Question

15. (a) Define ionization potential. (b)(i) State the three types of emission spectra. (ii) Name one source each which produces each of the spectra stated in (b)(i). (c) In an X-ray tube, electrons are accelerated to the target by a potential difference of 80kV. Calculate the: (i) speed of the electron; (ii) threshold wavelength of the electron. [h=6.6×10⁻³⁴Js; e=1.6×10⁻¹⁹C; Me=9.1×10⁻³¹kg]
(d) An x-ray photon of frequency 4.5×10¹⁸ Hz strikes an electron, assumed to be at rest. If the electron absorbs all the photon energy, calculate the speed acquired by the electron.

Accepted Answer

(a) Ionization potential is the potential absorbed by an atom on its ground state which just removes an electron completely from the atom.

(b)(i) Types of emission spectra: (i) line spectrum, (ii) band spectrum, (iii) continuous spectrum.
(ii) Sources: Line spectrum — atoms in gases (e.g. hydrogen and neon at low pressure in a discharge tube). Band spectrum — from molecules, e.g. CO₂ in a discharged tube. Continuous spectrum — from the sun or from solids and liquids.

(c) Given: Potential V=80kV=80×10³V, mass of electron Me=9.1×10⁻³¹kg, charge q on electron=1.6×10⁻¹⁹C, Planck's constant h=6.6×10⁻³⁴Js.
(i) qV = ½mv², so v² = 2qV/m = (2×1.6×10⁻¹⁹×80×10³)/(9.1×10⁻³¹) = 2.8×10¹⁶; v = √(2.8×10¹⁶) = 1.68×10⁸ ms⁻¹.
(ii) Threshold wavelength λ = h/p, where p = mass × velocity: λ = h/(Me×v) = (6.6×10⁻³⁴)/(9.1×10⁻³¹×1.68×10⁸) = 4.32×10⁻¹² m.

(d) Photon frequency f=4.5×10¹⁸Hz. Since the electron absorbs all the energy of the photon, energy of photon = kinetic energy of the electron: hf = ½mv². v² = 2hf/m = (2×6.6×10⁻³⁴×4.5×10¹⁸)/(9.1×10⁻³¹) = 6.53×10¹⁵; v = √(6.53×10¹⁵) = 8.08×10⁷ ms⁻¹.

WAEC Physics 2013 Theory — Question 15

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