(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 ]
Verified Correct Answer: .
Physical Principles & Numerical Solution
a) Ionization potential V is the potential energy required to remove an electron from an atom completely to infinity from ground state.
(b)(i) The three types of emission are
- band spectrum
- continuous spectrum and
- line spectrum.
(ii) Sources of each emmision spectrum are:
- Band spectrum
- from molecules or carbon(IV) oxides in a discharge tube.
- Continuous spectrum
- from sun, solids and liquids.
- Line spectrum
- from atoms in gases such as hydrogen or neon in a discharge tube.
(c)(i) eV =\(\frac{1}{2} mV^2\)
v = \(\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 80 \times 10^3}{9.1 \times 10^{-31}}}\)
= \(\sqrt{2.81 \times 10^{16}}\)
= 1.68 x 10\(^8 ms^{-1}\)
(ii) \(\pi = \frac{h}{mv} = \frac{6.6 \times 10^{-31}}{9.1 \times 10^{-31} \times 1.68 \times 10^8}\)
= 4.32 x 10\(^{-12}\)m
(d) hf = \(\frac{1}{2}mv^2\)
6.6 x 10\(^{-34} \times 4.5 \times 10^{18}\)
= \(\frac{1}{2} \times 9.1 \times 10^{-31} \times v^2\)
\(v^2 = \frac{2.97 \times 10^{-15}}{4.55 \times 10^{-3}}\)
\(v^2 = \sqrt{6.53 \times 10^{15}}\)
= 8.08 x 10\(^7 ms^{-1}\)
Source: Myschool Past Questions Archive (Record #53156)