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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.
| 551. |
A radioactive X decays to Y. In the radionuclide, the ratio of mass of element X to that of Y is `n` at `t=0`. It is observed that at time `t=t_(0)`, this ratio becomes equal to `1//n`. Assuming that all the decay products (except `gamma-` photons) remain in the sample, calculate half life of X. |
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Answer» Correct Answer - `[(log_(2))/(log_(n)).t_(0)]` |
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| 552. |
The mass number of a nucleus is.A. always less than its atomic numberB. always more than its atomic numberC. equal to its atomic numberD. sometimes more than and sometimes equal to its atomic number |
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Answer» Correct Answer - D In `._(1)H^(1),A = Z` |
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| 553. |
The mass number of a nucleus is equal toA. the number of neutron in the nucleusB. the number of protons in the nucleusC. the number of nucleous in the nucleusD. none of them |
| Answer» Correct Answer - C | |
| 554. |
Choose the correct optionA. Atomic mass unit `(u)` is `1//12` of the mass of a neutral carbon atom `._(6)C^(12)` in its lowest enegry state `1amu = 931MeV`B. `Z:` Atomic number (number of protons) `N:` Number of neutrons `A:` mass number `(A=Z+N)`C. Isotopes: nuclei having same number of protons but differenct number of neutrons Isobars: nuclei with same number of `A` Isotones: nuclei with same number of neutrons but different are correctD. All options are correct |
| Answer» Correct Answer - D | |
| 555. |
Size of nucleus is of the order ofA. `10^(-10)m`B. `10^(-15)m`C. `10^(-12)m`D. `10^(-19)m` |
| Answer» Correct Answer - B | |
| 556. |
During a nuclear fission reaction,A. a heavy nucleus breaks into two gragments by itselfB. a light nucleus bombarded by thermal neutrons breaks upC. a heavy nucleus bombared by thermal neutrons breaks upD. two light nuclei combine to given a heavier nuclear reactor possibly other products |
| Answer» Correct Answer - C | |
| 557. |
Which of the following is correct regarding binding enegry of nucleus?A. The nucleons are bound together in a nucleus and energy must be supplied to the nucleus to separate the consituent nucleons to large distance. The amount of energy needed to do this is called the binding enegry o fthe nucleus. If the nucleons are initially well separated and are brought to from the nucleus, this much energy is released `B.E. = Deltamc^(2), Deltam:` mass defect `B.E. = (Zm_(p)+Nm_(n) -M)c^(2)`B. `B.E.//`nucleon is maximum for `._(26)Fe^(56)`, hence iron is most stable elementC. For a nuclear reaction, `alpha`-value of the reaction is the difference between the rest mass enegry of the initial consituents and that of the final products `Q =U_(i) -U_(f)` `A+B rarr C+D` `Q-`value `={(m_(A) +m_(B)) - (m_(C) +m_(D))}c^(2)`D. All options are correct |
| Answer» Correct Answer - D | |
| 558. |
Which of the following is correct regarding nuclear fission? (i) The process in which a heavy nucleus is broken into two nearly equal fragments, is called ‘Nuclear Fission’ (ii) Nautral uranium has two istopes,`._(92)U^(238)`and `._(92)U^(235)` in the ratio `1 : 0.07`. the fission of `U^(235)` takes place only by fats neutrons, whereast that of `U^(235)` is possible by slow neutrons (iii) One of the reaction for fission of `U^(235)` `._(92)^(236)U+._(0)^(1)nrarr._(92)^(236)Urarr._(52)^(144)Ba+._(36)^(89)Kr+3_(0)^(1)n +` energy (iv) In the fission of one `U^(235)` nucleus, about `200MeV` of enegry of obtained. Most of this enegry is obtained from of `K.E.` of fragments obtained by fission, the rest of obained as `K.E.` of emitted neutrons, `gamma`-rays, heat and light radiationA. `(i),(ii),(iii)`B. `(ii),(iii),(iv)`C. `(i),(ii),(iv)`D. all |
| Answer» Correct Answer - D | |
| 559. |
Which of the following is corect regarding nucleus? (i) The radius of nucleus `R = R_(0)A^(1//3)` where `R_(0) = 1.1xx10^(-15)m, A =` mass number (ii) the density of nucleus `= rho=(M)/(V) = (mA)/((4)/(3)piR^(3))` `m:` mass of proton or neutron (iii) Nucleus stability depends on neutron-proton ratio `N//Z` `N//Z ~= `.6` for heaviest stable nucleus (iv) The total angular momentum of the nucleus is the resultant of all the spin and orbital angular momenta of the individual nucleon (nucleon means either proton or neutron). The total angular momentum of a nucleus is called the nuclear spin of that nucleus.A. `(i),(ii),(iii)`B. `(ii),(iii),(iv)`C. `(i),(ii),(iv)`D. all |
| Answer» Correct Answer - D | |
| 560. |
Which of the following is correct regarding nucleus forces?A. Nuclear force are attractive and charge independent `F_("pp") = F_("nn") = F_("pn")`B. Nuclear forces are short-range forces. When separtion is about `1fm`, nuclear forces are `50` to `60` times greater than electromagnetic forces. When separation exceeds `10fm`, nuclear forces are negligibleC. Nuclear forces are not central forces (like gravitational or electro-static), but spin dependent. For parallel spin, is higher.D. All options are correct |
| Answer» Correct Answer - D | |
| 561. |
The power obtained in a reactor using `U^(235)` disintergration is `1000kW`. The mass decay of `U^(235)` per hour isA. `1` microgramB. `10` microgramC. `20`microgramD. `40`microgram |
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Answer» Correct Answer - D `E = Pt = mc^(2)` `1000 xx 10^(3) xx 3600 = mxx(3xx10^(8))^(2)` `m = (36xx10^(8))/(9xx10^(16)) = 4xx10^(-8) kg = 4xx10^(-5)g` `= 40 xx 10^(-6)g = 40` microgram |
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| 562. |
The excitation energy of a hydrogen -like ion in its first excited state is `40.8 eV` Find the energy needed to remain the electron from the ionA. 54.4 eVB. 13.6 eVC. 40.8 eVD. 27.2 eV |
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Answer» Correct Answer - A |
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| 563. |
An element `A` decays into element `C` by a two-step process : `A rarr B + ._2 He^4` `B rarr C + 2 e overline` Then.A. A and C are isotopesB. A and C are isobarsC. A and B are isotopesD. A and B are isobars |
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Answer» Correct Answer - A |
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| 564. |
An element `A` decays into element `C` by a two-step process : `A rarr B + ._2 He^4` `B rarr C + 2 e overline` Then.A. `A` and `C^(2)` are isomersB. `A` and `C` are isotopesC. `A` and `B` are isobarsD. `A` and `B` are isotopes |
| Answer» Correct Answer - B | |
| 565. |
An element `A` decays into element `C` by a two-step process : `A rarr B + ._2 He^4` `B rarr C + 2 e overline` Then.A. `A` and `C` are isotopesB. `A` and `C` are isobarsC. `A` and `B` are isotopesD. `A` and `B` are isobars |
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Answer» Correct Answer - A |
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| 566. |
If a proton and anti-proton come close to each other and annihilate, how much energy will be released ?A. `1.5xx10^(-10)J`B. `3xx10^(-10)J`C. `4.5xx10^(-10)J`D. None of these |
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Answer» Correct Answer - B |
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| 567. |
The mass defect in a particular nuclear reaction is `0.3` grams. The amont of energy liberated in kilowatt hours is. (Velocity of light `= 3 xx 10^8 m//s`).A. `1.5xx10^(6)`B. `2.5xx10^(6)`C. `3xx10^(6)`D. `7.5xx10^(6)` |
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Answer» Correct Answer - D |
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| 568. |
Over what distance in free space will the intensity of a `5 eV` neutron beam reduced by a factor one - half ? `[T_(1//2) = 12.8 " min"]` |
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Answer» Correct Answer - [23808 km] |
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| 569. |
The table that follows shows some measurementf of the decay rate of a sample of `.^(128)I`, a radio nuclide often used medically as a tracer to measure the rate at which iodine is absorbed by the thyroid gland. `|{:("Time(min)",A (counts//s),"Time(min)",A(counts)//s,),(4,392.2,132,10.9,),(36,161.4,164,4.56,),(68,65.5,196,1.86,),(10,26.8,218,1.00,):}|` The half life `t_(1//2)` for this radio nuclide.A. `25 min`B. `50 min`C. `2.5 min`D. `5 min` |
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Answer» Correct Answer - A `A=A_(0)e^(-lambda t)rArrlnA=lnA_(0)-lambda t` `:. lambda=(lnA_(0)-lnA)/(t)=(6.2-0)/(225)=0.0275 min^(-1)` `:. T_(1//2)=(0.693)/(lambda)=25 min` |
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| 570. |
A nuclear reaction is given as `A + B rarr C+ D` Binding energies of `A,B,C,` and `D`, are given as `B_1,B_2,B_3` and `B_4`. Find the energy released in the reaction. |
| Answer» `(B_3 _B_4) -(B_1 + B_2)` | |
| 571. |
Most of energy released in the fission is carried byA. neutronsB. fission fragmentsC. neutrons and fragments carry equallyD. positrons |
| Answer» Correct Answer - B | |
| 572. |
Regarding Prompt neutronsA. They are highly energeticB. They constitute `99%`C. Cannot initiate chain reactionD. `1,2,3` are correct |
| Answer» Correct Answer - D | |
| 573. |
The critical mass of a fissionable material isA. `0.1 kg` equivalentB. The minimum mass needed for chain reactionC. The rest mass equivalent to `1020` jouleD. `0.5 kg` |
| Answer» Correct Answer - B | |
| 574. |
Nuclear reactios obey the law of conservation ofA. Mass and energyB. ChargeC. MomentumD. All of the above |
| Answer» Correct Answer - D | |
| 575. |
The reactor in which the number of fissionable nuclides produced are more than the used it calledA. breeder reactorB. Pressurised reactorC. Heterogenous reactorD. Homogenerous reactor |
| Answer» Correct Answer - A | |
| 576. |
The most abundant isotope of helium has a `._(2)^(4)H` nucleus whose mass is `6.6447 xx 10^(-27) kg` . For this nucleus, find (a) the mass defect and (b) the binding energy. Given: Mass of the electron: `m_e =5.485799 xx 10^(-4) u`, mass of the proton: `m_(P) =1.007276 u` and mass of the neutron: `m_(n) =1.008 665 u`. |
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Answer» The symbol `._2^4He` indicates that the helium nucleus contains `Z = 2` protons and `N = 4-2=2` neutrons. To obtain the mass defect `Delta m`, we first determine the sum of the individual masses of the separated protons and neutrons. Then, we subtract from this sum the mass of the mass of the nucleus. (a) We find that the sum of the individual masses of the nucleons is `underbrace (2(1.6726 xx 10^(-27) kg))_("two protons") + underbrace (2(1.6749 xx 10^(-27) kg))_("two neutrons")` `=6.6950 xx 10^(-27) kg` This value is greater than the mass of the intact He necleus, the mass defect is `Delta m=6.6950 xx 10^(-27) kg -6.6447 xx 10^(-27) kg` `=0.0503 xx 10^(-27) kg` (b) According to Eq. (iv), the binding energy is given by `Delta E_(BE) =(Delta m)c^(2) =(0.0503 xx 10^(-27) kg) (3.00 xx 10^(8) ms^(-1))^(2)` ` =4.53 xx 10^(-12) J` Usually, binding energies are expressed in energy units of electrons volt instead of joule `(1 eV =1.60 xx 10^(-19) J)`. because Binding energy `=(4.53 xx 10^(-12) J)((l eV)/(1.60 xx 10^(-19) J ))` `=2.83 xx 10^(7) eV =28.3 MeV` In this result, one million electron volt is denoted by the unit MeV. The value of `28.3 MeV` is more than two million times greater than the energy required to remove an orbital electron from an atom. |
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| 577. |
The radius of germanium `(Ge)` nuclide is measured to be twice the radius of `._(4)^(9)Be`. The number of nucleons in `Ge` areA. `73`B. `74`C. `75`D. `72` |
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Answer» Correct Answer - D |
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| 578. |
Identify the correct ascending order of `alpha, beta` and `gamma` with reference to their ioninzing power (I) `alpha`-ray (II) `gamma`-ray (III) `beta`-rayA. II,III,IB. I,III,IIC. II,I,IID. I,II,III |
| Answer» Correct Answer - A | |
| 579. |
Identify the correct ascending order of `alpha, beta` and `gamma` with reference to their ioninzing power (I) `alpha`-ray (II) `gamma`-ray (III) `beta`-ray |
| Answer» Correct Answer - B | |
| 580. |
After a certain lapse of time, fraction of radioactive polonium undecayed is found to be `12.5%` of the initial quantity. What is the duration of this time lapsed if the half life of polonium is 138 days ?A. `414` daysB. `407` daysC. `421` daysD. `410` days |
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Answer» Correct Answer - A `(dN)/(dt)=lambda N`, here `N="avagadrono"//238, lambda=0.693//T_((1)/(2))` |
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| 581. |
A sample of `.^(18)F` is used internally as a medical diagnostic tool to look for the effects of the positron decay `(T_(1//2) =110 min)` . How long does it take for `99%` of the `.^(18)F` to decay? |
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Answer» Correct Answer - `12.2 h` Radioactive deacy equation is `N =N_0 e^(- lambda t)=N_0 e^(-1n(2)t//T)" "( because lambda=(In2)/(T))` After decay of `99%` of the initial sample, only `1%` will be left, i.e., `N//N_(0)=1%` `:. (N)/(N_(0))=(1)/(100)=e^(-1n(2)t//T)` If we take the natural logarithm, we have `-1n 100= -1n 2 xx(t)/(T)` Which on solving for t yields `t=(1 n100)/(1n2)xx T =(log 100)/(log2) xxT` `=(2)/(0.3010) xx 10 =731 min =12.2 h`. |
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| 582. |
In a radioactive substance at `t = 0`, the number of atoms is `8 xx 10^4`. Its half-life period is `3` years. The number of atoms `1 xx 10^4` will remain after interval.A. 9 yearsB. 8 yearsC. 6 yearsD. 24 years |
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Answer» Correct Answer - A (a) By formula `N = N_0((1)/(2))^(t//T)` or `10^4 = 8 xx 10^4 ((1)/(2))^(t//3)` or `((1)/(8)) = ((1)/(2))^(t//3)` or `((1)/(2))^3 = ((1)/(2))^(t//3) rArr 3 = (t)/(3)` Hence `t = 9 years`. |
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| 583. |
Consider the following nuclear reaction, `X^200rarrA^110+B^90+En ergy` If the binding energy per nucleon for X, A and B are `7.4 MeV`, `8.2 MeV` and `8.2 MeV` respectively, the energy released will beA. `200 MeV`B. `160 MeV`C. `110 MeV`D. `90 MeV` |
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Answer» Correct Answer - B Energy released=BE of products-BE of reactants |
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| 584. |
The atomic mass of thorium `._(90)^(234)Th` is `234.04359 u`, while that of protactinium `._(91)^(234)Pa` is `234.04330 u`. Find the energy released when `beta` decay changes `overset(234)(90)Th` into |
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Answer» To find the energy released, we follows the usual procedure of determining how much the mass has decreased because of the decay and then calculating the equivalent energy. The decay and the masses are shown below: `underset(234.04359 u) (._(90)^(234)Th) rarrubrace(._(91)^(234)Pa+._(1)^(0)e)_("234.04330 u")` When the `._(90)^(234)Th` nucleus of a thorioum atom is converted into a `._(91)^(234)Pa ` nucleus, the number of orbital electrons remains the same, so the resulting protactinium atom is missing one orbital electron. However, the given mass includes all `91` electrons of a neutral protactinium atom. In effect, then, the value of `234.04330 u` for `._(91)^(234)Pa` already includes the mass of the `beta` particle. Tha mass decrease that accompanies the `beta^(-)` decay is `234.04359 u -234.04330u = 0.00029 u`. The equivalent energy `(1 u=931.5MeV)` is` 0.27 MeV`. This is the maximum kinetic energy that the emittted electron can have. |
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| 585. |
The radius of the oxygen nucleus `._(8)^(16)O` is `2.8 xx 10^(-15)m`. Find the radius of lead nucleus `._(82)^(205)Pb`. |
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Answer» `R_(0) = 2.8 xx 10^(-15)m, A_(0) = 16, A_(Pb) = 205 R prop A^(1//3)` `(R_(0))/(R_(Pb)) = ((A_(0))/(A_(Pb)))^(1//3) = (2.8 xx 10^(-15))/(R_(Pb)) = ((16)/(205))^(1//3)` `R_(Pb) = 6.55 xx 10^(-15)m`. |
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| 586. |
`alpha`-particles of enegry `400 KeV` are boumbardel on nucleus of `._(82)Pb` . In scattering of `alpha`-particles, it minimum distance from nucleus will beA. 0.59 nmB. 0.59 ÅC. 5.9 ÅD. 0.59 pm |
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Answer» Correct Answer - D |
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| 587. |
The element Curium `_(96)^(248Cm` has a mean life of `10^(13)` second. Its primary decay mode are spontaneous fission and a `alpha-decay` , the former with a probability of `8%` and the latter with a probability of `92%` Each fission released `200 MeV` of energy . The masses involved in `alpha - decay` are as follows `_(96)^(248) Cm = 248.072220 u, _(94^(244) Pu = 244.064100 u and _(2)^(4) He = 4 .002603 u ` calculate the power output from a sample of `10^(20)` Cm atom `(1 u = 931 MeV//e^(2))` |
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Answer» Correct Answer - `3.32xx10^(-5)Js^(-1)` Activity of fission `=A=(dN)/(dt)=lambdaN=(1)/(tau)N=(10^(20))/(10^(13))=10^(7)` as probability is `8%` so reactivity `=(8)/(100)xx10^(7)=8xx10^(5)Mev//sec`. Rate of decay of `alpha`-particle `=(92)/(100)xx10^(7)=92xx10^(5)` `Delta M` for `alpha` decay `=0.00517 "amu"` Energy released per `alpha`-decay `=92xx10^(5)xx5.1363 Mev//sec` Total power output `=16xx10^(7)+4.725xx10^(7)=33.16 mu W` |
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| 588. |
In radioactive decay of a radioactive atom, its stability increases. It is a spontaneous process.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - A (a) In the spontaneous process potential energy decreases therefore stability increases. |
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| 589. |
The `beta`-decay process, discovered around `1900`, is basically the decay of a neutron `(n)`, In the laboratory, a proton `(p)` and an electron `(e^(-))` are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a tro-body dcay process, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process i.e., `n rarr p + e^(-)+overset(-)v_(e )`, around `1930`, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino `(overset(-)V_(e ))` to be massless and possessing negligible energy, and neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is `0.8xx10^(6)eV`. The kinetic energy carried by the proton is only the recoil energy. What is the maximum energy of the anti-neutrino?A. zeroB. Much less than `0.8xx10^(6)eV`C. Nearly `0.8xx10^(6)eV`D. Much larger than `0.8xx10^(6)eV` |
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Answer» Correct Answer - C `KE_(max)="of"beta^(-)` `Q=0.8xx10^(6)eV` `KE_(P)+KE_(beta^(-))+KE_(overset(-)v)=Q` `KE_(P)` is almost zero When `KE_(beta^(-))=0` then `KE_(overset(-)v)=Q-KE_(P)` `~=Q` |
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| 590. |
Which one of the following is a possible nuclear reaction ?A. `._5^10 B + ._2^4 He rarr ._7^13N + ._1^1H`B. `._93^239Np rarr ._94^239Pu + beta^- + overline v`C. `._11^23Na + ._1^1 H rarr ._10^20 Ne + ._2^4 He`D. `._7^11N + ._1^1H rarr ._6^12 C + beta^- + v` |
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Answer» Correct Answer - B (b) In any nucear reaction mass number and atomic number should remain conserved. Reaction ( c) satisfies this condition. Also for `._93^239 Np`. Neutron to proton ratio is greater than `1.52` which makes it unstable. |
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| 591. |
Starting with a sample of pure `.^66 Cu, 7//8` of it decays into `Zn` in `15 min`. The corresponding half-life is.A. 10 minB. 5 minC. 15 minD. `7 (1)/(2) min` |
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Answer» Correct Answer - B (b) `N = N_0 (1 - e^(-lamda t))` `rArr (N_0 - N)/(N_0) = e^(-lamda t)` `:. (1)/(8) = e^(-lamda t)` `rArr 8 = e^(lamda t)` `rArr 3 1n 2 = lamda t` `rArr lamda = (3 xx 0.693)/(15)` Half-life period `t_(1//2) = (0.693)/(3 xx 0.693) xx 15` `t_(1//2) = 5 min`. |
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| 592. |
Starting with a sample of pure `.^(66)Cu, 7//8` of it decays into `Zn` in `15` minute. The corresponding half-life is:A. `10` minuteB. `15` minuteC. `5` minuteD. `7(1)/(2)`minute |
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Answer» Correct Answer - C `N=N_(0)(1-e^(-lambdat))` `rArr (N_(0)-N)/(N_(0))=e^(-lambdat)` `:. (1)/(8)=e^(-lambdat)` `rArr 8=e^(lambdat t)" "rArr 3 ln 2=lambda t " "rArr=(3xx0.693)/(15)` `t_(1//2)=(0.693)/(3xx0.693)xx15" "t_(1//2)=5 min` |
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| 593. |
If radius of the `._(13)^(27)Al` nucleus is estimated to be `3.6` Fermi, then the radius of `._(52)^(125)Te` nucleus be nerarly:A. `6` FermiB. `8` FermiC. `4` FermiD. `5` Fermi |
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Answer» Correct Answer - A `R R_(0)(A)^(1//3)` `(R_(Al))/(R_(Te))=(R_(0)(A_(Al))^(1//3))/(R_(0)(A_(Te))^(1//3))=(3)/(5)` `:. R_(Te)=(5)/(3)xx3.6` `R_(Te)=6 "Fermi"` |
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| 594. |
When high energy alpha particles `(._(2)^(4)He)` pass through nitrogen gas, an isotope of oxygen is formed with the emission of particles named x, the nuclar reaction is `._(7)^(14)N+._(2)^(4)Herarr._(8)^(17)O+x` What is the name of x?A. ElectronB. ProtonC. NeutronD. Positron |
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Answer» Correct Answer - B |
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| 595. |
How many `beta`-particles are emitted during one hour by `1.0 mu g` of `Na^(24)` radionuclide whose half-life is `15` hours? [Take `e^((-0.693//15))=0.955=0.9555`, and avagadro number `=6xx10^(23)`] |
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Answer» Correct Answer - `(6xx10^(23)xx10^(-6))/(24)[1-e^(-0.693//15)]=1.128xx10^(15)` No. of particle in time `t` is `N=N_(0)(1-e^(-lambdat))=N_(0)(1-e-t(ln2)/(T_(1//2)))` `N_(0) =` No. of nuclei in `1mug` of `Na^(24)` `N_(0)=(6xx10^(23)xx10^(-6))/(24)` `:. N=(6xx10^(23)xx10^(-6))/(24)xx(1-e^(-(0.693)/(15)xx1))=1.128xx10^(15)` |
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| 596. |
The radio active nuclides `A` and `B` have half lives `t` and `2t` respectivey. If we start an experiment with one mole of each of them, the mole ratio after time interval of `6t` will beA. `1:2`B. `1:8`C. `1:6`D. `1:1` |
| Answer» Correct Answer - B | |
| 597. |
Why does the fussion occur at high temeprature?A. Atoms are inised at high temperatureB. Molecuels break up at high temperatureC. Nuclei break up at high temperatureD. Kinetic energy is high enough to overcome repulsion between nuclei |
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Answer» Correct Answer - D |
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| 598. |
Mass defect of an atom refers toA. Inaccurate measurement of nucleonsB. Mass annihilated to produce energy to bind the nucleusC. Packing fractionD. Difference in number of neutrons and protons in the nucleus |
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Answer» Correct Answer - B |
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| 599. |
Calculate the specific activities of `Na^(24) & U^(235)` nuclides whose half lives are `15` hours and `7.1xx10^(8)` years respectively. |
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Answer» Correct Answer - `(N_(A))/(24)xx(0.693)/(15xx60xx60)= 3.2 xx10^(17) dps` & `(N_(A))/(235)xx(0.693)/(7.1xx10^(8)xx365xx86400)= 0.8xx10^(5) dps` |
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| 600. |
In which of the following processes, the number of protons in the nuleus increase ?A. `alpha-decay`B. `beta^(-)-decay`C. `bata^(+)-decay`D. k-capture |
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Answer» Correct Answer - B |
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