- Tardigrade
- Question
- Physics
- A hydrogen-like atom (atomic number Z ) is in a higher excited state of quantum number n . By successively emitting two photons of energies 10.20eV and 17.00eV respectively this excited atom can transition to the first excited state. By successively emitting two photons of energies 4.25eV and 5.95eV respectively the atom can also transition to the second excited state. If the value of (Z/n) is (3/x) then find x . (ionization energy of hydrogen atom =13.6eV )
Q. A hydrogen-like atom (atomic number ) is in a higher excited state of quantum number . By successively emitting two photons of energies and respectively this excited atom can transition to the first excited state. By successively emitting two photons of energies and respectively the atom can also transition to the second excited state. If the value of is then find . (ionization energy of hydrogen atom )
Answer: 6
Solution: