(a) Define ionization potential. (b)(i) State the three types of emission spectra. (ii) Name one...

PHYSICSWAEC 2013

(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 ]

Explanation

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\(^7 ms^{-1}\)

Back to WAEC 2013 QuestionsPrevious QuestionNext Question


Post an Explanation Or Report an Error

If you see any wrong question or answer, please leave a comment below and we'll take a look. If you doubt why the selected answer is correct or need additional more details? Please drop a comment or Contact us directly.

Your email address will not be published. Required fields are marked *

Don't want to keep filling in name and email whenever you make a contribution? Register or login to make contributing easier.