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KVPY2013PhysicsAtomic Physics

A singly ionized helium atom in an excited state (n=4) emits a photon of energy 2.6 eV . Given that the ground state energy of hydrogen atom is -13.6 eV , the energy ( E _ t ) and quantum number (n) of the resulting state are respectively,

Options

  1. AE_ t =-13.6 eV , n=1
  2. BE_ t =-6.0 eV , n =3
  3. CE_ t =-6.0 eV , n =2
  4. DE_ t =-13.6 eV , n =2

Correct answer

B. E_ t =-6.0 eV , n =3

Step-by-step solution

array l 2.6=13.6 z² [ 1 n² - 1 4² ] 2.6 13.6 4 =2² [ 1 n² - 1 4² ] array 2.6 13.6 4 = 1 n² - 1 16 n=3 now Energy array l E= 13.6 Z² n² e V - 13.6 4 9 e V=-6 e V array

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