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
B
n=4, z=3
C
n=2, z=2
D
n=3, z=6

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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