A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. Thisexcited atom can make a transition to the first excited state by successively emitting two photonsof energies 10.20 and 17.00 eV, respectively. Alternatively, the atoms from the same excited statecan make a transition to the second excited state by successively emitting two photons of energies4.25 and 5.95 eV, respectively.a) The value of ‘n’ and ‘Z’ respectively are –b) How much energy will be emitted when an electron in this atom moves from (n + 1) toground state?
Question
A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. Thisexcited atom can make a transition to the first excited state by successively emitting two photonsof energies 10.20 and 17.00 eV, respectively. Alternatively, the atoms from the same excited statecan make a transition to the second excited state by successively emitting two photons of energies4.25 and 5.95 eV, respectively.a) The value of ‘n’ and ‘Z’ respectively are –b) How much energy will be emitted when an electron in this atom moves from (n + 1) toground state?
Solution
The problem involves the use of the Rydberg formula for hydrogen-like atoms:
1/λ = RZ^2(1/n1^2 - 1/n2^2)
where R is the Rydberg constant (approximately 13.6 eV), Z is the atomic number, n1 and n2 are the principal quantum numbers of the two states involved in the transition, and λ is the wavelength of the emitted photon.
a) The energy of the emitted photon is given by the difference in energy between the two states, which is given by the Rydberg formula. We have two sets of transitions, so we can set up two equations:
10.20 eV = 13.6 Z^2 (1/2^2 - 1/n^2) 17.00 eV = 13.6 Z^2 (1/1^2 - 1/2^2)
and
4.25 eV = 13.6 Z^2 (1/3^2 - 1/n^2) 5.95 eV = 13.6 Z^2 (1/2^2 - 1/3^2)
Solving these equations simultaneously gives Z = 1 and n = 3.
b) The energy emitted when an electron moves from the (n + 1) state to the ground state is given by the Rydberg formula:
E = 13.6 Z^2 (1/1^2 - 1/4^2) = 12.75 eV.
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