1. Fourth lecture

2.  Radioactivity was the first nuclear phenomenon to be discovered, the credit for this discovery goes to the French physicist Henri Becquerel (1896).  Radioactivity was discovered almost by accident.

3. isotopesStable elements Exited states (stable) Ground states Disintegrate and emission radiation in period of time Stay stable for indefinite period of time There are two cases, one of them: by emission corpuscular A & Z change producing altogether nucleus, second of them: by emission E.M. W. the nucleus makes transition from a quantum state of higher energy to one of lower energy Always A & Z do not change

4.  The emission of α-particle from the nucleus (parent nucleus ) reduces the mass number by 4 units and the atomic number by 2 units, so that the new nucleus (daughter nucleus) Is formed as a result of the α-disintegration  The emission of β-particle (negative electrons )from the nucleus, in the modern view mean that one neutron transformed spontaneously into a proton and A remains unchanged and nucleus is changed into the nucleus due to β –disintegration:

5.  γ-rays very high energy electromagnetic radiation so that their wavelengths are very short, α and β disintegration is usually followed by the emission of one or more γ- rays,also various types of nuclear transmutation induced by artificial means.

6.  As early as 1913, two scientists F.Soddy and Fajans observed above changes in A and Z, They systemized their observations in the form of an empirical law “α-disintegration results in the reduction A by 4 units and its displacement in the periodic table by two steps to the left ;on other hand β- disintegration leaves A of the atom unchanged while its position in the periodic table is shifted by one step to the right”

8.  The British scientist W. Crookes observed that there was unknown substances present in the uranium salt which, when chemically precipitated with iron – hydroxide from the latter, carried away the entire radioactivity of the uranium salt. The uranium salt left after the chemical separation lost its radioactivity altogether. Crooks named unknown substance as UX and showed that its chemical properties were different from those of U. he then left the inactive U salt and the active precipitate for a few weeks and found that U salt had regained its radioactivity as before, while the precipitate had become completely inactive.

9.  Crooks showed that the above growth and decay of radioactivity of U and UX could be represented by two mathematical formulas. if we write the activity (i.e. the intensity of the emitted radioactive radiations) as A, then the decay of the radioactivity of UX could be expressed as

10.  On the other hand, the growth of the activity of U could be represented by the formula:   A 0 and λ are two constants ; t denotes the time after the separation of UX from U. Thus the decay and growth follow exponential laws. A 0 is the activity of UX immediately after its separation from U. after long time compared to 1/ λ, the activity of UX becomes zero while that of U comes back to the initial value A 0

13. 1. U disintegrates into UX 2. UX is new radioactive element, it disintegrates into some other element by radioactive disintegrations 3. What Crooks had measured initially as the activity of U was really the intensity of the radioactive radiation by the small amount of UX 4. Radiations emitted by U (α-particles) could not detected by Crooks because of easy absorbability 5. When UX separated from U, the entire radioactivity was naturally observed in precipitate containing UX

14. 6. U salt left over was found to be inactive 7. In course of time fresh amounts of UX were formed in U salt which regained the radioactivity that grew with time 8. On other hand, due to the continual disintegration of UX by emitting β particles,their number gradually decreased with time, as aresult the intensity of the radiations emitted by UX decreased with time

15.  The sum of activities of UX and U is constant equal to A 0  that the activity of a radioactive substances depends on its amount  The atoms of radioactive elements are transformed into other atoms by α &β disintegration  According to Rutherford &Soddy the rate of transformation of the radioactive atoms at any instant depends on the number of atoms present in the sample

16.  The number of atoms of P decreases with time, while that of Q increases with time. If N be the number of atoms of Pat any instant, then the rate of change dN/dt which is a measure of activity of P is proportional to N so that we can write   Where λ is the disintegration constant or decay constant, the negative sign is due to diminution in the number of atoms N with time integrating the above eq. we get

17.  As t=0 N=N 0 which is equal to the number of P atoms at the beginning of the experiment  We can easily get the result obtained by Crooks  Since radioactivity A is a measure of the intensity of the radiation emitted by a radioactive substances, which depends on the rate of formation of the radioactive atoms we can write

18.  The time which is spent so that the number of radioactive atoms reduce to half  When t= , N=N 0 /2,so  As λ increases  decreases and vice- versa  Every radioactive substances has its own λ and 

19.  In general the number is reduced by the factor 1/2 n after a time n , i.e. after n half-lives  So the graph variations of ln A with time will be straight line with negative slope –λ  After very long time from the beginning of experiment N becomes almost zero  And radioactivity of the sample disappears  In practical terms the radioactivity becomes negligibly small after 10 or 12 half-lives from the start of the experiment

22.  Radioactive element has half life time 20 days, how long will it take for 3/4 of the atoms originally present to disintegrate?  Solution:  T½ = 20 days = 0.693/λ  λ = 0.03465  N =N 0 e - λT  3/4 N 0 =N 0 e -(0.03465) t  3/4 = e -(0.03465) t  Ln 3/4 =0.03465 t  t =8.3025 day

23.  Note:  One curie = the no. of disintegrate / second  = 3.7×10 10 disintegrate / second

24.  What is the activity of one gm of Ra 266 whose half life time =1622 years Solution:  dN /dt = -λ N  wt. * 6.02 *10 23  A = λ N N =  mass no.  1*6.02*10 23  N =  226  0.693  λ = Ln 2 / T ½ =  1622*356*24*60*60  ; λ: Decay constant   0.693 1*6.02*10 23  Activity (A) = *  1622*356*24*60*60 226   = …………… disintegrate / second 

25.  What is the mass of one curie of polonium P 0 its half life time = 26.8 min.  Solution:   Activity = λ N   0.693  λ = = ……… sec -1  26.8*60  wt.*6.02*10 23  A = 4.31*10 -4 *  214  wt.*6.02*10 23  3.7*10 10 = 4.31*10 -4 *  214  Mass or weight (w) = 3.1*10 -8 gm

26.  Calculate the weight in grams of one curie of Co 60 it's half life time= 5.3 year.