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Question

A Hydrogen like atom of atomic number Z is in an excited state 2n. It can emit a maximum energy photon of 204 eV. If  it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Identity correct statement(s). 
  1. Z = 4 
  2. n = 4
  3. n = 2
  4. Ground state energy of this atom is -217.6 eV


Solution

The correct options are
A Z = 4 
C n = 2
D Ground state energy of this atom is -217.6 eV
204=13.6z2(114n2)         ...(1)
40.8=13.6z2(1n214n2)     ...(2)
Dividing (1) by (2)5=4n213n=2
using (1)
20413.6=z21516z=4
energy of ground state is,
E=13.6 z2=13.6×16=217.6 eV

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