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CBSE Class 12 Physics 2014 Delhi Set 1 Paper

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Question : 19 of 30
Marks: +1, -0
A 12.5 eV12.5\ \mathrm{eV} electron beam is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited? Calculate the wavelength of the first member of Lyman and first member of Balmer series.
Solution:  
For Hydrogen atom
The 1  st   1^{\;\text{st }\;} excited energy =E2−E1=10.2 eV=E_2-E_1=10.2\ \mathrm{eV}
The 2  nd   2^{\;\text{nd }\;} excited energy =E3−E1=12.09 eV=E_3-E_1=12.09\ \mathrm{eV}
The 3  rd   3^{\;\text{rd }\;} excited energy =E4−E1=12.75 eV=E_4-E_1=12.75\ \mathrm{eV}
Hence, hydrogen atoms will be excited to 4  th   4^{\;\text{th }\;} energy level or 3  rd   3^{\;\text{rd }\;} excited state.
For Balmer series
  1λ  =R(  122−  132)\;\frac{1}{\lambda}\;=R \left(\;\frac{1}{2^2}-\;\frac{1}{3^2}\right)
  1λ  =R(  536)\;\frac{1}{\lambda}\;=R \left(\;\frac{5}{36}\right)
λ  =  365R\lambda\;=\;\frac{36}{5R}
λ  =  365×1.09×107\lambda\;=\;\frac{36}{5 \times 1.09 \times 10^7}
  =6.605×10−7 m\;=6.605 \times 10^{-7}\ \mathrm{m}
For Lyman series
  1λ  =R(  112−  122)\;\frac{1}{\lambda}\;=R \left(\;\frac{1}{1^2}-\;\frac{1}{2^2}\right)
  1λ  =R(  34)\;\frac{1}{\lambda}\;=R \left(\;\frac{3}{4}\right)
λ  =  43R\lambda\;=\;\frac{4}{3R}
  =1.285×10−7 m\;=1.285 \times 10^{-7}\ \mathrm{m}
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