An atom with one electron has ionization energy of 24 eV. An electron in this atom makes a transition from an excited energy level, where E = -15 eV, to the ground state. What is the wavelength of the emitted photon from this transition?

Step-by-step solution

The ionization energy of the atom is 24 eV , which means the energy of the ground state is E₁ = -24 eV . The energy of the excited state is given as E₂ = -15 eV . When the electron transitions from the excited state to the ground state, the energy of the emitted photon is: E = E₂ - E₁ E = -15 eV - (-24 eV ) = 9 eV The wavelength of the emitted photon is given by: = hc E Using hc 1242 eV nm : = 1242 eV nm 9 eV = 138 nm Answer: 138 nm