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

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

The energy, E_n , of this atom, in its n^th allowed state, is E_n = \(-\frac{(13.6) z^2}{n^2} eV\)

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

For the ground state of atom, n = 1. Therefore, 

E = – 340 eV; K = |E| = 340 eV 

and U = 2E = – 680 eV