A particle in a box with zero point energy E₁ = 4.0 eV,undergoes a transition from the third excited state to the second excited state emitting a photon.What is the energy of the photon? A) 4.0 eV B) 16 eV C) 20 eV D) 28 eV E) 26 eV

Question

Quizplus · Accepted Answer

The energy difference between the third excited state and the second excited state is given by: E32 = E₁ (3^-2 - 2^-2) = 4.0 eV*(5/9) = 2.22 eV The energy of the emitted photon is equal to this energy difference, given by: Ephoton = E32 = 2.22 eV Converting this to electron volts (eV), we get: Ephoton = 2.22 eV * (1.602 x 10^-19 J/eV) = 3.55 x 10^-19 J Rounding this to two significant figures, we get: Ephoton = 3.6 x 10^-19 J This is equivalent to 22.5 eV, which is closest to the answer C) 20 eV.

A Particle in a Box with Zero Point Energy E1

A Particle in a Box with Zero Point Energy E₁