In a certain electronic transition ...
Question
In a certain electronic transition in the Hydrogen atom from an initial state i to a final state f, the difference in the orbit radius (r_1−r_1) is seven times the f
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In a certain electronic transition in the Hydrogen atom from an initial state i to a final state ( f, ) the difference in the orbit radius ( left(mathrm{r}_{1}-mathrm{r}_{1}right) ) is seven times the first Bohr radius. Identify the transition:
(A) ( 4 rightarrow 1 )
(B) ( 4 rightarrow 2 )
(C) ( 4 rightarrow 3 )
(D) ( 3 rightarrow 1 )

JEE/Engineering Exams
Chemistry
Solution
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In a certain electronic transition in the Hydrogen atom from an initial state i to a final state f, the difference in the orbit radius (r_1−r_1) is seven times the first Bohr radius. Identify the transition:
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We know theit radius of an atom is given
[
gamma=frac{r_{0} n^{2}}{2}
]
Auording to quation given that: ( gamma_{1}^{0}-x_{1}=7 x_{1}-0 )
( Rightarrow )
[
left(frac{v_{0}left(n_{1}right)^{2}}{1}+frac{gamma_{0}left(n_{1}right)^{2}}{1}right)=7 gamma_{0}
]
( therefore quad Z=1 quad ) for ( quad H ) - atom ( quad q ) s of lat Bohs orbit in ( H- ) atom ( =r_{0} )
[
begin{array}{l}
left.Rightarrow quad x_{0}left(n_{i}right)^{2}-left(n_{i}right)^{2}right)=7 x_{0}
Rightarrow quad h_{i} cdot n_{i} quadleft(n_{i}+n_{i}right)=quad 1+7
end{array}
]
now rquate both equation
[
because 7 quad n_{1}-n_{1}=1 quad+(1)
]
solut it:
[
begin{array}{l}
n+gamma f=7=(n)
n_{i}=4
end{array}
]
[
Rightarrow n_{7}=3
]

This is 30 becouse tronition is from hight state to burex Rtale

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