Test on Nuclear Pysics Examination Time - 40 minutes Answer all questions in te spaces provided Tis wole test involves an imaginary element called Bedlum wic as te isotope notation sown below: 47 11 Bd 1. Bedlum decays by alpa decay to anoter element, tantrum, Tt : a. Write a balanced equation in isotope notation to describe tis decay. () 47 Bd 43 119 Tt 4 He Q 11 b. Explain wy alpa decay is te most likely decay for tis element () It s a very large nucleus Electrostatic force from all te protons will be stronger tan strong force. Sow tis decay on te grap of neutron number (N) against proton number (Z). (3) 3. N Bd 14 Tt Dv Z 11
In te box, write down te correct number of neutrons in a tantrum nucleus. (1) 4. Tantrum is unstable, and decays by beta minus decay to an isotope of drivelium, Dv. a. Write down a balanced equation for tis in isotope notation. () Tt Dv 43 43 0 119 10 1 e e b. Sow te decay from tantrum to drivelium on te grap. () 5. A sample of bedlum consists of 3 10 11 atoms. It emits alpa particles and its activity is worked out to be 1 10 8 Bq. a. Calculate te decay constant,. () = N/N = 1 10 8 3 10 11 = 3.3 10-4 s -1 b. Calculate te alf life of bedlum. () T ½ = 0.693 3.3 10-4 T ½ = 080 s c. Wat is te activity of te bedlum sample after 40 minutes? (3) lna = ln(1.0 10 8 ) + -(3.3 10-4 400) = 18.4 0.79 = 17.63 A = e 17.63 = 4.53 10 7 Bq 6. Bedlum also emits gamma rays as it decays to tantrum. a. Wat is a gamma ray? (1) A poton of very sort wavelengt electromagnetic radiation b. Explain ow gamma rays are produced in te daugter nucleus. () Nucleus is excited after a decay event. It loses energy by emitting a gamma poton.
7. Complete te energy level diagram. (4) Bd Tt 5 kev Ground State 8. Calculate te wavelengt of te gamma ray produced. (3) 5 kev = 5000 1.6 10-19 = 3.6 10-14 J = c/e = (6.63 10-34 3 10 8 ) 3.6 10-14 = 5.53 10-1 m 9. Drivelium decays to pandemonium (Pm) by anoter alpa decay. Pandemonium is a metastable nuclide. Wat is meant by te term metastable? (3) Metastable refers to a nuclide tat can remain excited for a long period of time. Tey release gamma potons randomly
10. Pandemonium is used as a source of gamma. However it decays by an alpa decay to bolocsium (Bs). a. How can a source of pure gamma be acieved? (1) Put a seet of paper over te source. b. A Geiger-Müller tube as a cross-sectional area of 1.5 10-4 m. Wen te source is in its box, te counter records a count rate of 35 counts per minute. It is ten placed 10 cm from te pandemonium source. Te counter sows a count rate of 1350 counts per minute. i. Calculate te corrected count rate in Bq. () Corrected rate = 1350 35 = 1315 min -1 A = 1315 60 = Bq ii. Calculate te gamma activity of te pandemonium source. (3) Work out te area of te spere: A = 4r = 4 0.1 = 0.16 m Total activity = (0.16 1315) 1.5 10-4 A = 1.01 10 6 min -1 = 16800 Bq c. Te Geiger-Müller tube is now moved to a distance of 45 cm from te source. Calculate te number of counts per minute tat is sown by te counter. () Te tube is 4.5 times furter away, so count rate = 1315 4.5 = 65 min -1 Displayed count rate = 65 +35 = 100 min -1
11. Alpa particles are fired at bedlum atoms tat are in a tin foil in a vacuum: a. Wat is likely to appen to te alpa particles? () Most will go straigt troug But a very small number will be deflected b. Wy is te foil placed in a vacuum? (1) Alpa is stopped by 1 cm air, or less energy lost by collisions 1. Te diagram sows an alpa particle at its closest point to te bedlum nucleus. Bedlum nucleus Alpa particle stationary at P Alpa coming in P r C Te kinetic energy of te alpa particles is initially 7.68 MeV.
c. Explain te energy canges tat occur as te alpa particle approaces te nucleus. () Kinetic energy is converted into potential energy. All te energy at P is potential. d. Calculate te speed of te alpa particle. (3) Mass of alpa particle = 4 1.67 10-7 kg = 6.68 10-7 kg. Energy = 7.68 10 6 1.6 10-19 = 1.3 10-1 J v = ( 1.3 10-1 ) 6.68 10-7 kg = 3.68 10 14 v = 1.9 10 7 m s -1 e. Sow tat te Ruterford radius is given by te equation: r c Ze mv 0 Ek = Ep. were Z is te proton number. (3) 1 mv 1 Ze 1 4Ze Ze r c r c 4 r 4 mv mv 0 0 0 r c = f. Te mass of a nucleon = 1.67 10-7 kg. Calculate te radius of closest approac of an alpa particle to te bedlum nucleus. () 11 (1.6 10-19 ) 8.85 10-1 6.68 10-7 3.68 10 14 r c = 4.5 10-14 m
13. Electron scattering gives a more reliable estimate of te size of te bedlum nucleus. Explain wit reference to fundamental forces wy electrons are used in preference to alpa particles. (3) Alpa particles feel te strong nuclear force and te electromagnetic force. Strong nuclear force is not well understood. Electrons are leptons Tey only interact by te electromagnetic force. 14. Electrons are g. Wat is meant by te de Broglie wavelengt? () Moving particles can be sown to ave wave properties de Broglie wavelengt is inversely proportional to te momentum. Sow tat. () mev E k = ev, and mv ev m mev m ev v v v m m m mev i. Calculate te de Broglie wavelengt of electrons tat are accelerated by a potential difference of 100 kv. () = 6.63 10-34 ( 9.1 10-31 1.6 10-19 100 10 3 ) 1/ = 3.9 10-1 m
15. Calculate te nuclear radius of Bedlum (r 0 = 1.4 10-15 m) () R = 1.4 10-15 (47) 1/3 R = 8.8 10-15 m 16. Explain weter te wavelengt you worked out in 4(c) will resolve te bedlum nucleus. () Te wavelengt of 3.9 10-1 m is too long to resolve te nucleus. Te minimum size resolved by tis wavelengt is 4 10-1 m Tis is 450 times bigger tan te Bd nucleus. 17. Te angle of te first minimum of te electron diffraction pattern is given by te equation: 0.61 sin R R is te nuclear radius and is te de Broglie wavelengt. An electron diffraction pattern for bedlum sows te angle of te first minimum is 30 degrees. Calculate te de Broglie wavelengt of te electrons tat are diffracted. Hence work out te accelerating potential difference. (4) = R sin /0.61 = (8.8 10-15 0.5) 0.61 = 7. 10-15 m V me V = (6.63 10-34 ) 9.1 10-31 1.6 10-19 (7. 10-15 ) V =.9 10 10 V Total = 70 marks