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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.
| 751. |
`M, M_(n) & M_(p)` denotes the masses of a nucleous of `._(Z)X^(A)` a neutron, and a proton respectively. If the nucleus is separated in to its individual protons and neutrons thenA. `M=(A-Z)M_(n)+ZM_(p)`B. `M=ZM_(n)+(A-Z)M_(p)`C. `M gt (A-Z)M_(n)+ZM_(p)`D. `M lt (A-Z)M_(n)+ZM_(p)` |
| Answer» Correct Answer - D | |
| 752. |
A radioactive nuclide is produced at the constant rate of n per second (say, by bombarding a target with neutrons). The expected number `N` of nuclei in existence `t` s after the number is `N_(0)` is given byA. `N=N_(0)e^(-lambdat)`B. `N=n/lambda+N_(0)e^(-lambdat)`C. `N=n/lambda+(N_(0)-n/lambda)e^(-lambdat)`D. `N=n/lambda+(N_(0)+n/lambda)e^(-lambdat)` |
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Answer» Correct Answer - C |
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| 753. |
A nuclear decay is expressed as ltbr. `._(6)C^(11) rarr ._(5)B^(11)+beta^(+)+X` Then the unknown particle `X` isA. neutronB. antineutrinoC. protonD. neutrino |
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Answer» Correct Answer - D |
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| 754. |
In the given nuclear reaction `A, B, C, D, E` represents `._92 U^238 rarr^(alpha) ._BTh^A rarr^(beta) ._D Pa^C rarr^(E) ._92 U^234`.A. `A = 234, B = 90, C = 234, D = 91, E = beta`B. `A = 234, B = 90, C = 238, D = 94, E = alpha`C. `A = 238, B = 93, C = 234, D = 91, E = beta`D. `A = 234, B = 90, C = 234, D = 93, E =alpha` |
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Answer» Correct Answer - A (a) `._92U^238 overset (alpha)rarr ._90Th^234 overset (beta)rarr ._91 Pa^234` `overset (E(._1beta^0))rarr ._92 U^234`. |
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| 755. |
Consider the nuclear decay reaction. `._(z)^(A)Xrarrunderset(z-1)AY+_(+)._(1)^(0)e+v` For the above `beta^(+)` decay reaction to the feasible what should be the minimum difference in atomic masses of X and Y? Rest mass of v is nearly zero. |
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Answer» Correct Answer - Twice the mass of an electron. |
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| 756. |
In nuclear reaction `._(4)Be^(4)+._(z)X^(A)rarr._(z+2)Y^(A+3)+R , R` denotesA. electronB. positronC. protonD. neutron |
| Answer» Correct Answer - A | |
| 757. |
A radioactive nuclide is produced at the constant rate of n per second (say, by bombarding a target with neutrons). The expected number `N` of nuclei in existence `t` s after the number is `N_(0)` is given byA. `N=N_(0)e^(-lambda t)`B. `N=N_(0)e^(-lambda t)`C. `N=n/lambda+(N_0+n/lambda)e^(-lambda t)`D. `N=n/lambda+(N_0+n/lambda)e^(-lambda t)` |
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Answer» Correct Answer - c `(dN)/(dt) =n -lambda N` because the population `N` is simultaneously incereasing at rate n and decreasing due to decay at rate `lambda N`. ` underset(N_(0))overset(N) (int)(dN)/(n-lambdaN)=underset(0) overset(t)(int)dt` `(1)/(lambda)In ((n-lambdaN_(0))/(n-lambdaN))=t` `N=(n)/(lambda)+(N_(0)-(n)/(lambda))e^(-lambdat)`. |
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| 758. |
In the nuclear process , `C_(6)^(11) rarr _(2)B^(11) + beta^(+) + X , X` stands for……. |
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Answer» `._6C^(11) rarr ._2B^(11) +beta^(+) +X rArr `._6C^(11) rarr ._2B^(11) +._(+1)^0e +X` (neutrino) The balancing of atomic number and mass number is OK. Therefore X stands for energy. |
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| 759. |
Consider the following reaction : `H_(1)^(2) + H_(1)^(2) = He_(2)^(4) + Q` Mass of the deuterium atom ` = 2.0141 u` Mass of helium atom `= 4.0024u ` This is a nuclear ……… reaction in which the energy `Q` released is …….. MeV`.A. `12`B. `6`C. `24`D. `48` |
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Answer» Correct Answer - c `._1H^2 + .`_1H^(2) rarr `._1He^4 + Q` `Delta m=m(._2He^4) -2 m.(_1H^2)` `Delta m=4.0024 -2(2.0141)` `Delta m= -0.0258u` Now, `Q=c^(2)Delta m` `=(0.0258)(931.5)MeV` `~~24 MeV`. |
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| 760. |
Determine the average `.^(14)C` activity in decays per minute per gram of natural carbon found in living organisms if the concentration of `.^(14)C` relative to that of `.^(12)C` is `1.4 xx10^(-12)` and half -life of `.^(14)C` is `T_(1//2)=57.30` years. |
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Answer» The decay constant is `lambda =(0.693)/(T_(1//2)) =(0.693)/(5730 xx 5.26 xx10^(5))` As `1` year `=3.01 xx 10^(9)` min and for `1.0 g` of carbon, the number of moles is `n=1//2`, so `N=nN_(A) =((1)/(2))(6.02 xx 10^(23))=5.0 xx10.^(22) "nulei" g^(-1)` . The given concentration factor is `(.^(12)C)/(.^(12)C) =1.4 xx10^(-12)` Thus, the number of `((.^(14)C)/(.^(12)C)) =(5.0 xx10^(22))(1.4 xx10^(-12))=7.0 xx10^(10) .^(14)C "nuclei" g^(-1)` The activity,number of deacys per minute, is `(Delta N)/(Delta t) lambda N=(2.30 xx 10^(-10))(7.0 xx10^(10))` `16 .^(14)C "decays" g.^(-1) min.^(-1)` . |
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| 761. |
Consider the reaction `._1^2H+_1^2H=_2^4He+Q`. Mass of the deuterium atom`=2.0141u`. Mass of helium atom `=4.0024u`. This is a nuclear………reaction in which the energy `Q` released is…………MeV. |
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Answer» This is a nuclear fusion reaction. Energy released`=(Delta m)[(931.5 MeV//u]` `=[4.0024 -2 xx 2.0141]xx931.5 MeV` ` = -24.03 MeV` (heat release) |
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| 762. |
Determine the product of the reaction: `._3^7Li +._2^4He rarr ? +n` What is the Q value of the reaction? |
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Answer» Given: reaction Find:` Q=`? In order to balance the reaction, the total amount of nucleons (sum of A-number) must be the same on both sides. Same for the Z-number. Number of nucleons (A): `7+4 =X+1 rArrX =10` Number of nucleons (Z): `3+2 =X +0 rArr Y =5` Thus,it is B, i.e., `._3^7Li +._2^4He rarr. _5^(10)B +._0^1n` . |
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| 763. |
Find the binding energy of a nucleus consisting of equal numbers of protons and neutrons and having the radius one and a half time smaller than of `Al^(27)` nucleus. |
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Answer» Correct Answer - `Be^(8),E_(b)=56.5MeV` The mass of defect is given by `E_(b)=Deltamc^(2)` where `Delta m =` mass defect If `m_(n)=` mass of neutron, `m_(p)=` mass of proton, `m_(N)=` mass of formed nucleus For binding energy of `._(z)X^(A)`, Here the number of proton `=Z`, The number of neutron `=A-Z` `:. Deltam =Zm_(p)+(A-Z)m_(n)-m_(N)` If atomic masses of the neucleus is given, then `:. Delta m=Zm_(p)+Zm_(e )+(A-Z)m_(n)-(m_(N)+Zm_(e ))` `=Z(m_(p)+m_(e ))+(A-Z)m_(n)-M_(N)` `=ZM_(H)+(A-Z)m_(n)-M_(N)` Here `M_(H)=` atomic mass of hydrogen `M_(N)=` atomic mass of formed atom. `:." "R=R_(0)A^(1//3)` `R_(Al)=R_(0)(27)^(1//3)` `R=R_(0)A^(1//3)` But `R=(R_(Al))/((3)/(2))=(2)/(3)R_(Al)` or `R_(0)A^(1//3)=(2)/(3)R_(0)(27)^(1//3)` `:. " " A=((2)/(3))^(3)=27=(8)/(27)xx27=8` From periodic table, the element is `Be^(8)` Atomic masses are `1.007825` for hydrogen, `1.008665` for nuetron, `8.00531` for `Be` `Deltam=(4m_(H)+4m_(n)-M_(N))"amu"` `=(4xx1.007825+4xx1.008665-8.00531)"amu"` `=(4.0313+4.034660-8.00531)"amu"` `E_(b)=Delta mc^(2)=0.06065xx931 MeV` `=56.4651 MeV=56.5 MeV` |
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| 764. |
The number of protons, electrons and neutrons in the nucleus of `._(13)Al^(27)` isA. `13,`13,`14`B. `13,0,14`C. `14,14,13`D. `14,0,13` |
| Answer» Correct Answer - B | |
| 765. |
Calculate the average binding energy per nucleon of `._(41)^(93)Nb` having mass 9.2.906 u.. |
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Answer» In order to compute binding energy, let us first find the total mass of all protons and neutrosn in Nb and substarct mass of the Nb: Given: `m_(p) =1.007276 u` and `m_(n)=1.008665 u` Number of protons: `N_(p) =41` Number of neutrons: `N_(n) =93 -41=52` mass differences: `Delta m=41 m_(p) + 52 m_(n) -m_(Nb)` `=41(1.007825 u) +52(1.008665 u) -(92.9063768 u)` `=0.865028 u ` Thus, binding energy per nuclear is . `(E_(b))/(A)=((Delta)c^(2))/(A)=((0.865028u)(931.5MeV//u))/(93)` `=8. 66 MeV ` nucleon^(-1)`. |
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| 766. |
Cabalt-60 is useful in cancer therapy. Cabalt-60 is source of `Y-radiations` capable of killing cancerous cell/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) Radiotherapy is exposing tumours//cancer cells to radiations, as from radioactive cobalt `-60`. Radon `-120`, `I -131`. Carbalt `-60` is a source of `gamma-raioactions`. These because cancer cells are more rapidly than others because cancer cells are undifferentiated. However, healthy cells in vicinity of cancer cells are also damaged. Hence, option (a) is true. |
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| 767. |
The binding energy per nucleon, for nuclei with atomic mass number `A gt 100`, decreases with `A`. The nuclear forces are weak for heavier nuclei.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) A better measure of the binding between the constituents of the nucleus is the binding energy per nucleon, which is the ratio of the binding energy of a nucleus to the number of the nucleons in that nucleus. For heavy nuclei `(A gt 100)` Coulomb repulsion between the protons and inside the nucleus increasing. This results in decrease in binding energy per nucleon. Now, nuclear force is the familiar Coulomb force that determines the motion of atomic electrons. Nuclear force is nearly same for all nuclei. Hence, option ( c) is true. |
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| 768. |
The strong interaction exists inA. Gravitational forcesB. Elcetrostatic force of attractionC. Nuclear forcesD. Magnetic force on a moving charge |
| Answer» Correct Answer - C | |
| 769. |
The short range attractive nuclear forces that are responsible for the binding of nucleons in a nucleus ae supposed to be caused by the role played by the particles calledA. PositronB. `m`-MesonC. `K`-MesonD. `pi`-Meson |
| Answer» Correct Answer - D | |
| 770. |
Two protons are kept at a separation of `10 nm`. Let `F_n and F_e` be the nuclear force and the electromagnetic force between them.A. `F_(n) gt gt F_(e)`B. `F_(n)=F_(e)`C. `F_(n)lt lt F_(e)`D. `F_(n) ~~ F_(e)` |
| Answer» Correct Answer - C | |
| 771. |
The graph between the binding energy per nucleon (E) and atomic mass number (A) is as-A. B. C. D. |
| Answer» Correct Answer - D | |
| 772. |
The wrong statement about the binding energy isA. It is the sum of the rest mass energies of nucleus minus the energy of the nucleusB. it is the energy released when the nucleons combine to from nucleusC. it is the energy required to break a given nucleus into its constitutent nucleusD. it is the sum of the kinetic energy of all the nucleons in the nucleus |
| Answer» Correct Answer - B | |
| 773. |
Which of the following is a wrong description of binding energy of a nucleus?A. It is the energy required to break a nucleus into its constituent nucleons.B. It is the energy released when free nucleons combine to from a nucleusC. It is the sum of the rest mass energies of its nucleons minus the rest energy of the nucleusD. It is the sum of the kinetic energy of all the nucleons in the nucleus |
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Answer» Correct Answer - D It is the sum of the kinetic energy of all the nucleons in the nucleus |
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| 774. |
The binding energy per nucleon is maximum at `A=56` and its value is around `_MeV//"Nculeon"`A. `8.4`B. `8.7`C. `9`D. `7.8` |
| Answer» Correct Answer - B | |
| 775. |
The wrong statement about the binding energy isA. It is the sum of the rest mass energies of nucleous minus the rest mass energy of the nucleus.B. It is the energy released when the nucleons combine to form a nucleus.C. It is the energy required to break a given nucleus into its constituent nucleons.D. It is the sum of the kinetic energies of all the nucleons in the nucleus. |
| Answer» Correct Answer - D | |
| 776. |
If the binding energy per nucleon in `L i^7` and `He^4` nuclei are respectively `5.60 MeV` and `7.06 MeV`. Then energy of reaction `L i^7 + p rarr 2_2 He^4` is.A. `19.6MeV`B. `2.4MeV`C. `8.4MeV`D. `17.3MeV` |
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Answer» Correct Answer - D `._(3)Li^(7)+_(1)H^(1)rarr 2_(2)He^(4)` Energy of reaction `= 8xx7.06 - 7xx5.60 = 17.28MeV` |
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| 777. |
Average binding energy per nucleon over a wide range isA. `8 MeV`B. `8.8 Me V`C. `7.6 Me V`D. `1.1 Me V` |
| Answer» Correct Answer - A | |
| 778. |
Average binding energy per nucleon over a wide range isA. `8eV`B. `8MeV`C. `8 J`D. `8ergs` |
| Answer» Correct Answer - B | |
| 779. |
Average binding energy per nucleon over a wide range isA. `8` MeVB. `8` eVC. `8 J`D. `8` erg |
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Answer» Correct Answer - A |
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| 780. |
The nuclei `._(6)C^(13) & ._(7)N^(14)` can be described asA. isotonesB. isobarsC. isotopes of carbonD. isotopes of nitrogen |
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Answer» Correct Answer - A |
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| 781. |
The nuclei `._(6)C^(13) & ._(7)N^(14)` can be described asA. isotonesB. isobarsC. isomersD. isotopes |
| Answer» Correct Answer - A | |
| 782. |
The graph of `1n (R/R_0)` versus `1n A (R = radius` of a nucleus and `A =` its mass number) isA. Straight lineB. ParabolaC. EllipseD. Circle |
| Answer» Correct Answer - A | |
| 783. |
The graph of `1n (R/R_0)` versus `1n A (R = radius` of a nucleus and `A =` its mass number) isA. a straight lineB. a parabolaC. an ellipseD. none of these |
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Answer» Correct Answer - A `R = R_(0)A^(1//3)` `(R )/(R_(0)) = A^(1//3)` `ln((R )/(R_(0))) = (1)/(3) lnA, y = (1)/(3)x ("straight line")` In`((R )/(R_(0)))` versus `ln A` will be a straight line |
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| 784. |
Half-life of a radioactive substance` A` and `B` are, respectively, `20 min` and `40min`. Initially, the samples of `A` and `B` have equal number of nuclei. After `80 min`, the ratio of the ramaining number of `A` and `B` nuclei isA. `1 : 16`B. `4 : 1`C. `1 : 4`D. `1 : 1` |
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Answer» Correct Answer - C |
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| 785. |
Atomic weight of boron is `10.81` and it has two isotopes `._5 B^10` and `._5 B^11`. Then ratio of `._5 B^10` in nature would be.A. `19:81`B. `10:11`C. `15:16`D. `81:19` |
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Answer» Correct Answer - A |
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| 786. |
Atomic weight of boron is `10.81` and it has two isotopes `._5 B^10` and `._5 B^11`. Then ratio of `._5 B^10` in nature would be.A. `19 : 81`B. `10 : 11`C. `15 : 16`D. `81 : 19` |
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Answer» Correct Answer - A |
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| 787. |
A `280 `days old radioactive substance shown an activity of 6000 dps, 100 days later its activity between 3000 dps ,what was its initial activity ?A. `20000 dps`B. `24000 dps`C. `120000 dps`D. 6000 dps` |
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Answer» Correct Answer - b We know that `lambda=(2.303)/(t) log .A_(0)/(A)` where `A_(0)` is the initial activity. A is the activity at time t. `:. lambda=(2.303)/(280)log.(A_(0))/(600) =(2.303)/(420) log.(A_(0))/(3000)` On solving, we get `A_(0) =24000 dps` |
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| 788. |
The decay constant for the radioactive nuclide `64Cu` is `1.516 xx 10^(-5) s^(-1)` . Find the activity of a sample containing `1mug` of `^64Cu`. Atomic weight of copper = `63.5 g (mol)^-1`. Neglect the mass difference between the given radioisotope and normal coper. |
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Answer» `63.5 g `of copper has `6 xx 10^(23)` atoms. Thus, the number of atoms in `1 mg` of `Cu` is `N=(6 xx 10^(23) xx 1 mu g)/(63.5g) =9.45 xx10^(15)` The activity = `lambda N` `=(1.516 xx10^(-5) s^(-1)) xx (9.45 xx 10^(15))` `=1.43 xx 10^(11)` disintegrations `s^(-1)` `=(1.43 xx 10^(11))/(3.7 xx 10^(10)) Ci =3.86 Ci` Activity after n half - lives is `A_(0)//2^(n)` . `T_(avg) =("Sum agees of all the nuclei")/(N_(0))=(underset(0)overset(infty)(int)N_(0)e^(-lambdat)dt)/(N_(0))=(1)/(lambda)`. |
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| 789. |
Consider the beta decay `^198 Au rarr ^198 Hg ** + Beta^(-1) + vec v`. where `^198 Hg^**` represents a mercury nucleus in an excited state at energy `1.088 MeV` above the ground state. What can be the maximum kinetic energy of the electron emitted? The atomic mass of `^198 Au` is `197.968233 u` and that of `^198 Hg` is `197.966760 u`. |
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Answer» Correct Answer - [0.2834 MeV] |
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| 790. |
The half-life of `^198 Au` is `2.7 days`. Calculate (a) the decay constant, (b) the average-life and (C ) the activity of `1.00 mg` of `^198 Au`. Take atomic weight of `^198 Au` to be `198 g mol^(-1)`. |
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Answer» The half-life and the decay constant are related as ` T_(1//2)=(In2)/(lambda)=(0.693)/(lambda)` or `lambda=(0.693)/(t_(1//2))=(0.693)/(2.7 days)` `A=lambda N=(2.9xx10^(-6)s^(-1))(3.03xx10^(18))` `=8.8xx10^(12)` distntegrations `s^(-1)` `=(8.8 xx10^(12))/(3.7xx10^(10)) Ci=140 Ci`. |
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| 791. |
`1 mg` gold undergoes decay with `2.7` days half-life period, amount left after `8.1` days isA. `0.91 mg`B. `0.25 mg`C. `0.5 mg`D. `0.125 mg` |
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Answer» Correct Answer - D (d) `N = N_0((1)/(2))^n rArr N = N_0 ((1)/(2))^(t//T_(1//2))` `rArr N = 1 xx ((1)/(2))^((8.1)/(2.7)) = ((1)/(2))^3 = (1)/(8)` `rArr N = (1)/(8) mg = 0.125 mg`. |
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| 792. |
A radioactive nucleus `._92 X^235` decays to `._91 Y^231`. Which of following particles are emitted ?A. One alpha and one electronB. Two deutrons and one positionC. One alpha and one protonD. One proton and four neutrons |
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Answer» Correct Answer - A (a) `._92 X^235 overset (alpha)rarr ._90 X^231 overset (._-1 e^0)rarr ._91 Y^231`. |
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| 793. |
`N` atoms of a radioactive element emit `n` alpha particles per second. The half-life of tge element is.A. `(n)/(N) sec`B. `(N)/(n)sec`C. `(0.693 N)/(n) sec`D. `(0.693 n)/(N) sec` |
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Answer» Correct Answer - C ( c) `(dN)/(dt) = -lamda N rArr -lamda N(Given (dN)/(dt) = n)` `:. lamda = -(n)/(N)` `:.` Hal life `= (0.693)/(lamda) = (0.693)/(lamda) = (0.693)/(n) sec`. |
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| 794. |
Two samples A and B of same radioactive nuclide are prepared. Sample A has twice the initial activity of sample B. For this situation, mark out the correct statement (s).A. The half-lives of both the samples would be same.B. The half-lives of the samples are different.C. After each has passed through `5` half-lives, ratio of activity of `A` to `B` is `2:1`.D. After each has passed through `5` half-lives, ratio of activity of `A` to` B` is `64:1`. |
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Answer» Correct Answer - a,c Half -lives of both the samples would be same as half-life is the property of radioactive material and is independent of number of nuclei present or its activity. Let `R_(0) B =R_(0)`, then `R_(0) A=2B_(0)`, where `R_(0)` denotes initial activity. Activity of `A` after `5` half-lives is `R_(A) =(R_(0A))/(2^(5))=(2R_(0))/(2^(5))a` Activity of `B` after half-lives is `R_(B)=(R_(0B))/(2^(5))=(R_(0))/(2^(5))` `:. (R_(A))/(R_(B))=(2)/(1)` . |
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| 795. |
A body of mass `m_(0)` is placed on a smooth horizontal surface . The mass of the body is decreasing exponentially with disintegration constant , `lambda` . Assuming that the mass is ejected backwards with a relative velocity u . If initially the body was at rest , the speed of body at time t is |
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Answer» Correct Answer - `[u lambda t]` |
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| 796. |
In a nuclear reactor `.^235U` undergoes fission liberating `200 MeV` of energy. The reactor has a `10%` efficiency and produces `1000 MW` power. If the reactor is to function for `10 yr`, find the total mass of uranium required. |
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Answer» Total energy produced by the reactor in time `t = 10 years` `E =1000 xx 10^(6) xx 10 xx 3.15 xx 10^(7) J` `=3.15 xx 10^(7) J` `Efficiency = ("output energy")/("input energy")` ` rarr` Input energy caused by fission `=(output energy)/(efficiency)` `=(3.15 xx10^(17))/((10//100))=3.15xx10^(18) J` Energy produced by `1` fission of `.^(135)U=200 MeV` `=200 xx1.6xx10^(-13) J` ` =3.2 xx 10^(-11) J` Therefore, Number of fissions required `=(Total energy )/(Energy per fission)` `=(3.15 xx 10^(18))/(3.2 xx10^(28))` `~~9.8 xx 10^(28)` Hence, mass of nranium required is given by ` m=(N)/(N_(a)) xx 235 kg =(9.8xx 10^(28))/(6.02 xx 10^(26))` `=38.2 xx 10^(3) kg`. |
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| 797. |
The element curium `._96^248 Cm` has a mean life of `10^13s`. Its primary decay modes are spontaneous fission and `alpha`-decay, the former with a probability of `8%` and the later with a probability of `92%`, each fission releases `200 MeV` of energy. The masses involved in decay are as follows `._96^248 Cm=248.072220 u`, `._94^244 P_u=244.064100 u` and `._2^4 He=4.002603u`. Calculate the power output from a sample of `10^20` Cm atoms. (`1u=931 MeV//c^2`) |
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Answer» Correct Answer - `[3.32 xx 10^(-5) Js^(-1)]` |
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| 798. |
Potassium-40 can decay in three modes .It can decay by `beta^(-)` - emission,`beta^(+)` -emission or electron capature. (a) Write the equation showing the end products. (b) Find the Q-value in each of the three cases. Atomic masses of `Ar_18^40`,`K_19^40`and `Ca_20^40` are 39.9624 u,39.9640 u,and 39.9626 u respectively. |
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Answer» Correct Answer - [(a) `._(19)^(40)K rarr ._(20)^(40)Ca + e^(-) + bar(v), ._(19)^(40)K rarr ._(18)^(40)Ar + e^(+) +v, ._(19)^(40)K + e^(-1) rarr ._(18)^(40)Ar+v` (b) 1.3034 MeV, 0.4676 MeV, 1.490 MeV] |
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| 799. |
`P^32` beta-decays to `S^32`.Find the sum of the energy of the antineutrino and the kinetic energy of the `beta`-particle. Neglect the recoil of the daughter nucleus. Atomic mass of `P^32=31.974 u` and that of `S^32=31.972 u`. |
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Answer» Correct Answer - [1.86 MeV] |
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| 800. |
During `beta`-decay (beta minus), the emission of antineutrino particle is supported by which of the following statement (s)?A. Angular momnetum conservation holds good in any nuclear ractionB. Linear momnetum conservation holds good in any nuclear ractionC. The KE of emitted `beta`-particel is varying conitnuously to a maximum value.D. none of these |
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Answer» Correct Answer - a,b,c If the nuclear reaction involving `beta`-decay is `n rarr p + e^(-1)`, the spins on two sides are not equals as all the three (neutron, proton and electron) have spins of `+1//2`. So, to conserve angular momentum (spin), some other particle must be emitted. Through experiments it has been observed that direction of emitted electron and recoiling nuclei are almost never exactly oppsite as required for linear momentum to be conserved. During `beta`-decay, the energy of electron is found to vary continuously from `0` to a maximum value (this maximum value is a characteristic of nuclide). To explain this experimental observation, we also need some other particle. |
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