Half-life of a radioactive substance is T. At time t_1 activity of a radioactive substance is `R_1` and at time `t_2` it is `R_2`. Find the number of nuclei decayed in this interval of time.

Half-life of a radioactive substance is T. At time t_1 activity of a r

1.	Half-life of a radioactive substance is T. At time t_1 activity of a radioactive substance is `R_1` and at time `t_2` it is `R_2`. Find the number of nuclei decayed in this interval of time.
Answer» Correct Answer - A::B Half-life is given by `t_(1//2) = (1n2)/(lambda)` :. `lambda = (1n2)/(t_(1//2))=1n2/T` Activity `R= lambdaN` :. `N=R/lambda=(RT)/(1n2)` When activity is `R_1`, numbers of nuclei are `N_1 =(R_1T)/(1n2)` Similarly, `N_2 = (R_2T)/(1n2)` :. Numbers decayed `=N_1-N_2 = ((R_1-R_2)T)/(1n2)`

Discussion

No Comment Found

Related InterviewSolutions

Statement I : Threshold frequency supports wave nature of light. Statement II : Photo electrons are not emitted when incident light has frequency lesser than threshold frequency.A. If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement– I.B. If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement – I.C. If Statement - I is true but Statement - II is false.D. If Statement - I is false but Statement - II is true.
A radioactive substance is being consumed at a constant of `1 s^(-1)`. After what time will the number of radioactive nuclei becoem `100`. Initially, there were 200 nuceli present.A. 1 sB. `(1)/(ln(2))s`C. `ln (2) s`D. 2 s
Assuming the splitting of `U^235` nucleus liberates 200 MeV energy, find (a) the energy liberated in the fission of 1 kg of `U^235` and (b) the mass of the coal with calorific value of `30 kJ//g` which is equivalent to 1 kg of `U^235`.
The energy released by the fission of a single uranium nucleus is 200 MeV. The number of fission of uranium nucleus per second required to produce 16 MW of power is (Assume efficiency of the reactor is 50%)A. (a) `2xx10^6`B. (b) `2.5xx10^6`C. (c) `5xx10^6`D. (d) None of these
A radioactive isotope is being produced at a constant rate `dN//dt=R` in an experiment. The isotope has a half-life `t_(1//2)`.Show that after a time `t gtgt t_(1//2)`,the number of active nuclei will become constant. Find the value of this constant.A. (a) ATB. (b) `A/T1n2`C. (c) AT 1n2D. (d) `(AT)/(1n2)`
A radiaocatice isotope is being produced at a constant rate `X`. Half-life of the radioactive substance is `Y`. After some time, the number of radioactive nuceli become constant. The value of this constant is .A. `(XY)/(ln(2))`B. XYC. (XY) ln (2)D. `(X)/(Y)`
A radioactive element` X` converts into another stable elemnet `Y`. Half-life of `X` is `2h`. Initally, only `X` is present. After time `t`, the ratio of atoms of `X` and `Y` is found to be `1:4` Then `t` in hours is .A. 2B. 4C. between 4 and 6D. 4
A nucleus X, initially at rest , undergoes alpha dacay according to the equation , ` _(92)^(A) X rarr _(2)^(228)Y + a ` (a) Find the value of `A` and `Z` in the above process. (b) The alpha particle producted in the above process is found to move in a circular track of radius `0.11 m` in a uniform mmagnatic field of `3` Tesia find the energy (in MeV) released during the process and the binding energy of the patent nucleus X Given that `: m (gamma) = 228.03 u, m(_(0)^(1)) = 1.0029 u. ` `m (_(2)^(4) He) = 4.003 u , m (_(1)^(1) H) = 1.008 u `
The half - life of `^(215)` At is `100 mu , s.`The time taken for the radioactivity of a sample of `^(215)` At to dacay to `1//16^(th)` of its initialy value isA. `400 mu s `B. `63 mu s `C. `40 mu s `D. `300 mu s `
X-rays are produced by impinging_______on a target.A. `alpha` particlesB. protonsC. electronsD. neutrons

Half-life of a radioactive substance is T. At time t_1 activity of a radioactive substance is `R_1` and at time `t_2` it is `R_2`. Find the number of nuclei decayed in this interval of time.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment