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consider a hypothetical atom with single electron . In this atom, when an electron de-excites from energy level `n=xx` to `n=2,` wavelength (lambda) of the radiation emitted is given by lambda =`(Ax^(2))/(x^(2)-4)`(where`A` is a eonstant). Choose the correct alternatives.A. Least energetic photon emitted during such a transition will have wavelength `1.8A.`B. Most energetic photon emitted in such trasition will have Wavelength A.C. Ionization potential of the atom in its ground state is `(hc)/(1.8eA) `D. Ionization potential of the atom in its first excited statei (hc)/(eA) |
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Answer» Correct Answer - A::B::D `lambda=(Ax^(2))/(x^(2)-4)Rightarrow1/lambda=4/A(1/2^(2)-1/2^(2))` ` Rightarrow1/lambda_(max)=4/A(1/4-1/9)Rightarrowlambda_(max)=1.8A` For most energetic photon `x=oo Rightarrow 1/lambda_(min)=4/A(1/4-1/oo)Rightarrowlambda_(min)=A` for ionization potential in its ground state `1/lambda=4/A(1/1^(2)-1/oo^(2))Rightarrow lambda=0.25A` ` i.e eV=(hc)/lambda Rightarrow=(hc)/(exx0.25A)` for ionization potential in its first excited state`1/lambda=4/A(1/2^(2)-1/oo^(2)) Rightarrow =lambda=Ai.e eV=(hc)/lambda=RightarrowV=(hc)/(eA)` |
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