Question

27.In the quantum theory of hydrogen, as an electron drops to lower EPE orbit, it emits a photon with missing energy equal to

A. ΔQ, i.e., themal equilibrium

B. (Δm)c2

C. Δ(EPE)

Answer 1

Answer:

According to the quantum theory of a hydrogen atom, the energy difference between the two energy states is exactly equal to the difference between the their individual energies.

That means $\Delta$ E = E₂ - E₁ = h$\nu$, where $\nu$ is the fequency of the emitted photon.

Therefore, an electron drops from an exited energy state to the ground state, then the energy of the emitted photon is equal to the Δ(EPE)

Hence, option(C) is correct.