(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...
Question 1 Report
(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 the target by a potential difference of 80 A Calculate the:
(i) speed of the electron;
ii) threshold wavelength of the electron. [h=6.6 x 10\(^{-34}\) Js; e = 1.6 x 10\(^{-19}\) C; Me = 9.1 x 10\(^{-31}\)
d) An x-ray photon of frequency 4.5 x 10\(^{-18}\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\(^{-34}\) Js; Me = 9.1 x 10\(^{-31}\) kg ]
Answer Details
(a) Ionization potential
The ionization potential of an atom is the minimum potential difference (energy per unit charge) needed to remove the most loosely bound (outermost) electron completely from the atom in its ground state.
(b)(i) Three types of emission spectra:
- Continuous (band-free) spectrum.
- Line (atomic) spectrum.
- Band (molecular) spectrum.
(b)(ii) One source of each:
- Continuous spectrum: a glowing solid such as the filament of an electric lamp (incandescent solid).
- Line spectrum: a glowing (excited) gas of a single element, e.g. a sodium/mercury vapour lamp or hydrogen discharge tube.
- Band spectrum: glowing (excited) molecular gases or vapours, e.g. a discharge in a molecular gas.
(c) X-ray tube (electrons accelerated through \(V = 80\,\text{kV}\))
(i) Speed of the electron from \(eV = \tfrac{1}{2}mv^{2}\):
\[ v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2(1.6\times10^{-19})(8.0\times10^{4})}{9.1\times10^{-31}}} = \sqrt{2.81\times10^{16}} = 1.68\times10^{8}\,\text{ms}^{-1} \](ii) Threshold (minimum) wavelength from \(eV = \dfrac{hc}{\lambda}\):
\[ \lambda = \frac{hc}{eV} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{(1.6\times10^{-19})(8.0\times10^{4})} = \frac{1.98\times10^{-25}}{1.28\times10^{-14}} = 1.55\times10^{-11}\,\text{m} \](d) X-ray photon of frequency \(4.5\times10^{18}\,\text{Hz}\) absorbed by a stationary electron
Photon energy: \(E = hf = (6.6\times10^{-34})(4.5\times10^{18}) = 2.97\times10^{-15}\,\text{J}\).
All of it becomes electron kinetic energy, \(\tfrac{1}{2}mv^{2} = E\):
\[ v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2(2.97\times10^{-15})}{9.1\times10^{-31}}} = \sqrt{6.53\times10^{15}} = 8.08\times10^{7}\,\text{ms}^{-1} \]Download The App On Google Playstore
Everything you need to excel in your exams
