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 ( left(mathrm{r}_{1}-mathrm{r}_{1}right) ) is seven times the first Bohr radius. Identify the transition:

# 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 )

Solution

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