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CBSE Class 12 Physics 2017 Delhi Set 2 Paper

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Question : 1 of 8
Marks: +1, -0
SECTION -B
Find the wavelength of the electron orbiting in the first excited state in hydrogen atom.
Solution:  
Calculation of wavelength of electron in first excited state
Radius of n  th   n^{\;\text{th }\;} orbit
r  =r0n2=0.53n2×År\;=r_0 n^2=0.53 n^2 \times \text{Å}
  =0.53×4Å\;=0.53 \times 4 \text{Å}
  =2.12Å\;=2.12 \text{Å} [∵[ \because Here n=2]n=2]
For an electron revolving in n  th   n^{\;\text{th }\;} orbit, according to de Broglie relation
  2πrn=nλ  ,   \;2 \pi r_n = n \lambda \;\text{, }\;
     For   1  st      excited state   n=2\;\;\text{ For }\; 1^{\;\text{st }\;} \;\text{ excited state }\; n=2
  2×3.14×2.12×10−10=2λ\;2 \times 3.14 \times 2.12 \times 10^{-10} = 2 \lambda
  λ=3.14×2.12×10−10\;\lambda = 3.14 \times 2.12 \times 10^{-10}
  =6.67Å\;=6.67 \text{Å}
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