Question 12
Part II. Candidates are expected to answer any three questions from this part.
- State the terms:
- transmutation as it relates to radioactivity;
- stopping potential
A nucleus C, formed artificially from A and B, is radioactive and quickly decays to another nucleus E as indicated in the nuclear equations above.
Determine the values of p, q, r, and s.
- A certain metal of work function 1.6 is irradiated with an ultra-violet light of wavelength 3.6 x 10-7 m. Calculate the maximum:
- kinetic energy of an ejected electron in joules;
- speed of an emitted electron.
( 1 = 1.6 x 10-19 J, c = 3.0 x 108 ms-1, me = 9.1 x 10-31kg, h = 6.6 x 10-34 )
- If the source of the ultra-violet light in 12(c) is moved away from the surface of the metal, state its effect on the maximum speed of the ejected electron.
Observation
Part (a): Few candidates succeeded in defining transmutation and fewer number of candidates defined stopping potential correctly. Virtually all the candidates did not give additional information.
Part (b): Performance in this part was above average. A good number of the candidates gave the correct values of p,q,r and s. However, many of them forgot to indicate the charge and mass of the proton and beta and so lost some marks.
Part (c): Seems a familiar terrain for the candidates. Many candidates scored reasonably high marks in this part.
Part (d): This part was avoided by most respondent. Performance was poor.
The expected answers are:
(a)(i) Transmutation
The change of a nuclide into another one by radioactive disintegration/nuclear bombardment with particles .
Valid additional information
e.g. conversion of Hg to Au / nuclear equation
(ii) Stopping potential
The negative potential of the anode of a photocell that would just prevent electrons from reaching the anode.
Valid additional information
e.g. eV = ½ mv2 or Vs = hf - hfo
(b) + + P + 1 = 23 + 2 = 25
P = 24
q + 1 = 11 + 1 = 12
q = 11 + r+o = 24
r = 24
s - 1 = 11
s = 12
(c)(i) K.Emax = E - Wo
E = - Wo
= - 1.6 x 1.6 x 10-19
= 2.94 x J
(ii) K.E = ½ mv2max
OR
Vmax =
=
= 8.04 x 105 ms-1
(d) No effect