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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 701. |
A radioactive nucleus can decay by two differnet processess. The mean value period for the first process is `t_(1)` and that the second process is `t_(2)` .The effective mean value period for the two processes is .A. `(t_(1) + t_(2))/(2)`B. `t_(1) + t_(2)`C. `sqrt t_(1)t_(2)`D. `(t_(1) + t_(2))/(t_(1) + t_2)` |
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Answer» Correct Answer - d Let the decay constant for the first and second processes be `lambda_(1)` and `lambda_(2)` and the effective decay constant for the comined process be `lambda`. Then, `lambda_(1)=(log_(e) 2)/(t_(1)) , lambda_(2)=(log_(e) 2)/(t_(2))` and` lambda=(log_(e) 2)/(t)` Now, the probability for decay through first process in a small time interval dt is `lambda_(1) dt` and the probability for decay through second process in the same time interval dt is `lambda_(2) dt` . The probability for decay by the combined process in the same time interval dt is `lambda_(1) dt + lambda_(2) dt`. But this is also equal to `lambda dt`. ` :. lambda dt = lambda_(1) dt + lambda_(2) dt` `:. lambda=lambda_(1) + lambda_(2)` or `(log_(e) 2)/(t)=(log_(e) 2)/(t_(1))+ (log_(e) 2)/(t_(2))` or `(1)/(t) =(1)/(t_(1))+(1)/(t_(2))` or `t=(t_(1)t_(2))/(t_(1) +t_(2))`. |
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| 702. |
The radioactivity of a sample is `R_(1)` at a time `T_(1)` and `R_(2)` at time `T_(2)`. If the half-life of the specimen is T, the number of atoms that have disintegrated in the time `(T_(2) -T_(1))` is proporational toA. `R_(1) T_(1)=R_(2) T_(2)`B. `R_(1) - R_(2)`C. `(R_(1)-R_(2))/(T)`D. `(R_(1) -R_(2)) T` |
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Answer» Correct Answer - d `R_(1) =N_(1) lambda , R_(2)=N_(2) lambda` Also, `T=log_(e).(2)/(lambda)` or `lambda=log_(e). (2)/(T)` `:. R_(1) -R_(2) =(N_(1) -N_(2)) lambda` `=(N_(1) -N_(2))log_(e).(2)/(T)` `:. (N_(1) -N_(0))=((R_(1) -R_(2))T)/(log_(e) 2)` i.e., `(N_(1) -N_(2)) prop (R_(1) -R_(2))T`. |
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| 703. |
Assume that a neutron breaks into a proton and an electron . The energy reased during this process is (mass of neutron `= 1.6725 xx 10^(-27) kg` mass of proton `= 1.6725 xx 10^(-27) kg` mass of electron `= 9 xx 10^(-31) kg )` |
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Answer» Correct Answer - `0.73 MeV` Mass defect of the process is given by `Delta m=["Mass of neutron" -("mass of proton + mass of electron")]` ` =[1.6747 xx 10^(-27) -(1.6725 xx10^(-27) + 9 xx 10^(-31))]` `=0.0013 xx 10^(-27) kg` According to mass -energy relatinship, Energy released = `Delta m c^(2)` `E=(0.0013 xx10^(-27)) xx (3 xx 10^(8))^(2)` `(1.17 xx 10^(-3))/(l1.6 xx 10^(-19))` =0.73 xx10^(6) eV=0.73 MeV`. |
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| 704. |
In a fusion process a proton and neutron combine to give a deuterium nucleus. If `m_(o)` and `m_(p)` be the mass of neutron and proton respectively the mass of deuterium nucleus isA. equal to `m_(o)+m_(p)`B. more than `m_(o)+m_(p)`C. less than `m_(o)+m_(p)`D. can be less than or more than `(m_(o)+m_(p))` |
| Answer» Correct Answer - C | |
| 705. |
A nucleus, absorbing a neutron, emits an electron to go over to neptunium which on futher emitting an electron goes over to plutonium. How would you represent the resulting plutonium? |
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Answer» The successive processess may be expressed as .`_(92)U^(238) overset (._(0)n^(1)) rarr. _(92)U^(239) overset (beta (electron)) rarr. _(93)Np^(239)` overset(beta(electron)) rarr ._(94)Pu^(239)` Thus, the resulting plutonium has atomic number 94 and mass number `239` and is expressed as `._(94)Pu^(239)`. |
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| 706. |
In the Radioactive transformation `R overset(alpha)rarrXoverset(beta)rarrYoverset(beta)rarrZ`, the nucllii `R` and `Z` areA. IsotopesB. isobarsC. isomersD. Isotones |
| Answer» Correct Answer - A | |
| 707. |
Name the reaction which takes place when a slow neutron beam strikes `._(92)^(235)U` nuclei. Write the nuclear reaction involved. |
| Answer» `._(92)^(235)U+._(o)n^(1)rarr._(92)^(236)Urarr._(56)^(141)Ba+._(36)^(92)Ke+3._(o)n^(1)+"energy"` | |
| 708. |
Define decay constant |
| Answer» Decay constant of a radioactive element is the reciprocal of the time during which the number of atoms left in the sample reduces to `1//e` times the original number of atoms in the sample. | |
| 709. |
A raioactive sample contains `2.2 mg` of pure `._(6)^(11)C` which has half-life period of `1224` seconds. Calculate (i) the number of atoms present initially (ii) the activity when `5 mu g` of the sample will be left. |
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Answer» (i) Number of atoms present intially `=(6.023xx10^(23)xx2.2xx10^(-3))/(11)=1.2xx10^(20)"atoms"` (ii) No. of atoms present in `5 mu g` of the sample `N=(6.023xx10^(23)xx5xx10^(-6))/(11)=2.74xx10^(17)"atoms"` Activity of the sample `=lambdaN` `=(0.693)/(T_(1//2))xxN=(0.693xx2.74xx10^(17))/(1224)=1.55xx10^(14) "disintergration"//"seconds."` |
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| 710. |
Mark out the coreect statemnet (s).A. For an exothermic reaction, if `Q` value is` +12.56MeV` and the`KE` of incident particcle is `2.44 MeV`, then the totla `KE` of products of reaction is `15.00 MeV`.B. For an exothermic reaction, if `Q` value is `+12.56MeV` and the KE of incident particcle is `2.44 MeV`, then the totla `KE` of products of reaction is `12.56 MeV`.C. For an endothermic reaction, if we give the energy equal `|Q|` value of reaction, then the reaction will be carried out.D. For an exothermic reaction, the BE per nucleon of products should be greater than the BE per nucleon of reactants. |
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Answer» Correct Answer - a,d For an exothermic reaction `X + x rarr Y+y` If `K_(1)` is the kinetic energy of incident paricles x, then from energy conservation, `K_(i) (m_(x)+M_(x))c^(2)=k_(Y)+K_(y)+(M_(Y)+m_(y))C^(2)` `K_(Y)+K_(y)=k_(i)+(m_(x)+N_(X)-M_(Y)-m_(y))c^(2))` `K_(Y) +K_(y)=K_(i) +Q` In any exothermic reaction, mass of the products is less than the mass of reactants, i.e., in products, the nucleon are more tightly bound and hence have greater BE per nucleon as compared to BE per nucleon or reactants. For endohtermic reaction to be carried out, minimum energy given to the reactant must be greater than `|Q|` value. |
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| 711. |
Mark out the coreect statemnet (s).A. In alpha decay,the energy released is shared between alpha particle and daughter nulceus in the form of kinetic energy and share of alpha particle is more than that of the daughter nucleus.B. In beta decay,the energy released is in the form of kinetic energy of beta particles.C. In beta minus decay,the energy released is shared between electron and antineutrinoD. In gamma decay,the energy released is in the form of energy carried by photons termed as gamma rays. |
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Answer» Correct Answer - a,c,d All the statements are very conceptual statements related to different decays. |
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| 712. |
1.00 kg of `.^(235)U` undergoes fission process. If energy released per event is `200 MeV`, then the total energy released isA. `5.12 xx 10^(24)MeV`B. `6.02 xx 10^(23)MeV`C. `5.12 xx 10^(16)MeV`D. `6.02 xx 10^(6)MeV` |
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Answer» Correct Answer - c The number of nuclei in `1 kg .^(235)U` is `N=(N_(A))/(235)xx(1xx10^(23))` `N=(6.023 xx 10^(23))/(235) xx 10^(3) =2.56 xx 10^(24) "nuclei"` Total energy released is `E =N xx 200 MeV` `=5.12 xx10^(26)MeV`. |
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| 713. |
Calculate the total energy released if `1.0 kg` of `.^(235)U` undergoes fission, taking the disintergration energy per event to be `Q=208 MeV` (a more accurate value than the estimate given previously). |
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Answer» We need to know the number of nuclei in `1.00 kg` of uranium. Because `A=235`, the number of nuclei is `N=((6.02 xx 10^(23) "nuclei mol"^(-1))/(235 g mol^(-1))) (1.00 xx 10^(3) g)` `=2.56 xx10^(24) nuclei` Hence, the disintergration energy is ` E=nQ =(2,56 xx 10^(24) "nuclei") (("208 89MeV")/("nucleus"))` Because `1 MeV` is equivalent to `4. 45 xx 10^(20) kWh, E=2.3.7xx10^(7) kWh`. This is enough energy to keep a `100 W` light buth burning for about `30000 years`. |
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| 714. |
Consider one of fission reactions of `^(235)U` by thermal neutrons `._(92)^(235)U +n rarr ._(38)^(94)Sr +._(54)^(140)Xe+2n` . The fission fragments are however unstable and they undergo successive `beta`-decay until `._(38)^(94)Sr` becomes `._(40)^(94)Zr` and `._(54)^(140)Xe` becomes `._(58)^(140)Ce`. The energy released in this process is Given: `m(.^(235)U) =235.439u,m(n)=1.00866 u, m(.^(94)Zr)=93.9064 u, m(.^(140)Ce) =139.9055 u,1u=931 MeV]` .A. `156 MeV`B. `208 MeV`C. `465 MeV`D. cannot be computed |
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Answer» Correct Answer - b The complete fission is `._(92)^(235)U + n rarr ._(40)^(94)Zr +._(58)^(140)Ce + 2n+6e^(-1)` `Q =[m(.^(235)U) - m(.^(94)Zr) - m(.^(140)Ce) -m(n)]c^(2)` `=208 MeV`. |
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| 715. |
Two other possible ways by which `.^(235)U` can undergo fission when bombarded with a neutron are (1) by the release of `.^(140)Xe` and `.^(94)Sr` as fission fragments and (2) by the release of `.^(132)Sn` and `.^(101)Mo` as fission fragments. In each case, neutrons are also released. Find the number of neutrons released in each of these events. |
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Answer» By balancing mass numbers and atomic numbers, we find that these reaction can be written as `{:(._(0)^(1) n +._(92)^(235) U rarr. _(54)^(140)Xe +._(38)^(94) Sr + 2. _(0)^(1) n),(._(0)^(1) n +._(92)^(235) U rarr ._(50)^(132)Sn +._(42)^(101) Mo + 3._(0)^(1) n):}` Thus, two neutrons are released in the first event and three in the second. |
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| 716. |
The working principle in atom bomb isA. under-critical chain reactionB. Critical chain reactionC. super-critical chain reactionD. All the above. |
| Answer» Correct Answer - C | |
| 717. |
Heavy water isA. Water at `4^(@)C`B. Watercontaining various saltsC. Compound of heavy oxygen and hydrogenD. Compound of oxgen and deuterium. |
| Answer» Correct Answer - D | |
| 718. |
Nuclear reactor is surrounded by concrete walls toA. Strength the constructionB. Control the chain reactionC. from a protective shieldD. as moderator |
| Answer» Correct Answer - C | |
| 719. |
The operation of a nuclear reactor is said to be critical, if the multiplication factor (k) has a valueA. 1B. 1.5C. 2.1D. 2.5 |
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Answer» Correct Answer - A (a) The multiplication factor (k) is an important reactor parameter and is the ratio of number of neutrons present at the beginning of a particular generation to the number present at the beginning of the next generation. It is a measure of the growth rate of the neutrons in the reactor. For `k = 1`, the operation of the reactor is said to be critical. |
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| 720. |
`._(92)^(238)U` has 92 protons and `238` nucleons. It decays by emitting an alpha particle and becomes:A. `._92^234 U`B. `._92^235 U`C. `._90^234 Th`D. `._93^237 Np` |
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Answer» Correct Answer - C ( c) Let the daughter nucleus be `._Z^A X`. So, reaction can be shown as `._92^238 U rarr ._Z^A X + ._2^4 He` From conservation of atomic mass `238 = A + 4` `rArr A = 234` From conservation of atomic number `92 = Z + 2` `rArr Z = 90` So, the resultant nucleus is `._90^234 X`, i.e., `._90^234 Th`. |
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| 721. |
If radius of the `._(13)^(27)Al` nucleus is taken to be `R_(AI)`, then the radius of `._(53)^(125)Te` nucleus is nearlyA. `(53/13)^(1/3)R_(Al)`B. `5/3 R_(Al)`C. `3/5 R_(Al)`D. `(13/53)^(1/3)R_(Al)` |
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Answer» Correct Answer - B |
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| 722. |
The ratio of radii of nuclei `._13 A1^27` and `._52 X^A` is `3 : 5`. The number of neutrons in the nuclei of `X` will beA. 52B. 73C. 125D. 13 |
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Answer» Correct Answer - B |
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| 723. |
If radius of the `._13^27 A1` nucleus is taken to be `R_(A1)` then the radius of `._53^125 Te` nucleus is nearly.A. `((53)/(13))^(1//3) R_(A1)`B. `(5)/(3) R_(A1)`C. `(3)/(5) R_(A1)`D. `((13)/(53))^(1//3) R_(A1)` |
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Answer» Correct Answer - B (b) As we know, `R = R_0 (A)^(1//3)` Where `A` = mass number `R_(A1) = R_0 (27)^(1//3) = 3 R_0` `R_(Te) = R_0 (125)^(1//3) = 5 R_0 = (5)/(3) R_(A1)`. |
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| 724. |
A certain mass of hydrogen is changes to helium by the process of fusion. The mass defect in fusion reaction is `0.02866 u`. The energy liberated per `u` is (given `1 u = 931 MeV)`A. 2.67 MevB. 26.7 MeVC. 6.675 MeVD. 13.35 MeV |
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Answer» Correct Answer - C ( c) `(0.02866 xx 931)/(4) MeV = (26.7)/(4) MeV = 6.675 MeV = 6.675 MeV`. |
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| 725. |
How many disintegrations per second will occur in one gram of `._92U^(238)`, if its half life against alpha decay is `1.42xx10^(17)s` ? |
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Answer» Given Half-life period `(T)=(0.693)/(lambda)=1.42xx10^(17)s` `lambda=(0.693)/(1.42xx10^(17))=4.88xx10^(-18)` Avagadro number `(N)=6.023xx10^(23)` atoms `n=` Number of atoms present in `1 g` of `._(92)^(238)U = (N)/(A)` `=(6.023xx10^(23))/(238)=25.30xx10^(20)` Number of disintegrations`=(dN)/(dt)=lambdan` `=4.88xx10^(-18)xx25.30xx10^(20)` `=1.2346xx10^(4)` disintergrates/sec |
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| 726. |
The half life of `._92U^(238)` against `alpha` -decay is `4.5xx10^9` years. What is the activity of 1g sample of `._92U^(238)`?A. `1.23xx10^(4) Bq`B. `2.4xx10^(5) Bq`C. `1.82xx10^(6) Bq`D. `4.02xx10^(8) Bq` |
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Answer» Correct Answer - A Activity or rate of decay is `R= lambda N` where `lambda=(0.693)/(T_(1//2))` and `N=` No. of atom is `1 gm` of `._(92)^(238)U=(1xx6.025xx10^(23))/(238)=25.3xx10^(10)` atoms , `R= lambda N` |
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| 727. |
The half life of a radioactive substance is `5xx10^(3)`yrs. In how many years will its activity decay to 0.2 times its initial activity? Take `log_(10) 5=0.6990`. |
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Answer» `T= 5000` years, `(N)/(N_(0))=0.2=(2)/(10)=(1)/(5)` `lambda=(0.693)/(T)=(0.693)/(5000)` `(N)/(N_(0))=e^(-lambdat)` `(1)/(5)=(1)/(e^(lambdat))rArr5=e^(lambdat)` `log_(e )^(5)=lambdat` `t=(2.303xx0.6990xx5000)/(0.693)` `t=11614.6` years `=1.1615xx10^(4)` years |
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| 728. |
After emission of one `alpha` particle followed by one `beta`-particle from `""_(92)^(238)X`, the number of neutrons in the atom will beA. `140`B. `142`C. `144`D. `146` |
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Answer» Correct Answer - B `._(92)U^(238)rarr._(92)x^(234)+._(2)He^(4)+._(-1)e^(0)` no of neutrons `=234 - 92 = 142` |
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| 729. |
The half-life period of a radioactive element x is same as the mean life time of another radioactive element y. Initially, both of them have the same number of atoms. Then, (a) x and y have the same decay rate initially (b) x and y decay at the same rate always (c) y will decay at a faster rate than x (d) x will decay at a faster rate than yA. `X`and `Y` have the same decay rate initiallyB. `X` and `Y` decay at the same rate alwaysC. `Y` will decay at a faster rate than `X`D. `X` will decay at a faster rate than `Y` |
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Answer» Correct Answer - C `(T_(1//2))_(X) = (bar(T))_(Y) = 1.44(T_(1//2))_(Y)` `(T_(1//2))_(Y) lt (T_(1//2))_(X)` |
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| 730. |
The half-life period of a radioactive element x is same as the mean life time of another radioactive element y. Initially, both of them have the same number of atoms. Then, (a) x and y have the same decay rate initially (b) x and y decay at the same rate always (c) y will decay at a faster rate than x (d) x will decay at a faster rate than yA. `X` and `Y` have the same decay rate initiallyB. `X` and `Y` decay at the same rate alwaysC. `Y` will decay at a faster rate than `X`D. `X` will decay at a faster rate then `Y`. |
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Answer» Correct Answer - C ( c) `(T_(1//2))_x = (t_(mean))_y` `rArr (0.693)/(lamda_x) = (1)/(lamda_y) rArr lamda_x = 0.693 lamda_y` or `lamda_x lt lamda_y` Also rate of decay `= lamda N` Initially number of atoms (N) of both are equal but since `lamda_y gt lamda_x`, therefore, `y` will decay at a faster rate than `x`. |
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| 731. |
The isotope `._(92)U^(238)` successively undergoes eight `alpha`-decays and six `beta`-decays. The resulting isotope is found to be `._(92-5y)X^(238-4x)`. Find the value of ratio `x//y`. |
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Answer» Correct Answer - 4 `._(92)U^(238)overset( 8alpha+6 beta)rarr_(82)Pb^(206)` |
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| 732. |
A star initially has `10^40` deuterons. It produces energy via the processes `._1H^2+_1H^2rarr_1H^3+p` and `._1H^2+_1H^3rarr_2He^4+n`. If the average power radiated by the star is `10^16` W, the deuteron supply of the star is exhausted in a time of the order of (a) `10^6s` (b) `10^8s` (c) `10^12s` The masses of the nuclei are as follows `M(H^2)=2.014` amu, `M(n)=1.008` amu, `M(p)=1.007` amu,`M(He^4)=4.001`amuA. `10^6 sec`B. `10^8 sec`C. `10^12 sec`D. `10^16 sec` |
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Answer» Correct Answer - C ( c) Mass defect `= 3 xx 2.014 - 4.001 - 1.007 - 1.008 = 0.026 amu = 0.026 xx 931 xx 10^6 xx 1.6 xx 10^-19 J` `=3.82 xx 10^-12 J` Power of star `= 10^16 W` Number of deuterons used `= (10^16)/(Delta M) = 0.26 xx 10^28` Deuteron supply exhausts in `(10^40)/(0.26 xx 10^28) = 10^12 s`. |
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| 733. |
A star initially has `10^40` deuterons. It produces energy via the processes `._1H^2+_1H^2rarr_1H^3+p` and `._1H^2+_1H^3rarr_2He^4+n`. If the average power radiated by the star is `10^16` W, the deuteron supply of the star is exhausted in a time of the order of (a) `10^6s` (b) `10^8s` (c) `10^12s` The masses of the nuclei are as follows `M(H^2)=2.014` amu, `M(n)=1.008` amu, `M(p)=1.007` amu,`M(He^4)=4.001`amuA. `10^(6)s`B. `10^(8)s`C. `10^(12)s`D. `10^(16)s` |
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Answer» Correct Answer - c The given ractions are `{:(._(1)H_(2)+._(1)H^(2)rarr._(1)h^(3)+p),(._(1H^(2)+._1H2)rarr._2He^4_n),(3.h^3 rarr ._2He^4+n+p):}` Mass defect, `Delta m=(3xx2.014 -4.001 -1.007 -1.008)a.m.u.` = `0.026 a.m.u.` Energy released `=0.026 xx 931 MeV` `=0.026xx931 xx1.6 xx10^(-13) J` `=3.87 xx 10^(-12) J` This is the energy produced by the consumption of three deutron atoms. Therefore, total energy released by `10^(40)` deuterons is `(10^(40))/(3) xx 3.87 xx 10^(-12) J=1.29 xx 10^(28) J` The average power radiated is `P=10^(6)W` or `10^(16) J s^(-1)`. Therefore, total time is exhaust all deutrons of the star will be `t=1.29 xx(10^(28))/(10^(16))=1.29 xx 10^(12) s ~~ 10^(12)s`. |
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| 734. |
A star initially has `10^40` deuterons. It produces energy via the processes `._1^2H+_1^2Hrarr_1^3H+p` and `._1^2H+_1^3Hrarr_2^4He+n`, where the masses of the nuclei are `m(.^2H)=2.014` amu, `m(p)=1.007` amu, `m(n)=1.008` amu and `m(.^4He)=4.001` amu. If the average power radiated by the star is `10^16 W`, the deuteron supply of the star is exhausted in a time of the order ofA. `10^6 sec`B. `10^8 sec`C. `10^12 sec`D. `10^16 sec` |
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Answer» Correct Answer - C ( c) Mass defect `= 3 xx 2.014 - 4.001 - 1.007 - 1.008 = 0.026 amu = 0.026 xx 931 xx 10^6 xx 1.6 xx 10^-19 J` =`3.82 xx 10^-12 J` Power of star `= 10^16 W` Number of deuterons used `= (10^16)/(Delta M) = 0.26 xx 10^28` Deuteron supply exhausts in `(10^40)/(0.26 xx 10^28) = 10^12 s`. |
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| 735. |
The half-life of a radioactive substance is 3h and its activity is `1mu Ci`. Then the activity after 9h will be (in `mu Ci`)-A. `1/9`B. `1/27`C. `1/3`D. `1/8` |
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Answer» Correct Answer - D `A=A_(0)e^(-lambdat)` `lambda=(ln2)/(t_(1//2)) t=9h` then `A=(A_(0))/8` |
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| 736. |
At a given instant there are `25%` undecayed ratio-activity nuclei in sample. After `10` second, the number of undecyed nuclei reduces to `12.5%`. Calculate (a) mean-life of the nuclei and (a) the time in which the number undecayed nuclei will further reduce to `6.25%` of the reduced number. |
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Answer» (a) Half-like `T_(1//2) = 10 sec` Meanlife `bar(T) = 1.44T_(1//2) = 1.44 xx 10 = 1.44 sec` (b) `(N_(0))/(4) overset(T_(1//2))rarr(N_(0))/(8)overset(T_(1//2))rarr(N_(0))/(16)` `t = T_(1//2) = 10 sec` |
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| 737. |
Two radioactive substances `A` and `B` have half lives of `T` and `2T` respectively. Samples of `a` and `b` contain equal number of nuclei initially. After a time `4T`, the ratio of the number of undecayed nuclei of `A` to the number if undecayed nuclei of `A` to the number of undeacyed nuclei of `B` isA. `1 : 4`B. `1 : 2`C. `2 : 1`D. `4 : 1` |
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Answer» Correct Answer - A `t = 4T,(T_(1//2))_(A) = T, (T_(1//2))_(B) = 2T` `A: t = n(T_(1//2))_(A) rArr 4T = nT rArr n = 4` `N_(A) = N_(0) ((1)/(2))^(n) = N_(0) ((1)/(2))^(4) = (N_(0))/(16)` `B: t = n(T_(1//2))_(B) rArr 4T = n xx 2T rArr n = 2` `N_(B) = N_(0)((1)/(2))^(n) = N_(0)((1)/(2))^(2) = (N_(0))/(4)` `(N_(A))/(N_(B)) = (N_(0)//16)/(N_(0)//4) = (1)/(4)` |
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| 738. |
Half-lives of two radioactive substances `A` and `B` are respectively `20` minutes and `40` minutes. Initially, he sample of `A` and `B` have equal number of nuclei. After `80` minutes the ratio of the remaining number of `A` and `B` nuclei is :A. `1 : 16`B. `4 : 1`C. `1 : 4`D. `1 : 1` |
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Answer» Correct Answer - C ( c) We know that `N = N_0 ((1)/(2))^(n_A)` For `A`, `N = N_0 ((1)/(2))^(n_A) = N_0 ((1)/(2))^4 = (N_0)/(16)` `[because n_A = (t)/(T_A) = (80)/(20) = 4]` For `B`, `N_B = N_0 ((1)/(2))^(n_B) = N_0 ((1)/(2))^2 = (N_0)/(4)` `:. (N_A)/(N_B) = (1)/(4)` or `N_A : N_B = 1: 4`. |
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| 739. |
Half-lives of two radioactive substances `A` and `B` are respectively `20` minutes and `40` minutes. Initially, he sample of `A` and `B` have equal number of nuclei. After `80` minutes the ratio of the remaining number of `A` and `B` nuclei is :A. `1:16`B. `4:1`C. `1:4`D. `1:1` |
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Answer» Correct Answer - C `(N_(A))/(N_(B)) = (N_(0)((1)/(2))^(80//20))/(N_(0)((1)/(2))^(80//40)) =(((1)/(2))^(4))/(((1)/(2))^(2))=((1)/(2))^(2) = (1)/(4)` |
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| 740. |
Radioactive isotopes are produced in a nuclear physics experiment at a constant rate `dN//dt=R` .An inductor of inductance 100 mH, a resistor of resistance `100 Omega` and a battery are connected to form a series circuit.the circuit is switched on at the instant the production of radioactive isotope starts. It is found that `i//N` remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t. Find the half-life of the isotope. |
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Answer» Correct Answer - `[6.93 xx 10^(-4)s]` |
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| 741. |
For a radioactive material, its activity `A` and rate of change of its activity of `R` are defined as `A=(-dN)/(dt)` and `R=(-dA)/(dt)`, where `N(t)` is the number of nuclei at time `t`. Two radioactive source `P` (mean life `tau`) and `Q` (mean life `2 tau`) have the same activity at `t=0`. Their rates of activities at `t=2 tau` are `R_(p)` and `R_(Q)`, respectively. If `(R_(P))/(R_(Q))=(n)/(e )`, then the value of `n` is: |
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Answer» Correct Answer - `2` `A_(P)=A_(0)e^((-t)/(tau)), A_(Q)=A_(0)e^((-t)/(2tau))` `R_(P)=(A_(0))/(tau)e^((-t)/(tau)), R_(Q)=(A_(0))/(2tau)e^((-t)/(2tau))` at `t=2tau` `(R_(P))/(R_(Q))=((A_(0)e^(-2))/(tau))/((A_(0))/(2tau)e^(-1))=(2)/(e )` |
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| 742. |
Let `m_(p)` be the mass of a poton , `M_(1)` the mass of a `_(10)^(20) Ne` nucleus and `M_(2)` the mass of a `_(20)^(40) Ca` nucleus . ThenA. `M_(2) =2M_(1)`B. `M_(2)gt2M_(1)`C. `M_(2)lt2M_(1)`D. `M_(1)lt10(m_(n)+m_(p))` |
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Answer» Correct Answer - C::D |
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| 743. |
The rest energy of an electron is.A. 510 KeVB. 931 KeVC. 510 MeVD. 931 MeV |
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Answer» Correct Answer - A |
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| 744. |
The half-life of a radioactive nuclide is 20 hours. What fraction of original activity will remain after 40 hours? |
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Answer» `40` hours means `2` half lives. Thus, `A=(A_(0))/(2^(2))=(A_(0))/(4)` or `(A)/(A_(0))=(1)/(4)` So one fourth of the original activity will remain after `40` afters. Specific activity: The activity per unit mass is called specific activity. |
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| 745. |
The relationship between distintegration constant `(lambda)` and half-life `(T)` will beA. `lambda=(log_(10) 2)/(T)`B. `lambda=(log_(e)2)/(T)`C. `lambda=(T)/(log_(e)2)`D. `lambda=(log_(2)e)/(T)` |
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Answer» Correct Answer - B |
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| 746. |
A radioactive substance has a half-life of `1` year. The fraction of this material, that would remain after `5` years will be.A. `(1)/(32)`B. `(1)/(5)`C. `(1)/(2)`D. `(4)/(5)` |
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Answer» Correct Answer - A (a) Number of half lives `n = (5)/(1) = 5` Now `(N)/(N_0) = ((1)/(2))^n rArr (N)/(N_0) = ((1)/(2))^5 = (1)/(32)`. |
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| 747. |
In the given reaction `._z X^A rarr ._(z+1)Y^A rarr ._(z-1) K^(A - 4) rarr ._(z-1) K^(A - 4)` Radioactive radiations are emitted in the sequence.A. `alpha,beta,gamma`B. `beta,alpha,gamma`C. `gamma,alpha,beta`D. `beta,gamma,alpha` |
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Answer» Correct Answer - B |
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| 748. |
Which of the following is positively charged?A. `alpha`-particleB. `beta`-particleC. `gamma`-raysD. `X`-rays |
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Answer» Correct Answer - A |
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| 749. |
The half-life of radioactive material is `3 h`. If the initial amount is `300 g`, then after `18 h`, it will remainA. `4.68 g`B. `46.8 g`C. `9.375 g`D. `93.75 g` |
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Answer» Correct Answer - A |
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| 750. |
In the nuclear decay given below `._Z^A X rarr ._(Z-1).^A Y rarr ._(Z-1)^(A - 4) B^** rarr ._(Z-1)^(A-1) B`, the particle emitted in the sequence areA. `beta, alpha, gamma`B. `gamma, beta, alpha`C. `beta, gamma, alpha`D. `alpha, beta, gamma` |
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Answer» Correct Answer - A |
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