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A 12.75 eV electron beam is used to excitea gaseous hydrogen atom at room temperature.Determine the wavelengths and the corresponding series of the lines emitted. |
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Answer» Solution :We know that energy in the n-th orbit of HYDROGEN atom, `E_n=(13.6)/(n^2)eV` Energy of anelectron in the excited state after absorbing energy 12.75 eV becomes -13.6 +12.75 =-0.85 eV Thus , `n^2=(13.6)/(E_n)=(-13.6)/(-0.85)=16` or, n=4 Therefore, the electron getsexcited to n=4 state. `therefore` Total number of wavelengths in the spectrum `=(n(n-1))/(2)=(4xx3)/(2)=6` `therefore` During transition from the fourth Bohr orbit to the ground state, the decrease in energy of the atom may occur in6 different ways. The POSSIBLE emission LINES are shown in  Emitted wavelength , for the jump from initial energy LEVEL `E_i` to final energy level `E_f`, `lambda_(if)=(hc)/(E_i-E_f)=(6.6xx10^(-34)xx3xx10^(8))/(E_i-E_f)` `=(19.8xx10^(-26))/(E_i-E_f)m` `=(19.8xx10^(-26))/((E_i-E_f)xx10^(-10))"Å"` Wavelength emitted for the transition from n=3 to n=2 , `lambda_(32) =6547.6 "Å"` Wavelength emitted for the transition from n=3 to n=1 , `lambda_(31) =1023.6 "Å"` Wavelength emitted for the transition from n=2 to n=1 `lambda_(21)=1213.2 "Å"` Wavelength emitted for thetransition fromn=4to n=3 , `lambda_(43)=19110 "Å"` Lyman series -`lambda_(21)(1213"Å") and lambda_(31)(1024"Å")` Balmer series `-lambda_(32)(6548"Å")` `"Paschen series" -lambda_(43)(19110"Å")` |
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