NTA Abhyas JEE Main2020PhysicsAtomic PhysicsPractice
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 successively emitting two photons of energies 10.20 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy 4.25 eV and 5.95 eV respecti
Correct answer
3
Step-by-step solution
10.2 + 17 = 13.6 × Z 2 1 2 2 - 1 n 2 ;             4.25 + 5.95 = 13.6 × Z 2 1 3 2 - 1 n 2 Solving the above two equation we get, Z = 3 and n = 6 .