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Question

A hydrogen-like atom with the atomic number $$Z$$ is in the higher excited state of the quantum number $$'n'$$. This excited state atom can make a transition to the first excited state by successively emitted two photons of energy 10.2 eV and 17eV respectively. 
    Alternatively, the atom from the same excited state can make a transition to the 2nd excited state by emitting photons of energy 4.25 eV and 5.95 eV, respectively.  Determine $$Z$$ and $$n$$?


A
n=6, z=3
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B
n=4, z=3
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C
n=2, z=2
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D
n=3, z=6
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Solution

The correct option is D $$n = 6,\ z = 3$$
According to the given condition,

The electron makes a transition from state $$n_2$$ is the first excited state releasing $$ 10.2+17=27.2eV$$10.2=17=27.2eV  energy

$$27.2eV=13.6{ \left( z \right)  }^{ 2 }\left[ \cfrac { 1 }{ { \left( 2 \right)  }^{ 2 } } -\cfrac { 1 }{ { n }^{ 2 } }  \right] \Rightarrow 2={ z }^{ 2 }\left[ \cfrac { 1 }{ 4 } -\cfrac { 1 }{ { n }^{ 2 } }  \right] $$  ------ (1)

When the electron comes to the second excited state =4.25+5.95=10.2eV 
(n1=the energy is 4.2+5.95=10.2eV  4.25=5.95=10.2e 

$$10.2eV=13.6({ z }^{ 2 })\left[ \cfrac { 1 }{ { (3) }^{ 2 } } -\cfrac { 1 }{ { n }^{ 2 } }  \right] \Rightarrow 0.75={ 2 }^{ 2 }\left[ \cfrac { 1 }{ 9 } -\cfrac { 1 }{ { n }^{ 2 } }  \right] $$ ------- (2)

On solving we get, $$z=3,n=6$$.

Chemistry

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