×9.1×10−31×3.4×1.6×10−19\;= \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.4 \times 1.6 \times 10^{-19}}}
  =0.067 nm\;= 0.067\ \mathrm{nm}
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