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.

Options

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)

Answer

A, B, D

Explanation

`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)`