Advertisements
Advertisements
प्रश्न
In a given sample, two radioisotopes, A and B, are initially present in the ration of 1 : 4. The half lives of A and B are respectively 100 years and 50 years. Find the time after which the amounts of A and B become equal.
उत्तर
Let NA be the concentration of A after tA time and NB be the concentration of B after tB time.
So, NA = N0e−λAtA
NB = 4N0e−λBtB (as N0B = 4N0A)
Now half-life of A is 100 years and B is 50 years.
`So lambda_A = (ln^2)/100 and lambda_B = (ln^2)/50`
Dividing we get
`(lambda_A)/(lambda_B)= 1/2 or lambda_B =2lambda_A`
Now let after t years NA = NB
So`(N_A)/(N_B) = e^(-lambdaA')/(4e^(lambdaB')`
`N_A =N_B`
`4e^(-lambdaB') =e^(-lambdaA') `
`4= e^-(lambda_A -lambda_B)`
`ln4 = -(+lambda_A - 2lambda_A)t (because lambda_B = 2lambda_A)`
`ln4 = lambda_At`
`t = (ln4)/(ln2) xx 100 (because lambda_A = (ln2)/100)`
= 200 years
संबंधित प्रश्न
(a) Write the basic nuclear process involved in the emission of β+ in a symbolic form, by a radioactive nucleus.
(b) In the reactions given below:
(i)`""_16^11C->_y^zB+x+v`
(ii)`""_6^12C+_6^12C->_a^20 Ne + _b^c He`
Find the values of x, y, and z and a, b and c.
Derive the mathematical expression for law of radioactive decay for a sample of a radioactive nucleus
The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive `""_6^14"C"` present with the stable carbon isotope `""_6^12"C"`. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of `""_6^14"C"` and the measured activity, the age of the specimen can be approximately estimated. This is the principle of `""_6^14"C"` dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
A source contains two phosphorous radio nuclides `""_15^32"P"` (T1/2 = 14.3d) and `""_15^33"P"` (T1/2 = 25.3d). Initially, 10% of the decays come from `""_15^33"P"`. How long one must wait until 90% do so?
Why is it experimentally found difficult to detect neutrinos in this process ?
Define the activity of a given radioactive substance. Write its S.I. unit.
A radioactive isotope is being produced at a constant rate dN/dt = R in an experiment. The isotope has a half-life t1/2. Show that after a time t >> t1/2 the number of active nuclei will become constant. Find the value of this constant.
The isotope \[\ce{^57Co}\] decays by electron capture to \[\ce{^57Fe}\] with a half-life of 272 d. The \[\ce{^57Fe}\] nucleus is produced in an excited state, and it almost instantaneously emits gamma rays.
(a) Find the mean lifetime and decay constant for 57Co.
(b) If the activity of a radiation source 57Co is 2.0 µCi now, how many 57Co nuclei does the source contain?
c) What will be the activity after one year?
A source contains two species of phosphorous nuclei, \[\ce{_15^32P}\] (T1/2 = 14.3 d) and \[\ce{_15^33P}\] (T1/2 = 25.3 d). At time t = 0, 90% of the decays are from \[\ce{_15^32P}\]. How much time has to elapse for only 15% of the decays to be from \[\ce{_15^32P}\]?
Two radioactive materials X1 and X2 have decay constants 10λ and λ respectively. If initially, they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will belie after a time.
