Natural radio activity and the laws of radioactive transformation

Fourth lecture

Discovery of radio activity

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.

Comparison between stable elements and isotopes

isotopes Stable 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

Radioactive disintegration

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:

YX AZ

AZ

42

YAZ42

XAZ

XAZYAZ 1

YX A1Z

AZ

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.

displacement law

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”

Growth and decay of radioactivity

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.

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

tAAX exp0

On the other hand , the growth of the activity of U could be represented by the formula:

A0 and λ are two constants ; t denotes the time

after the separation of UX from U . Thus the decay and growth follow exponential laws. A0 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 A0

tAAU exp10

The above variations of the activities of U , UX with time can be understood

1. U disintegrates into UX2. 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

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

Note: The experiments of Crooks showed

The sum of activities of UX and U is constant equal to A0

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

Consider the disintegration of P to Q

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

tdt

dN

Ndt

dN

tNN exp0

As t=0 N=N0 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

tAtNNdt

dNA expexp 00

Half life of disintegration

The time which is spent so that the number of radioactive atoms reduce to half

When t= , N=N0/2 ,so

As λ increases decreases and vice- versa

Every radioactive substances has its own λ and

693.02ln

)exp(2 00

NN

In general the number is reduced by the factor 1/2n 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

0lnln AA

Problems 

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 =N0 e- λT 3/4 N0 =N0 e-(0.03465) t

3/4 = e-(0.03465) t Ln 3/4 =0.03465 t t =8.3025 day

note

Note: One curie = the no. of

disintegrate / second = 3.7×1010

disintegrate / second

What is the activity of one gm of Ra266 whose half life time =1622 years Solution:

dN /dt = -λ N wt. * 6.02 *1023 A = λ N N = mass no. 1*6.02*1023 N = 226 0.693 λ = Ln 2 / T ½ = 1622*356*24*60*60 ; λ: Decay constant 0.693

1*6.02*1023

Activity (A) = * 1622*356*24*60*60 226   = …………… disintegrate / second  

What is the mass of one curie of polonium P0 its half life time = 26.8 min.

Solution: Activity = λ N   0.693 λ = = ……… sec-1

26.8*60 wt.*6.02*1023 A = 4.31*10-4 * 214 wt.*6.02*1023

3.7*1010 = 4.31*10-4 * 214 Mass or weight (w) = 3.1*10-8 gm

Calculate the weight in grams of one curie of Co60 it's half life time= 5.3 year.