Question

12(b) The table below gives some of the energy levels of a hydrogen atom (N=1,2,3,4,5,∞ with E in eV = -13.60,-3.39,-1.51,-0.85,-0.54,0.00). (i) Draw the energy level diagram for the atom; (ii) Determine the wavelength of the photon emitted when the atom goes from the energy state n=3 to the ground state. [h=6.6x10^-34Js, C=3.0x10^8ms^-1, e=1.6x10^-19C]

Accepted Answer

(i) The energy level diagram is drawn with horizontal lines at the given eV values (E1=-13.60eV, E2=-3.39eV, E3=-1.51eV, E4=-0.85eV, E5=-0.54eV, E∞=0.00eV), with the ground state (n=1) at the bottom and n=∞ at the top.
(ii) ΔE=hc/λ ⇒ λ=hc/ΔE. Transition from n=3 to n=1: ΔE=E3-E1=-1.51-(-13.60)=12.09eV. Converting to Joules: ΔE=12.09x1.6x10^-19 = 1.9344x10^-18J. λ = (6.63x10^-34 x 3x10^8)/(1.9344x10^-18) = 1.03x10^-7m.

WAEC Physics 2017 Theory — Question 32

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