Question

A single electron revolves around a stationary nucleus of charge ze; (e = magnitude of charge on electron). A photon, of energy 47.2 eV, can excite this atom from its first excited state to its second excited state. The kinetic energy, K; the potential energy U, and the total energy, E, of this atom, in its ground state, are: 

(1) K = 340 eV, U = –170 eV, E = –340 eV 

(2) K = 340 eV, U = –680 eV, E = –340 eV 

(3) K = 85 eV, U = –170 eV, E = –85 eV 

(4) K = 170 eV, U = –340 eV, E = –170 eV

Answer 1

Correct option : (2) K = 340 eV, U = –680 eV, E = –340 eV

Explanation:

The energy, E_n, of this atom, in its n^th allowed state, is 

When this atom goes, from its first excited to its second excited state; n₁ = 2 and n₂ = 3. We are given that 

E₃ – E₂ = 47.2 eV