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Question

A hydrogen like atom is in its 5th excited state. This excited atom can make transition to the 1st excited state by successively emitting two photons of energy 10.2 eV and 17 eV respectively. Which of the following option(s) is/are correct?

A
This hydrogen like atom is Lithium.
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B
For 10.2 eV, photon transition is from n=6 to n=3.
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C
For 17 eV, photon transition is from n=4 to n=2.
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D
For 10.2 eV, photon transition is from n=2 to n=1.
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Solution

The correct options are
A This hydrogen like atom is Lithium.
B For 10.2 eV, photon transition is from n=6 to n=3.
The total energy emitted when the electron makes transition from 5th excited state to 1th excited state is given by
13.6eV Z2(14136)=10.2eV+17eV
13.6 Z2 (91)36=27.2

13.5×836 Z2=27.2

Z=272×36136×8Z=3

13.6×9 (1n2136)=10.2

By hit and trial we get, for n=3 equation is satisfied.
So for 10.2 eV photon, transition is made from n=6 to n=3.
and for the photon emission of 17 eV, transition is from n=3 to n=2.

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