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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.
| 851. |
On an average, a neutron loses half of its energy per collision with a quesi-free proton. To reduce a `2 MeV` neutron to a thermal neutron having energy `0.04 eV`, the number of collisions requaired is nearly . |
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Answer» Correct Answer - `[N cong 26]` |
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| 852. |
Which of the following assertions are correct? (i) A neutron can decay to a proton only inside a nucleus (ii) A proton can change to a neutron only inside a nucleus (iii) An isolated neutron can change into a proton (iv) An isolated proton can change into a neutronA. A neutron can decay to a proton only inside a nucleusB. A proton can change to a neutron only inside a nucleusC. An isolated neutron can change into a protonD. An isolated proton can change into a neutron |
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Answer» Correct Answer - B, C |
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| 853. |
A `Bi^(210)` radionuclide decays via the chain `Bi^(210)(beta^(-)-"decay")/(lambda_(1))Po^(210)(alpha-decay)/(lambda_(2)) Pb^(206)` (stable), where the decay constants are `lambda_(1) = 1.6 xx 10^(-6)s^(-1), T_(1//2) ~~ 5 days , lambda_(2) = = 5.8 xx 10^(-8)s^(-1), T_(1//2) ~~ 4.6` months. `alpha & beta` activites of the `1.00 mg` a month after its manufacture is `(x)/(5) xx 10^(11)` Find `x.2^((1)/(4.6)) = 0.86` |
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Answer» Correct Answer - `7` ` Bi^(250)underset(lambda_(1))rarrPo^(210)underset(lambda_(2))rarrPb^(208)` Let `N_(A), N_(B)` and `N_(C )` be the no of nuclei of `Bi^(210), Po^(210)` and `Pb^(208)` respectively. `N_(0)` be the no of nuclei of `Bi^(210)` at `t=0` `N_(0)=(1xx10^(-3))/(210)xx6.02xx10^(23) " "~~2.9xx10^(18)` `(dN_(A))/(dt)= -lambda_(1)N_(A)"_____"`I `(dN_(B))/(dt)=-lambda_(2)N_(B)+lambda_(1)N_(A)"______"`(II) `(dN_(c ))/(dt)= lambda_(2)N_(B)"_____"`III Solving equ. I , we get `N_(A)= "No" e^(-)lambda_(1)t` From equation II, `(dN_(B))/(dt)=-lambda_(2)N_(B)+lambda_(1)N_(o)e^(-lambda_(1)t)` This is first order line differential equation. On solving we get `N_(B)=e^(-lambda_(2)t)[(lambda_(1)N_(0))/(lambda_(2)-lambda_(1))e^((lambda_(2)-lambda_(2))t)+C]` where `C` is a constant The value of `C` can be obtained by taking `t-0, N_(B)=0` `rArrC=(lambda_(1)N_(0))/(lambda_(1)-lambda_(2))` `:. N_(B)=(lambda_(1)N_(o))/(lambda_(1)-lambda_(2))[e^(-lambda_(2)t)-e^(-lambda_(1)t)] :. beta`-activity `A_(B)=|(dN_(A))/(dt)|=N_(0)lambda_(1)e^(-lambda_(1)t)=2.9xx10^(18)xx1.6xx10^(-6)xx(1)/(2^(6))S^(-1)=0.72xx10^(11)S^(-1)` `alpha`-activity `A_(alpha)=lambda_(2)N_(B)=(lambda_(1)lambda_(2)N_(0))/(lambda_(1)-lambda_(2))(e^(-lambda_(2)t)-e^(-lambda_(1)t))~~(lambda_(1)lambda_(2)N_(0))/(lambda_(1))e^(-lambda_(2)t)=(lambda_(2)N_(0))/((1)/(2^(4xx6)))=1.4xx10^(11)s^(-1)` |
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| 854. |
To investigate the beta-decay of `Mg^(23)` radionuclide, a counter was activated at the moment `t=0`. It registered `N_(1)` beta particles by a moment `t_(1)=2 sec`, and by a moment `t_(2)=3t_(1)` the number of registered beta particles was `2.66` times greater. the mean life of the given nuclei isA. `3 sec`B. `7 sec`C. `16 sec`D. `14 sec` |
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Answer» Correct Answer - C `Mg^(23)` give one `beta`-particle in one disintergration. Number of `beta`-particle `=` number if decayed nuclei `=` number of disintergrations`=N_(0)(1-e^(-lambda t))` `:. N_(0)(1-e^(-lambda t))=N_(1) &` `N_(0)(1-e^(- lambda t_(2)))=N_(2)=2.66N_(1), 3t_(1)` after simplification `e^(-2 lambda)=0.882, rArr(1)/(lambda)= tau=16 sec` |
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| 855. |
A radionuclide has a half life of 10 s. Which of the following statements are correct? (a) In a sample, probability that a nucleus will decay in next 10 s is 0.5. (b) In a sample, the probability that a nucleus which has survived first 10 s, will decay in next 10 s is 0.5. (c) The probability that a nucleus which has survived first 10 s, will decay in next 10 s is 0.25. (d) If a sample of radionuclide has 4 nuclei, two nuclei will decay in next 10 s. |
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Answer» Correct Answer - (a),(b) |
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| 856. |
A free neutron at rest, decays into three particles: a proton, an electron and an anti neutrino. `._(0)^(1)nrarr._(1)^(1)P+._(-1)^(0)e+vecv` The rest masses are: `m_(n)=939.5656 MeV//c^(2)` `m_(p)=938.2723 MeV//C^(2) m_(e)=0.5109 MeV//c^(2)` In a particular decay, the antineutrino was found to have a total energy (including rest mass energy) of 0.0004 MeV and the momentum of proton was found to be equal to the momentum of electron. Find the kinetic energy of the electron. |
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Answer» Correct Answer - 0.7820 MeV |
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| 857. |
A free neutron decays into a proton, an electron andA. a beta particleB. an alpha particleC. an antineutrinoD. a neutrino |
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Answer» Correct Answer - C |
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| 858. |
The nucleus of `._(90)^(230)Th` is unstable against `alpha`-decay with a half-life of `7.6xx10^(3)"years"`. Write down the equation of the decay and estimate the kinetic energy of the emitted `alpha`-particle from the following data: `m(._(90)^(230)Th)=230.0381 "amu", m(._(68)^(226)Ra)=226.0254 "amu" m(._(2)^(4)He)=4.0026"amu"`. |
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Answer» The equation of the decay is `._(90)^(230)Th rarr_(88)^(225)Ra+_(2)^(4)He+Q` The energy `Q` is given by `Q=[m(Th)-m(Ra)-m(He)]c^(2)` Using the given data and `c^(2)=931.5 MeV//"amu"`, we get `Q=9.41 MeV`. This energy is shared by `Ra` and `He`. If the original nucleus `Th` is at rest, i.e. if the momentum of the system before `alpha`-decay is zero, the total momentum after the decay will also be zero. Thus `Ra` and `He` will have equal and opposite linear momenta. `:. m_(He)v_(He)= -m_(Ra)V_(Ra)` or `m_(He)v_(He)=m_(Ra)^(2)v_(ra)^(2)` ......(i) Now `(K.E.(HE))/(K.E.(Ra))=(1//2m_(He)v_(He)^(2))/(1//2m_(Ra)v_(Ra)^(2))=((m_(He)v_(He))/(m_(ra)v_(Ra)))xx((m_(Ra))/(m_(He)))` `=(m_(Ra))/(m_(He))`" " [use Eq.(i)] `~~(226)/(4)~~56.6` ......(ii) i.e. the kinetic energy of `He` is `56.5` times that of `Ra`, the total energy being `K.E.(He)+K.E.(Ra)=9.41 MeV` " "(ii) From eqs (ii) and (iii) we have `K.E. (Ra)=0.16 MeV` and `K.E.(He)~~9.25 MeV`. |
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| 859. |
(a) Why is an alpha particle, rather than neutrons and protons, emitted from an unstable nucleus? (b) A nucleus `.^(A)X` alpha decays to produce Y. The vol- ume of nucleus Y is nearly 56 times the volume of alpha particle. Find the mass number (A) of X. |
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Answer» Correct Answer - (b) A=228 |
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| 860. |
(A): Free Neutron decays into proton, electron, and antinuetrino (R ): Neutron is unstable outide the nucleusA. `A` and `R` are true and `R` is the correct explanation of `A`B. `A` and `R` are true and `R` is not the correct explanation of `A`C. `A` is true, `R` is falseD. `A` is false but `R` is true. |
| Answer» Correct Answer - A | |
| 861. |
Which of the following is most unstable ?A. NeutronB. ProtonC. ElectronD. `alpha`-particle |
| Answer» Correct Answer - A | |
| 862. |
All the radioactive elements are ultimately converted in lead. All the elements above lead are unstable.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 - C ( c) All those elements which are heavier than lead are radioactive. This is because in the nuclei of heavy atoms, besides the nuclear attractive forces, repulsive forces between the protons are also effective and these forces reduce the stability of the nucleus. Hence, the nuclei of heavier elements are being converted into lighter and lighter elements by emission of radioactive radiation. When they are converted into lead, the emission is stopped because the nucleus of lead is stable (or leaf is most stable elements in radioactive series). |
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| 863. |
Staements I: `._zX^4` undergoes `2 alpha`-decays, `2 beta`-decays (negative `beta`) and `2 gamma`-decays. As a result, the daughter product is `._z._-2X^(A-B)`. Staements II: In `alpha`-decay, the mass number decreases by 4 unit and atomic number decreases by `2` unit. In `beta`-decay (negative `beta`), the mass number remains unchanged and atomic number increases by `1` unit. In `gamma`-decay, mass number and atomic number remain unchanged.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) Reason is true by definition and correctly explains the assertion. Namely, `._Z X^A` undergoes `2 alpha` decays, `2 beta` decay (negative `beta`) and `2 gamma` decays. As a result the daughter product is `._(Z -2) y^(A-8)`. |
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| 864. |
Thorium `._(90)^(298)Bi` The produces a daughter nucleus that is radioactive. The augther in turn, produces it own radioactive daugher and so on. This process continues until bismuth `._(83)^(212)Bi` is reaches. What are the total number `N_(alpha)` of `alpha`-particle and the total number `N_(beta)` of `beta^(-)`-paricles that are generated in this series or radioactive decays? |
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Answer» `._(90)^(228)Th rarr ._(83)^(212)Bi` Decrease in mass number `= 228 - 212 = 16` Number of `alpha`-particles = `4` decrease in atomic number due to `alpha`-particles `= 8` Actual decrease in atomic number `= 7` i.e on e`beta^(-)` aprticle. |
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| 865. |
The mass density of a nucleus varies with mass number `A` asA. `A^(2)`B. `A`C. constantD. `1/A` |
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Answer» Correct Answer - C |
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| 866. |
If the nuclear force between two protons, two neutrons and between proton and neutron is denoted by `F_(p p), F_(n n)` and `F_(pn)` respectively, thenA. `F_(p p) ~~F_(n n)~~F_(pn)`B. `F_(p p) ne F_(n n)` and `F_(p p)=F_(n n)`C. `F_(p p)=F_(n n)=F_(pn)`D. `F_(p p) ne F_(n n) ne F_(pn)` |
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Answer» Correct Answer - C |
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| 867. |
The energy equivalent of one atomic mass unit isA. `1.6xx10^(-19) J`B. `6.02xx10^(23) J`C. `931` MeVD. `9.31` MeV |
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Answer» Correct Answer - C |
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| 868. |
Solar energy is mainly caused due toA. fusion reactionB. fission reactionC. combustion reactionD. chemical reaction |
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Answer» Correct Answer - A |
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| 869. |
Solar energy is mainly caused due toA. fusion of protons during sythesis of heavier elementsB. gravitational contractionC. burning of hydrogen in the oxygenD. fission of uranium present in the sun |
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Answer» Correct Answer - A |
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| 870. |
Solar energy is mainly caused due toA. nuclear fissionB. nuclear fusionC. gravitational contractionD. combustion |
| Answer» Correct Answer - B | |
| 871. |
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` isotopesD. `A` and `B` are isobars |
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Answer» Correct Answer - A (a) `alpha` decay decrease the mass number by `4` and atomic number by `1`. Here atomic number of `C` is same as that of `A`. |
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| 872. |
If `N_(0)` is the original mass of the substance of half - life period `t_(1//2) = 5 year` then the amount of substance left after `15` year isA. `N_0//8`B. `N_0//16`C. `N_0//2`D. `N_0//4` |
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Answer» Correct Answer - A (a) `N = N_0 ((1)/(2))^((t)/(T_1//2)) = N_0 ((1)/(2))^((15)/(5)) = (N_0)/(8)`. |
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| 873. |
When a deuteron of mass 2.0141 a.m.u and negligible K.E. is absorbed by a Lithium `(._(3)Li^(6))` nucleus of mass 6.0155 a.m.u. the compound nucleus disintegration spontaneously into two alpha particles, each of mass 4.0026 a.m.u. Calculate the energy carried by each `alpha` particle. |
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Answer» `._(3)^(6)Ll+._(1)^(2)Drarr._(2)^(4)He+._(2)^(4)He+Q` `m(._(3)^(6)Li)=6.0155u` `m(._(1)^(2)H)=2.0141 u` Total initial mass `=8.0296 u` TOtal final mass `=2m(._(2)^(4)He)=2xx4.0026` `=8.0052u` Mass defect, `Delta m=8.296-8.0052` `=0.0244u=0.024xx1.66xx10^(-27)kg` Energy released, `Q-Deltamxxc^(2)=0.244xx1.66xx10^(27)xx(3xx10^(8))^(2)` `=3.645xx10^(-12)J` Energy of each `alpha`-particle=`1.8225xx10^(-12)J` |
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