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Q.

A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number ‘n’. This excited atom can make a transition to the first excited state by emitting a photon of energy 27.2eV. Alternatively the atom from the same excited state can make a transition to second excited state by emitting a photon of energy 10.20 eV. The value of n and z are given (Ionization energy of hydrogen atom is 13.6eV)

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a

n=3 and z=6

b

n=6 and z=3

c

n=4 and z=8

d

n=8 and z=4

answer is A.

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Detailed Solution

ΔE=13.6Z21n121n2227.2=13.6Z21221n2110.2=13.6Z21321n22
eq(1)eq(2), solving n=6→3
put eq (3) in eq(1) , solving Z=3

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