Suppose the I.P. of a hypothetical hydrogen-like atom is 640 eV. The electron is in the orbit where its potential energy is -320 eV. The energy of the photon emitted when it jumps to the ground state

Question

Suppose the I.P. of a hypothetical hydrogen-like atom is 640 eV. The electron is in the orbit where its potential energy is -320 eV. The energy of the photon emitted when it jumps to the ground state is :-

TestHub · Accepted Answer

En = − 640/ n 2 eV and PE = 2E, so PE = − 1280/ n 2 = − 320 gives n 2 = 4 , n = 2 . E 2 ​ = − 160 eV and E 1 ​ = − 640 eV. Photon energy = 640 - 160 = 480 eV. Traps: 160 eV is the energy of the level; 320 eV is the PE. Answer: (A).

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