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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.
| 651. |
In nuclear reactions, we have the conservation ofA. Mass onlyB. Energy onlyC. Momentum onlyD. Mass, energy and momentum |
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Answer» Correct Answer - D (d) No energy and mass enters or goes out of the system of the reaction and no external force is assumed to act. |
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| 652. |
In nuclear reactions, we have the conservation ofA. mass onlyB. momentum onlyC. energy onlyD. mass,energy and momentum |
| Answer» Correct Answer - D | |
| 653. |
The decay constant of a radioactive substance is `0.693` per minute. What will be the half-life and mean life of it? |
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Answer» Given that: `lamabda=0.693//"minute"` Half-life `rArr T_(1//2)=(0.693)/(lambda)=1"minute"=60 sec`. |
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| 654. |
The half-life of a radioactive substance against `alpha-`decay is `1.2 xx 10^7 s`. What is the decay rate for `4 xx 10^15` atoms of the substance ?A. `4.6 xx 10^12` atoms/sB. `2.3 xx 10^11` atoms/sC. `4.6 xx 10^10` atoms/sD. `2.3 xx 10^8` atoms/s |
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Answer» Correct Answer - D (d) `(dN)/(dt) = -lamda N rArr |(dN)/(dt)| = (0.693)/(T_(1//2)) xx N` =`(0.693)/(1.2 xx 10^7) xx 4 xx 10^15 = 2.3 xx 10^8 `atoms/sec. |
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| 655. |
When Boron nucleus `B_(3)^(10)` is bombarded by neutrons , a- particle are emitted . The resulting nucleus is of the element …….. and has the mass number.... |
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Answer» `._5^(10)B +._0^1n rarr ._2^4He +._3^7Li` The resulting nucleus is of element lithium and mass number is 7. |
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| 656. |
Two radioactive substances `X` and `Y` initially contain an equal number of atoms. Their half-lives are `1` hour and `2` hours respectively. Then the ratio of their rates of disintergration after two hours isA. `1:1`B. `2:1`C. `1:2`D. `2:3` |
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Answer» Correct Answer - C `A=lambdaN` |
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| 657. |
A nuclear reactor generates power at 50% efficiency by fission of `._(92)^(235)U` into two equal fragments of `._(46)^(116)U` into two equal fragments of `._(46)^(116)Pd` with the emission of two gamma rays of 5.2 MeV each and three neutrons. The average binding energies per particle of `._(92)^(235)U` and `._(46)^(116)Pd` are 7.2 MeV and 8.2MeV respectiveley. Calculate the energy released in one fission event. Also-estimate the amount to `.^(235)U` consumed per hour to produce 1600 megawatt power. |
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Answer» `._(92)^(235)Urarr2._(46)^(116)Pd +3._(0)^(1)n+2gamma` `B.E.` of `._(92)^(235)U = 235xx7.2 = 1692MeV` `B.E.` of `._(46)^(116)Pd = 2[116 xx 8.2] = 1902.4MeV` Energy of `2gamma` photons `= 2xx5.2 = 10.4MeV` Enegry released per fission `= 1692-[1902.4+10.4]` `= 200MeV` NUmber of atoms in `m gram` of `._(92)^(235)U` is `(m)/(235) xx6.023xx10^(23)` Energy produced due to fission of `m` gram of `._(92)^(235)U` is `(m)/(235)xx6.023xx10^(23)xx200xx1.6xx10^(-13) J` Output power `= 1600MW = 1600 xx 10^(6) = 1.6 xx 10^(9)J//sec` Input power `= (1.6xx10^(9))/((50//100)) = 3.2 xx 10^(9)J//sec` Input energy `= 3.2 xx 10^(9)xx60xx 60J` `(m)/(235)xx6.023xx106(23)xx200xx1.6xx10^(-13)` `= 3.2 xx 10^(9) xx 3600` `m = 140.5 g` |
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| 658. |
Calculate the energy released by fission from `2 g` of `.^(235)._(92)U` in `kWh`. Given that the energy released per fission is `200 MeV`. |
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Answer» Mass of uranium `= 2g` Energy released per fission `= 200 MeV` `=200xx10^(6)xx1.6xx10^(-19)=3.2xx10^(-11)J` Number of atoms in `2` gram of uranium is `n=(2xx6.023xx10^(23))/(235)=5.125xx10^(21)"atoms"` Total energy released `=`No. of atoms `x` energy released per fission `=5.125xx10^(21)xx3.2xx10^(-11)=16.4xx10^(10)J` `=0.455xx10^(5)Kwh=4.55xx10^(4)Kwh` |
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| 659. |
In the fusion reaction `._1^2H+_1^2Hrarr_2^3He+_0^1n`, the masses of deuteron, helium and neutron expressed in amu are `2.015, 3.017 and 1.009` respectively. If `1 kg` of deuterium undergoes complete fusion, find the amount of total energy released. 1 amu `=931.5 MeV//c^2`. |
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Answer» `._(1)^(2)H+._(1)^(2)Herarr._(2)^(3)He+._(0)^(1)n` Mass defect: `Deltam = [(2.015xx2)-(3.017 +1.009)]` `=4 xx 10^(-3)"amu"` Equivalent enegry `= 4xx10^(-3) xx 931.5 = 3.726MeV` Number of atoms in `1kg` of `._(1)^(2)H= (1000)/(2) xx 6.023 xx 10^(23)` In one reaction, two atoms of `._(1)^(2)H` are used. Total energy released `=(1)/(2) ((100)/(2)xx6.023xx10^(23)) xx3.726xx1.6xx10^(-13)` `=9xx10^(13) J` |
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| 660. |
(a) Calculate the energy released by the fission of `2g` of `._(92)U^(235)` in `kWh`. Given that the energy released per fission is `200MeV`. (b) Assuming that `200MeV` of enrgy is released per fission of uranium atom, find the number of fissions per second required to released `1` kilowatt power. (c) Find the amount of energy produced in joules due to fission of `1g` of `._(92)U^(235)`assuming that `0.1%` of mass is transformed into enrgy.`._(92)U^(235) = 235 amu`, Avogadro number `= 6.023 xx 10^(23)` |
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Answer» (a) Number of fissions `=` Number of atoms `= (m)/(M)N_(A) = (2)/(235) xx 6.023 xx 10^(23)` Total energy released `=(2)/(235) xx 6.023 xx 10^(23) xx 200 xx 10^(6) xx 1.6 xx10^(-19)J` `= 16.4 xx 10^(10) J` `= (16.4xx10^(10))/(3.6xx10^(6)) = 4.55 xx 10^(4)kWh` `(1kWh = 3.6 xx 10^(6)J)` (b) `E = 1kW = 10^(3) J//sec` Number of fissions/sec `= (10^(3))/(200xx1.6xx10^(13)) = 3.125xx10^(13)` (c) Mass used `m = (0.1)/(100) xx 1=10^(-3) g` Number of fissions `= (10^(-3))/(235) xx 6.023 xx 10^(23)` Enegry produced `= (6.023xx10^(20))/(235) xx 200 xx 1.6 xx 10^(-13)` `= 8.2 xx 10^(7) J` |
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| 661. |
When `._92 U^235` undergoes fission, `0.1 %` of its original mass is changed into energy. How much energy is released if `1 kg` of `._92 U^235` undergoes fission ?A. `9 xx 10^(10) J`B. `9 xx 10^(11) J`C. `9 xx 10^(12) J`D. `9 xx 10^(13) J` |
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Answer» Correct Answer - D `E = mc^(2) = (1xx0.1)/(100)xx(3xx10^(8))^(2) = 9xx10^(13) J` |
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| 662. |
A sample of radioactive material has mass `m`, decay constant `lambda`, and molecular weight `M`. Avogadro constant `=N_(A)`. The initial activity of the sample is:A. `lambda m`B. `(lambda m)/(M)`C. `(lambda m N_(A))/(M)`D. `mN_(A^(e^(lambda)))` |
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Answer» Correct Answer - C Activity `= lambda N` where `N=(N_(A)m)/(M)` |
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| 663. |
Find the amount of energy released when `1` atom of Uranium `._(92)U^(235)(235.0439 "amu")` undergoes fission by slow neturon `(1.0087 "amu")` and is splitted into Krypton `._(36)Kr^(92)(91.8973 "amu")` and Barium `._(56)Br^(141)(140.9139 "amu")` assuming no energy is lost. Hence find the enrgy in `kWh`, when `1g` of it undergoes fission. |
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Answer» `._(92)^(235)U+._(0)^(235)nrarr._(56)^(141)Ba+._(36)^(92)Kr+3(._(0)^(1)n)` `Deltam = [{235.0439+1.0087}-{140.9139+91.8973+3xx1.0087}]` `= 0.2135amu` Equilvalent enrgy `= 0.2153xx931 = 200MeV` Number of atoms/fission in `1g` of atom is `(1)/(235) xx 6.023xx10^(23) = 2.56xx10^(21)` Energy released in fission of `1g` of `._(92)^(235)U` `= 2.56xx10^(21) xx 200MeV` `= 2.56 xx 10^(21) xx 200 xx 1.6 xx 10^(-13)J` `= (2.56xx10^(21)xx200xx1.6xx10^(-13))/(3.6xx10^(6))kWh` `= 2.28 xx 10^(4) kWh` |
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| 664. |
`92^(U^(235)` mucleus absorbs a slow neutron and undergoes fission into `54^(X^139)` and `38^(sr^94)` nuclie The other particies produced in this fission process areA. `1 beta` and `1 alpha`B. `2 beta` and `1` neutronC. `2` neturonsD. `3` neutrons |
| Answer» Correct Answer - D | |
| 665. |
A slow neutron`(n)` is captured by a `._(92)U^(235)` nucleus forming a highly unstable nucleus `._(92)U^(236**)` (where `**` ddenotes that the nucleus is in an excited state). The fission of the nucleus occurs byA. `._(92)^(236)U^(**)rarr._(50)^(140)Sn + ._(42)^(89)Mo + 6n+Q`B. `._(236)^(92)U^(**)rarr._(54)^(140)Sn + ._(38)^(94)Sr + 4n+Q`C. `._(92)^(236)U^(**)rarr._(52)^(144)Sn + ._(42)^(89)Mo + 3n+Q`D. `._(92)^(236)U^(**)rarr._(56)^(144)Ba + ._(36)^(89)Kr + 3n+Q` |
| Answer» Correct Answer - D | |
| 666. |
An artifical radioactive decay series begins with unstable `._94 ^241 Pu`. The stable nuclide obtained after eight `alpha-`decays and five `beta^+ -` decays is.A. `._83^209 Bi`B. `._82^209 Pb`C. `._82^205 Ti`D. `._82^201 Hg` |
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Answer» Correct Answer - A (a) Mass number decreases by `8 xx 4 = 32` Atomic number decreases by `8 xx 2 - 5 = 11`. |
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| 667. |
The `._(92)U^(235)` absorbs a slow neturon (thermal neutron) & undergoes a fission represented by `._(92)U^(235)+._(0)n^(1)rarr._(92)U^(236)rarr._(56)Ba^(141)+_(36)Kr^(92)+3_(0)n^(1)+E`. Calculate: The energy released when `1 g` of `._(92)U^(235)` undergoes complete fission in `N` if `m=[N]` then find `(m-2)/(5)`. `[N]` greatest integer Given `._(92)U^(235)=235.1175"amu (atom)"`, `._(56)Ba^(141)=140.9577 "amu (atom)"`, `._(36)r^(92)=91.9263 "amu(atom)" , ._(0)n^(1)=1.00898 "amu", 1 "amu"=931 MeV//C^(2)` |
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Answer» Correct Answer - `4` (i) `E=[M_(U)+m_(n)-M_(Ba)-M_(kr)-3m_(n)]931=200.57 MeV` (ii) `(N_(A))/(235)xxE=22.84 MWh` |
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| 668. |
A radioactive material decays by simulataneous emission of two particle from the with respective half - lives `1620` and `810` year . The time , in year , after which one - fourth of the material remains isA. 4860B. 2430C. 3240D. 1080 |
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Answer» Correct Answer - D (d) `lamda = lamda_1 + lamda_2` `rArr (1)/(T) = (1)/(T_1) +(1)/(T_2)` `:. T = (T_1 xx T_2)/(T_1 + T_2) = (810 xx 1620)/(810 + 1620) = 540 years` Hence, `(1)/(4) th` of material remain after `1080 years`. |
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| 669. |
Find the decay constant of `.^(55)Co` radio nuclide if its activity is known to decrease `4%` per hour. The decay product is non-radioactive. |
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Answer» Correct Answer - `[lambda ~= 1.1 xx 10^(-5)s^(-1)]` |
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| 670. |
A nuclide `A` undergoes `alpha`-decay and another nuclide `B` undergoed `beta`-decay. Then,A. All the `alpha`-paricles emitted by A will have almost the same speed.B. the `alpha`-particles emitted by B will have widely different speed.C. the `beta`-particles emitted by B will have almost the same speed.D. the `beta`-particles emitted by B may have almost the same speed. |
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Answer» Correct Answer - a,d In `alpha`-decay, the entire energy is carried away by the `alpha`-particles as its kinetc energy. In `beta^(-)` -decay, the energy, the energy is shared between the `beta`-particles and the anti-neutrino. Hence, the speed of the `beta`-particle will vary, depending on the energy of the anti-neutrino. |
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| 671. |
A radioactive material decays by simulataneous emission of two particle from the with respective half - lives `1620` and `810` year . The time , in year , after which one - fourth of the material remains isA. 1080B. 2430C. 3240D. 4860 |
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Answer» Correct Answer - A |
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| 672. |
A nuclide `A` undergoes `alpha`-decay and another nuclides `B` undergoes `beta`-decayA. All the `alpha`-particles emitted by `A` will have almost the same speedB. The `alpha`-particles emitted by `A` may have widely different speed.C. All the `beta^(-)` particle emitted by `B` will have almost three same speedD. The `beta^(-)` particles emitted by `B` may have widely different speeds |
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Answer» Correct Answer - A::D The total kinetic energy in the process will be taken by `alpha`-particles while in `beta^(-)` particle. |
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| 673. |
A sample of radioactive material decays simultaneouly by two processes A and B with half-lives `(1)/(2)` and `(1)/(4)h`, respectively. For the first half hour it decays with the process A, next one hour with the proecess B, and for further half an hour with both A and B. If, origianlly, there were `N_0` nuceli, find the number of nuclei after 2 h of such decay.A. `N_0/(2)^(8)`B. `N_0/(2)^(4)`C. `N_0/(2)^(6)`D. `N_0/(2)^(5)` |
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Answer» Correct Answer - a After First half hours, `N=N_(0)(1)/(2)` For`t=(1)/(2)h" to "t=1(1)/(2),1h` = four half-lives Hence, `N=(N_(0)(1)/(2))[(1)/(2)]^(2)=N_(0)((1)/(2))^(5)` For `t=(1)/(2)" to "t=2h` `[" for both" `A` and `B`, `(1)/(t_(1//2))=(1)/(t_(1//2)) + (1)/(t_(1)//4)=2 +4 =6 rArr t_(1//2) =(1)/(6)]` `(1)/(2) h=` half -lives `:. N=[(N_(0)(1)/(2))^(5)](1)/(2^(3))=N_(0)(1)/(2^(8))`. |
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| 674. |
The half-lives of radioactive sample are `30` years and `60` years for two decay processes. If the sample decays by both the processes simultaneously. The time after which, only one-fourth of the sample will remain isA. `10` yearsB. `20` yearC. `40` yearD. `60` year |
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Answer» Correct Answer - C `lambda=lambda_(a)+lambda_(2)or (1)/(t)-(1)/(t_(1))+(1)/(t_(2))`, where t is `T_(1//2)` `T_(1//2)=20` years |
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| 675. |
The binding energy per nucleon of deuterium and helium atom is `1.1 MeV` and `7.0 MeV`. If two deuterium nuclei fuse to form helium atom, the energy released is.A. `19.2 MeV`B. `23.6 MeV`C. `26.9 MeV`D. `13.9 MeV` |
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Answer» Correct Answer - B |
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| 676. |
Binding energy per nucleons vs mass curve for nucleus is shown in the figure `W,X,Y`and `Z` are four nuclei indicated on the curve . The process that would release energy is A. `Y rarr 2 Z`B. `W rarr X + Z`C. `W rarr 2 Y`D. `X rarr Y + Z` |
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Answer» Correct Answer - C ( c) Energy is released in a process when total Binding energy `(B.E)` of the nucleus is increased or we can say when total `B.E` of products is more than the reactants. By calculation we can see that only in case of option ( c), this happens. Given `W rarr 2 Y` `B.E` of reactants `= 120 xx 75 = 900 MeV` and `B.E` of products `= 2xx (60 xx 85) = 1020 MeV` i.e., `B.E` of products `gt B.E` of reactants. |
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| 677. |
Binding energy per nucleons vs mass curve for nucleus is shown in the figure `W,X,Y`and `Z` are four nuclei indicated on the curve . The process that would release energy is A. `Y rarr2Z`B. `Wrarr X+Z`C. `W rarr2Y`D. `XrarrY+Z` |
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Answer» Correct Answer - C When total `B.E.` of products `gt` recatnts, energy is released. `Yrarr 2Z rArr 2xx5xx30-8.5xx60 =- 210 MeV` `W rarr X +Z rArr 8xx90 +5xx30 - 7.5xx120 =- 30 MeV` `W rarr 2Y rArr 2xx8.5 xx 60- 7.5 xx120 = 120MeV` `X rarr Y+Z rArr 8.5xx60 +5xx30 - 8xx90 =- 60MeV` |
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| 678. |
After two hours `1//6` th of the initial amount of a certain radioactive isotope remains undecayed. The half life of the isotope is :A. 17.3 minB. 34.7 minC. 46.4 minD. 1 hour |
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Answer» Correct Answer - C |
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| 679. |
Binding energy per nucleons vs mass curve for nucleus is shown in the figure `W,X,Y`and `Z` are four nuclei indicated on the curve . The process that would release energy is A. `Y to 2Z`B. `Wto X+Z`C. `W to 2Y`D. `X to Y+ Z` |
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Answer» Correct Answer - C |
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| 680. |
The rate of decay `(R )` of nuclei in a radioactive sample is plotted against time `(t)`. Which of the following best represents the resulting curve ?A. B. C. D. |
| Answer» Correct Answer - A | |
| 681. |
Tritium is an isotope of hydrogen whose nucleus triton contains 2 neutrons and 1 proton . Free neutrons decay into `p+bar(e) +bar(n).` If one of the neutrons in Triton decays , it would transform into `He^(3)` nucleus. This does not happen. This is becauseA. Triton energy is less than that of a `He^(3)` nucleus.B. the electron created in the beta decay process cannot remain in the nucleus.C. both the neutrons in Triton have to decay simultaneously resulting in a nucleus with `3` protons, which is not a `He^(3)` nucleus.D. Because free neutrons decay due to external perturbations which is absent in a triton nucleus. |
| Answer» Correct Answer - A | |
| 682. |
A nuclear of mass `M +deltam ` is at rest and decay into two daughter nuclei of equal mass `(M)/(2)` each speed is `c` The speed of daughter nuclei isA. `(Deltam)/(M+Deltam)`B. `sqrt((2Deltam)/(M))`C. `sqrt(Deltam)/(M)`D. `sqrt((Deltam)/(M+Deltam))` |
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Answer» Correct Answer - B `Q=Delta m c^(2)=(1)/(2)xx((M)/(2))v^(2)+(1)/(2)xx((M)/(2))v^(2)` `Deltamc^(2)=(1)/(2)xxMv^(2)" "rArrv=csqrt((2Deltam)/(M))` |
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| 683. |
A nuclear of mass `M +deltam ` is at rest and decay into two daughter nuclei of equal mass `(M)/(2)` each speed is `c` The speed of daughter nuclei isA. `c(Delta m)/(M+Delta m)`B. `csqrt((2 Delta m)/(M))`C. `csqrt(( Delta m)/(M))`D. `csqrt((Delta m)/(M+Delta m))` |
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Answer» According to conservation of momentum `0 = (M)/(2) V_(1) - (M)/(2)V_(2), V_(1) = V_(2)`, `Deltamc^(2) = (1)/(2) (M)/(2) V_(1)^(2) + (1)/(2).(M)/(2) V_(2)^(2)` |
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| 684. |
The mean density of the nuclei is proportinal to:A. Mass numberB. Atomic numberC. The number of nucleonsD. None of the above |
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Answer» Correct Answer - D |
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| 685. |
A source (s) of beta particles contains a radioactive isotope X. The emitted beta particles are passed through two parallel slits `s_(1)` and `s_(2)` to get a narrow parallel beam. This beam is made to enter a uniform magnetic field (B). The particles follow semicircular trajectories and are found to hit the screen at all points from A to B the distance AB is equal to 2a. The reaction for decay of a nucleus of X is `XrarrY+._(-1)^(0)e+vecv` Calculate the minimum difference in mass of a nucleus of X and Y. mass of an electron is m and charge on it its e. |
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Answer» Correct Answer - `(a^(2)e^(2)B^(2))/(2mc^(2))` |
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| 686. |
A diatomic molecule moving at a speed u absorbs a photon of wavelenght `lambda` and then dissociates into two identical atoms. One of the atoms is found to be moving with a speed v in a direction perpendicualr to the initial direction of motion of the molecule. Take mass of the molecule to be M and calculate the binding energy of the molecule. Assume that momentum of absorbed photon is negligible compared to that of the molecule. |
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Answer» Correct Answer - `BE=(hc)/(lambda)-(1)/(2)Mu^(2)-(1)/(2)Mv^(2)` |
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| 687. |
A small quantity of a solution containing `N^(24)` radio-nuclide of half-life `T` and activity `R_(0)` is injected into blood of a person. `1 cm^(3)` of sample of blood taken from the blood of the person shows activity `R_(1)`. If the total volume of the blood in the body of the person is `V`, find the timer after which sample is taken.A. `(T)/(1n(2))[1 n(R_(0))/(VR_(1))]`B. `(T)/(ln(2))[1n(VR_(0))/(R_(1))]`C. `(T)/(ln(2))[1n(VR_(1))/(R_(0))]`D. `(T)/(ln(2))[1n(VR_(1))/(VR_(0))]` |
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Answer» Correct Answer - A Total volume of blood, `(R_(0)e^(-lambdat))/(R_(1))`, (or) `(VR_(1))/(R_(0))=e^(lambdat)` `-ln((VR_(1))/(R_(0)))=lambda t=(ln(2)t)/(T), t=(T)/(ln(2))ln((R_(0))/(VR_(1)))` |
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| 688. |
Consider radioactive decay of `A` to `B` with which further decays either to `X` or `Y`, `lambda_(1), lambda_(2)` and `lambda_(3)` are decay constant for `A` to `B` decay, `B` to `X` decay and Bto `Y` decay respectively. At `t=0`, the number of nuclei of `A,B,X` and `Y` are `N_(0), N_(0)` zero and zero respectively. `N_(1),N_(2),N_(3)` and `N_(4)` are the number of nuclei of `A,B,X` and `Y` at any instant `t`. At `t=oo`, which of following is incorrect ?A. `N_(2)=0`B. `N_(3)=(N_(0)lambda_(2))/(lambda_(2)+lambda_(3))`C. `N_(4)=(2N_(0)lambda_(3))/(lambda_(2)+lambda_(3))`D. `N_(1)+N_(2)+N_(3)+N_(4)=2N_(0)` |
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Answer» Correct Answer - B At `t= oo` is `2N_(0)`. Also, the total number of nuclei at any instant remains the same `:. N_(1)+N_(2)+N_(3)+N_(4)=2N_(0)` The ratio of X and Y formed are in the ratio `lambda_(2): lambda_(3)` and the total number of nuclei of X and Y at `t=oo` is `2N_(0)`. |
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| 689. |
Assume that the nuclear binding energy per nucleus` (B//A)` versus mass number `(A)` is as shown in the figure Use this plot to choose the correct (s) choice given below A. Fusion of nuclei with mass numbers lying in the range of `1 lt A lt50` will release energy.B. Fusion of nuclei with mass numbers lying in the range of `51 lt A lt100` will release energy.C. Fission of a nucleus lying in the mass range of `100 lt A lt 260` will release energy when broken into equal fragments.D. Fission of a nucleus lying in the mass range of `200 lt A lt 260` will release energy when broken into equal fragments. |
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Answer» Correct Answer - b,d In fusion, two or more lighter nuclei combine to make a comparatively heavier nucleus. In fission, a heavy nucleus beaks into two or more comparatively lighter nuclei. Further, energy will be released in a nuclei process if total binding energy increases. Hence, correct options are (b) and (d). |
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| 690. |
Assume that the nuclear mass is of the order of `10^(-27) kg` and the nuclear radius is of the order of `10^(15)m`. The nuclear density is of the order ofA. `10^(2)Kg//m^(3)`B. `10^(10)Kg//m^(3)`C. `10^(17)Kg//m^(3)`D. `10^(31)Kg//m^(3)` |
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Answer» Correct Answer - C `d=(3M_(n))/(4 piR^(3)) R="radius"` |
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| 691. |
The fraction of a radioactive material which reamins active after time t is `9//16`. The fraction which remains active after time `t//2` will be .A. `(4)/(5)`B. `(7)/(8)`C. `(3)/(5)`D. `(3)/(4)` |
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Answer» Correct Answer - d `(9)/(16)=((1)/(2))^((t)/(T))` `(N)/(N_(0))=((1)/(2))^((t)/(2T))` `((N)/(N_(0)))^(2)=((1)/(2))^(t//T)` or `(N)/(N_(0)^(2))=(9)/(16)` or `(N)/(N_(0)) =(3)/(4)` Note the special techaniq used in the problem. |
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| 692. |
After an interval of one day , `1//16th` initial amount of a radioactive material remains in a sample. Then, its half-life is .A. `6h`B. `12 h`C. `1.5 h`D. `3h` |
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Answer» Correct Answer - a `N=(N_(0))/(2^(t//T))` `(N_(0))/(16) =(N_(0))/(2^(t//T))` `2^(t//T) =16=2^(4)` or `T=(t)=(4)` or `T=(t)/(T)=(24)/(4) h=6h` . |
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| 693. |
A stationery thorium nucleus `(A=200 , Z=90)` emits an alpha particle with kinetic energy `E_(alpha)`. What is the kinetic energy of the recoilling nucleusA. `(E_(alpha))/(108)`B. `(E_(alpha))/(110)`C. `(E_(alpha))/(55)`D. `(E_(alpha))/(54)` |
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Answer» Correct Answer - d The `alpha` -particle emitting radioactive gas, thoron `-220`, decays to radium `-216` an emits an `alpha`-particles. The reaction can be represented by `._(90)^(220)Th rarr ._2^4He +._(66)^(216)Ra` By conservation of momentum of `alpha`-particles = momentum of recoilling nucleus Ra `rArr m_(alpha) v_(alpha) =M_(R) v_(R)` `rArr (v_(R))/(v_(alpha)) =(m_(alpha))/( m_(R)) =(4)/(216) =(1)/(54)` The kinetic energy of Ra, `E_(R)`, is realted to the kinetic energy of alpha particle `E_(alpha)` by `(E_(R))/(E_(alpha))=((1)/(2)m_(R)v_(R)^(2))/((1)/(2)m_(alpha)v_(alpha)^(2))=((m_(R))/(m_(alpha)))((v_(R))/(m_(alpha)))^(2)=((m_(R))/(m_(alpha)))((m_(alpha))/(m_(R)))^(2)` `=(m_(alpha))/(m_(R))=(1)/(54)` `:. E_(R)=(E_(alpha))/(54)`. |
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| 694. |
A sample of radioactive meterial has mass `m`, decay constant `lambda`, and molecular weight `M`. Avogardo constant `=N_(A)`. The initial activity of the sample is:A. `lambda m`B. `(lambda m)/(M)`C. `(lambdamN_(A))/(M)`D. `mN_(A)e^(lambda)` |
| Answer» Correct Answer - C | |
| 695. |
The half-life of `.^215At` is `100mus`. The time taken for the activity of a sample of `.^215At` to decay to `1/16th` of its initial value isA. `400 mu s`B. `6.3 mu s`C. `40 mu s`D. `300 mu s` |
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Answer» Correct Answer - a `(1)/(16)=(1)/(2^(t/100))` or `(1)/(2^(4))=(1)/(2^(t//100)) =4 =(t)/(100)` or `t=400 m u s`. |
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| 696. |
Among the following one statement is not correct when a junction diode is forward biasA. ElectronsB. `beta`-raysC. PositronD. protons |
| Answer» Correct Answer - D | |
| 697. |
The one has maximum activityA. UraniumB. PlutoniumC. RadiumD. Thorium |
| Answer» Correct Answer - C | |
| 698. |
On the bombardment of Boron with neutron. `alpha`-particle is emitted and product nucleus formed is……A. `._(6)C^(12)`B. `._(2)Li^(6)`C. `._(3)Li^(8)`D. `._(4)Be^(9)` |
| Answer» Correct Answer - C | |
| 699. |
An Electric field can deflectA. `alpha`-particlesB. `X`-raysC. NeutronsD. `gamma`-rays |
| Answer» Correct Answer - A | |
| 700. |
A nucleus `._n X^m` emits one `alpha` and one `beta` particles. The resulting nucleus is.A. `._n X^(m-4)`B. `._(n -2) Y^(m - 4)`C. `._(n -4) Z^(m -4)`D. `._(n -1) Z^(m -4)` |
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Answer» Correct Answer - D (d) `._n X^m overset (alpha)rarr ._(n -2)X^(m -4) overset (-beta)rarr ._(n-1) X^(m-4)`. |
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