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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.
| 501. |
The penentrating powder of beta particle compared to alpha particle isA. LessB. MoreC. EqualD. Can be more or less |
| Answer» Correct Answer - B | |
| 502. |
The units of radioactivity isA. FermiB. FaradC. CurieD. Hertz |
| Answer» Correct Answer - C | |
| 503. |
In a nuclear reactor, heavy water is used as aA. Constrolling materialB. ModeratorC. FuelD. Heat exchanger |
| Answer» Correct Answer - B | |
| 504. |
The ratio of the radii of the nuclei `._(13)^(27)Al " and " ._(52)Te^(125)` is approximately-A. `6 : 10`B. `13 : 52`C. `40 : 177`D. `14 : 73` |
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Answer» Correct Answer - A |
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| 505. |
`._(235)^(239)Pu._(94)` is undergoing `alpha-decay` according to the equation `._(94)^(235)Pu rarr (._(97)^(235)U) +._2^4 He` . The energy released in the process is mostly kinetic energy of the `alpha`-particle. However, a part of the energy is released as `gamma` rays. What is the speed of the emiited `alpha`-particle if the `gamma` rays radiated out have energy of `0.90 MeV`? Given: Mass of `._(94)^(239)Pu =239.05122 u`, mass of `(._(97)^(235)U)=235.04299 u` and mass of `._1^4He =4.002602 u (1u =931 MeV)`. |
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Answer» Correct Answer - `144.10ms^(-1)` `Delta m=0.00563 u=5.24 MeV` KE of `alpha`-particle` =5.24 -0.9=4.34 MeV` `:. v=sqrt((2.434.10.1.6.10^(9))/(4.1.67.10))=1.44.10 m s^(-1)`. |
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| 506. |
A tritrium gas target is bombared with a beam of monoenergetic protons of kinetic energy `K_(1) =3 MeV` The `KE` of the neutron emiited at `30^(@)` to the inicdent beam is `K_(2)` ? Find the value of `K_(1)//K_(2)` (approximately in whole number). Atomic masses are `H^(1) =1.007276 amu`, `n^(1)=1.008665 amu`, `._1 H^3=3.016050 amu`, `._2He^3=3.016030 amu`. |
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Answer» Correct Answer - `2` Nuclear reaction is `._(1)H^(1) +._(1)H^(3) rarr ._(2)He^(3) +_(0)n^(1)+Q` `Q=-1.2745 MeV` Using conservation of linear momnetum, we get `K_(2) =1.44 MeV, K_(1) =3 MeV` . |
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| 507. |
Consider a nuclear reaction `A+B rarr C`. A nucleus `A` moving with kinetic energy of `5 MeV` collides with a nucleus `B` moving with kinetic energy of `3MeV` and forms a nucleus `C` in exicted state. Find the kinetic energy of nucleus `C` just after its fromation if it is formed in a state with excitation energy `10 MeV`. Take masses of nuclei of `A,B` and `C ` as `25.0,10.0,34.995 amu`, respectively. (1amu=`930 MeV//c^(2))`. |
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Answer» Correct Answer - B `m_(A)C^(2)+K_(A)+m_(B)C^(2)+k_(B)=m_(C)C^(2)+k_(c)+` exciation energy. `(m_(A)+m_(B)-m_(c))c^(2)+k_(A)+k_(B)-k_(C)=` excitation energy `:. k_(C)=2MeV` |
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| 508. |
Consider a nuclear reaction `A+B rarr C`. A nucleus `A` moving with kinetic energy of `5 MeV` collides with a nucleus `B` moving with kinetic energy of `3MeV` and forms a nucleus `C` in exicted state. Find the kinetic energy of nucleus `C` just after its fromation if it is formed in a state with excitation energy `10 MeV`. Take masses of nuclei of `A,B` and `C ` as `25.0,10.0,34.995 amu`, respectively. `(1amu=930 MeV//c^(2))`. |
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Answer» Correct Answer - `2.65 MeV` Applying conservation of energy: `m_(A)c^(2)+K_(A)+m_(B)c^(2)+K_(B)c^(2)+K_(B)=mc^(c^(2))+K_(C)+excitation nergy ` `(m_(A)+m_(B)+m_(C)C^(2)+K_(A)+K_(B)=K_(C)+`excitation nergy `4.65 +5+3=K_(C) +10` or `K_(C) =2.65 MeV`. |
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| 509. |
Which of the following is used as a moderator in nuclear reactors?A. PlutoniumB. CadmiumC. Heavy waterD. Uranium |
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Answer» Correct Answer - C |
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| 510. |
Calculate the energy released when three alpha particles combine to form a `^12 C` nucleus. The atomic mass of `_2^4 He` is `4.002603 u`. |
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Answer» The mass of `a^(12)C` atom is exactly `12 u`. The energy released in the reaction `3(._(2)^(4)Hr)rarr._(6)^(12)C` is `[3m(._(2)^(4)He)-m(._(6)^(12)C)]c^(2)` `[3xx4.002603u-12 u](931 MeV//u)=7.27 MeV`. |
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| 511. |
In a hydrogen like atom electron make transition from an energy level with quantum number `n` to another with quantum number `(n - 1)` if `n gtgt1` , the frequency of radiation emitted is proportional to :A. `(1)/(n)`B. `(1)/(n^(2))`C. `(1)/(n^(3)//2)`D. `(1)/(n^(3))` |
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Answer» Correct Answer - D `DeltaE=hv` `v=(DeltaE)/(h)=k[(1)/((n-1)^(2))-(1)/(n^(2))]=(2Kn)/(n^(2)(n-1)^(2))~~(2K)/(n^(3))prop(1)/(n^(3))` |
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| 512. |
The nuclear reaction `.^2H+.^2H rarr .^4 He` (mass of deuteron `= 2.0141 a.m.u` and mass of `He = 4.0024 a.m.u`) isA. Fusion reaction releasing `24 MeV` energyB. Fusion reaction absorbing `24 MeV` energyC. Fission reaction releasing `0.0258 MeV` energyD. Fission reaction absoring `0.0258 MeV` energy |
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Answer» Correct Answer - A (a) Total mass of reactions =`(2.0141) xx 2 = 4.0282` amu Total mass of products `= 4.0024 `amu Mass defect `= 4.0282 "amu" - 4.0024 "amu"` =`0.0258` amu `:.` Energy released `E = 931 xx 0.0258 = 24 MeV`. |
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| 513. |
Heavy water is used as moderator in a nuclear reactor. The function of the moderator is |
| Answer» Heavy water is used as a moderator because its mass is nearest to that of a neutron and it ahs negligible chances for neutron absorption. | |
| 514. |
The nuclear reaction `.^2H+.^2H rarr .^4 He` (mass of deuteron `= 2.0141 a.m.u` and mass of `He = 4.0024 a.m.u`) isA. fusion reaction releasing `24 MeV` energyB. fusion reaction absorbing `24 MeV` energyC. fission reaction releasing `0.0258 MeV` energyD. fission reaction absorbing`0.0258 MeV` energy |
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Answer» Correct Answer - A |
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| 515. |
The volume occupied by an atom is greater than the volume of the nucleus by factor of aboutA. `10^(10)`B. `10^(15)`C. `10^(1)`D. `10^(5)` |
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Answer» Correct Answer - B |
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| 516. |
Energy generation in starts is mainly due toA. Chemical reactionsB. Fission of heavy nucleiC. Fusion of light nucleiD. Fusion of heavy nuclei |
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Answer» Correct Answer - C ( c) Energy of starts is due to the fission of light by drogen nuclei into `He`. In this process much energy is released. |
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| 517. |
The mass number of a nucleus is.A. sometimes equal to its atomic numberB. sometimes less than and sometimes more than its atomic numberC. always less than its atomic numberD. always more than its atomic number |
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Answer» Correct Answer - A |
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| 518. |
The mass of proton is `1.0073 u` and that of neutron is `1.0087 u` (`u=` atomic mass unit). The binding energy of `._(2)He^(4)` is (mass of helium nucleus `=4.0015 u`)A. `28.4` MeVB. `0.061 u`C. `0.0305 J`D. `0.0305` erg |
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Answer» Correct Answer - A |
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| 519. |
The radionuclide `.^(56)Mn` is being produced in a cyclontron at a constant rate `P` by bombarding a manganese target with deutrons. `.^(56)Mn` has a half-life of `2.5 h` and the target contains large numbers of only the stable manganese isotopes `.^(56)Mn`. The reaction that produces `^(56)Mn` is `.^(56)Mn +d rarr .^(56)Mn +p` After being bombarded for a long time, the activity of `.^(56)Mn` becomes constant, equal to `13.86 xx 10^(10) s^(-1)`. (Use `1 n2=0.693`, Avagardo number` =6 xx 10^(2), atomic weight of `.^(56)Mn=56 g mol^(-1)`). After the activity of `.^(56)Mn` becomes constant, number of `.^(56)Mn` nuclei present in the target is equal to .A. ` 5xx 10^(11)`B. ` 20 xx 10^(11)`C. ` 1.2xx 10^(14)`D. ` 1.8 xx 10^(15)` |
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Answer» Correct Answer - d As rate of decay =Rate of production `P=lambdaN rArr N=(P)/(lambda)=(Pt_(1//2))/(In2)=1.8xx10^(15)`. |
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| 520. |
The radionuclide `.^(56)Mn` is being produced in a cyclontron at a constant rate `P` by bombarding a manganese target with deutrons. `.^(56)Mn` has a half-life of `2.5 h` and the target contains large numbers of only the stable manganese isotopes `.^(56)Mn`. The reaction that produces `.^(56)Mn` is `.^(56)Mn +d rarr .^(56)Mn +p` After being bombarded for a long time, the activity of `.^(56)Mn` becomes constant, equal to `13.86 xx 10^(10) s^(-1)`. (Use `1 n2=0.693`, Avagardo number `=6 xx 10^(2)`, atomic weight of `.^(56)Mn=56 g mol^(-1)`). At what constant rate P, `.^(56)Mn` nuclei are being produced in the cyclontron during the bombardment?A. `2 xx 10^(11)nuclei s^(-1)`B. `13.86 xx 10^(10)nuclei s^(-1)`C. `9.6 xx 10^(10)nuclei s^(-1)`D. `6.93 xx 10^(10)nuclei s^(-1)` |
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Answer» Correct Answer - b In equilibrium, rate of decay = rate of production |
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| 521. |
Various rules of thumb have seen proposed by the scientific community to expalin the mode of radioactive decay by various radioisotopes. One of the major rules is called the `n//p` ratio. If all the known isotopes of the elemnts are plotted on a graph of number of neutrons (n) versus number of protons (p), it is observed that all isotopes lying outside of a 'stable' `n//p` ratio region are radioactive as shown f The graph exhibits straight line behaviour with unit slope up to `p=25`. Above `p=25`, tgose isotopes with `n//p` ratios lying above the stable region usually undergo beta decay. Very heavy isotopes `(pgt83)` are unstable because of their relativley large nuclei and they undergo alpha decay. Gamma ray emission does not involve the release of a particle. It represnts a change in an atom from a higher energy level to a lower energy level. `Th-230` undergoes a series of radioactive decay processes resulting in `Bi-214` being the final product. What was the sequence of the processes that occured?A. `alpha,alpha,alpha,g,beta`B. `alpha,alpha,alpha,alpha,beta`C. `alpha,alpha,beta,beta`D. `alpha,beta,beta,beta,gamma` |
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Answer» Correct Answer - b `._(90)^(230)Th rarr ._(83)^(214)Bi + x._(1)^(0)e^(bar) +y._(2)^(4)He^(2+)+z ("gamma ray")` Since the sum of the atomic numbers and mass numbers on either side of the equation must be equal (matter cannot be created or destroyed),we get `90=83-x+2y` and `230=214+4y` Solving, we get `y=4` and `x=1` The order of the reactions is irrelevant (i.e., `alpha, beta`,...) Since gamma rays have no atomic number or mass number, the value of z does not affect this particular calculation. |
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| 522. |
Various rules of thumb have seen proposed by the scientific community to expalin the mode of radioactive decay by various radioisotopes. One of the major rules is called the `n//p` ratio. If all the known isotopes of the elemnts are plotted on a graph of number of neutrons (n) versus number of protons (p), it is observed that all isotopes lying outside of a 'stable' `n//p` ratio region are radioactive as shown fig. The graph exhibits straight line behaviour with unit slope up to `p=25`. Above `p=25`, tgose isotopes with `n//p` ratios lying above the stable region usually undergo beta decay. Very heavy isotopes `(pgt83)` are unstable because of their relativley large nuclei and they undergo alpha decay. Gamma ray emission does not involve the release of a particle. It represnts a change in an atom from a higher energy level to a lower energy level. How would the radioisotope of magnesium with atomic mass `27` undergo radioactive decay?.A. electron captureB. alpha decayC. beta decayD. gamma ray emission |
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Answer» Correct Answer - c From the graph and the fact that the `n//p ("=no. of neutrons"//"no.of protons")` ratio for magnesium is `27//12`, which is greater than `1("=unit slope")`. |
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| 523. |
Various rules of thumb have seen proposed by the scientific community to expalin the mode of radioactive decay by various radioisotopes. One of the major rules is called the `n//p` ratio. If all the known isotopes of the elemnts are plotted on a graph of number of neutrons (n) versus number of protons (p), it is observed that all isotopes lying outside of a 'stable' `n//p` ratio region are radioactive as shown fig.5.28. The graph exhibits straight line behaviour with unit slope up to `p=25`. Above `p=25`, tgose isotopes with `n//p` ratios lying above the stable region usually undergo beta decay. Very heavy isotopes `(pgt83)` are unstable because of their relativley large nuclei and they undergo alpha decay. Gamma ray emission does not involve the release of a particle. It represnts a change in an atom from a higher energy level to a lower energy level. Which of the following represents the relative penetrating power of the three types of radioactive emission in decreasing order?A. `betagtalphagtgamma`B. `betagtgammagtalpha`C. `gammagtalphagtbeta`D. `gammagtbetagtalpha` |
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Answer» Correct Answer - d Alpha radiation consists of the largest particles (helium nuclei with a mass number of `4`, thus the greatest inertia)and are slowest (about `1//3` times the speed of the light). Beta radiation consists of smaller (electrons, `1//1370` times lighter than a proton) and faster partilces (about `4//5` times the speed of light). Gamma radiation consits of the smallest particles (photons, no mass ) which travel at the greatest speed (at the speed of light). |
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| 524. |
Various rules of thumb have seen proposed by the scientific community to expalin the mode of radioactive decay by various radioisotopes. One of the major rules is called the `n//p` ratio. If all the known isotopes of the elemnts are plotted on a graph of number of neutrons (n) versus number of protons (p), it is observed that all isotopes lying outside of a 'stable' `n//p` ratio region are radioactive as shown fig. The graph exhibits straight line behaviour with unit slope up to `p=25`. Above `p=25`, tgose isotopes with `n//p` ratios lying above the stable region usually undergo beta decay. Very heavy isotopes `(pgt83)` are unstable because of their relativley large nuclei and they undergo alpha decay. Gamma ray emission does not involve the release of a particle. It represnts a change in an atom from a higher energy level to a lower energy level. How would the radioisotope of magnesium with atomic mass `27` undergo radioactive decay?.A. Electron captureB. alpha decayC. Beat decayD. Gamma ray emission |
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Answer» Correct Answer - C `n//p` ratio of magneisum is `27//12+z gamma` |
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| 525. |
Various rules of thumb have seen proposed by the scientific community to expalin the mode of radioactive decay by various radioisotopes. One of the major rules is called the `n//p` ratio. If all the known isotopes of the elemnts are plotted on a graph of number of neutrons (n) versus number of protons (p), it is observed that all isotopes lying outside of a 'stable' `n//p` ratio region are radioactive as shown f The graph exhibits straight line behaviour with unit slope up to `p=25`. Above `p=25`, tgose isotopes with `n//p` ratios lying above the stable region usually undergo beta decay. Very heavy isotopes `(pgt83)` are unstable because of their relativley large nuclei and they undergo alpha decay. Gamma ray emission does not involve the release of a particle. It represnts a change in an atom from a higher energy level to a lower energy level. `Th-230` undergoes a series of radioactive decay processes resulting in `Bi-214` being the final product. What was the sequence of the processes that occured?A. `alpha, alpha, alpha,gamma, beta`B. `alpha, alpha, alpha,alpha,beta`C. `alpha,alpha,beta,beta`D. `alpha, beta, beta, beta, gamma` |
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Answer» Correct Answer - B `._(90)^(230)Th rarr_(83)^(214)Bi+x_(-1)e^(0)+y_(2)He^(4)+z gamma` Hence, `90=83-x+2y` and `230=214+4y` Solving, we get `y=4` and `x=1`. The order of the reactions is irrelevent |
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| 526. |
Atomic mass of the most useful material for fusion reaction isA. `1`B. `4`C. `235`D. `292` |
| Answer» Correct Answer - A | |
| 527. |
The heavier nuclei tend to have larger `N//Z` ratio becauseA. `(i),(ii)`B. `(ii),(iv)`C. `(ii),(iii)`D. `(iii),(iv)` |
| Answer» Correct Answer - D | |
| 528. |
The heavier nuclei tend to have larger `N//Z` ratio becauseA. a neutron is heavier than a protonB. a neutron is an ustable particleC. a neutron does not exert electric repulsionD. Coulomb forces have longer range compared to nuclear forces |
| Answer» Correct Answer - C::D | |
| 529. |
The liquid drop model of nucleus was proposed byA. Bohr,WheelerB. FermiC. RutherfordD. Chadwick |
| Answer» Correct Answer - A | |
| 530. |
The material used to slow neutrons in a reactor is calledA. ControlrodB. ModeratorC. FuelD. Heat exchanger |
| Answer» Correct Answer - B | |
| 531. |
Cadmium and Broron rods are used in a nuclear reactor toA. FuelB. ModeratorC. Control RodsD. None |
| Answer» Correct Answer - C | |
| 532. |
In a certain hypothetical radioactive decay process, species A decays into spesies `B` and species `B` decays into` C` according to the reactions `{:(A rarr 2B +"particles +energy"),(B rarr 2C+"particles +energy"):}` The decay constant for species `B` is `lambda_(2)=100 s^(-1)`. Initially, `10^(4)` moles of species of `A` were present while there was no none of `B` and `C`. It was found that species `B` reaches its maximum number at a time `t_(0)=2 1n(10)s`. Calcualte the value of maximum number of moles of `B`. |
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Answer» Correct Answer - `2` `(dN_A)/(dt) = - lambda_(1) N_(A), (dN_(B))/(dt)=2 lambda_(1) N_(A)-lambda_(2)N_(B)` N_(B) =maximum rArr `(dN_(B))/(dt)=0` `rArr 2 lambda_(1) N_(A)=lambda_(2)N_(B_(max))` or `N_(B_(max))=(2 lambda_(1))/(lambda_(2)) N_(A)` or `N_(B_(max)) =(2 lambda_(1))/(lambda_(2)) M_(0)e^(- lambda_(2)t )=2`. |
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| 533. |
If `M_(o)` is the mass of an oxygen isotope `._(8)O^(17), M_(p)` and `M_(N)` are the masses of a proton and neutron respectively, the nuclear binding energy of the isotope is:A. `(M_(o)-8M_(p))C^(2)`B. `(M_(o)-8M_(p)-9M_(N))C^(2)`C. `M_(o)C^(2)`D. `(M_(o)-17M_(N))C^(2)` |
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Answer» Correct Answer - B Nuclear binding energy `=`[mass of nucleus`-`mass of nucleons] `C^(2)=(M_(o)-8M_(p)-9M_(N))C^(2)` |
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| 534. |
The mass number of He is `4` and that for suphur is `32`. The radius of sulphur nuclei is larger than that of helium byA. `sqrt(8)`B. `4`C. `2`D. `8` |
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Answer» Correct Answer - C |
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| 535. |
Take a sample of lead and oxygen. They contain different atoms and the density of solid lead is much greater than that of gaseosu oxygen. Decide whether the density of the nucleus in a lead atom is greater than, approximately equa to, or less than that in an oxygen atom. |
| Answer» The nuclear density is nearly same in all atoms. The difference in densities between solid and lead and gaseous oxygen arises mainly because of the difference in how closely the atoms themselves are packed together in sloid and gaseous phase. | |
| 536. |
Find the binding energy of `_26^56 Fe`. Atomic mass of `^56 Fe` is `55.9349 u` and that of `^1 H` is `1.00783 u`. Mass of neutron `= 1.00867 u`. |
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Answer» The number of protons in `(._(26)^(56)Fe)=26` and the number of neutrons `=56-26=30`. The binding energy of `(._(26)^(56)Fe) =26` and the number of neutrons `=56-26 =30`. The binding energy `(._(26)^(56)Fe)` `=[26 xx 1.0078 u+30 xx 1.00867 u-55.9349 u]c^(2)` `=(0.52878 u)c^(2)` `=(0.52878 u)(931 MeV//u) =492 MeV.` |
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| 537. |
In the following nuclear reaction `._(13)Al^(27)+._(2)He^(4)rarr._(15)P^(30)+X,X` will beA. `._(-1)^(0)e`B. `._(1)^(1)H`C. `._(2)^(4)He`D. `._(0)^(1)n` |
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Answer» Correct Answer - D |
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| 538. |
Complete the following nuclear reactions: (a) `._(7)^(14)N+._(2)^(4)He rarr ._(8)^(17)O+`_____. (b) `._(15)^(30)P rarr ._(14)^(30)Si+`_____. (c) `._(4)^(9)Be+._(2)^(4)He rarr ._(6)^(12)C+` _____. (d) `._(1)^(3)H+._(1)^(2)H rarr _(2)^(4) He +`_____. (e) `._(4)^(9) Be(p, alpha)` _____. (f) `._(20)^(43)Ca (alpha, "_____")._(21)^(46)Sc` |
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Answer» Correct Answer - `[(a) ._(1)^(1)H (b) ._(1)^(0)e (c) ._(0)^(1)n (d) ._(0)^(1)n (e) ._(3)^(6)Li (f) ._(1)^(1)p]` |
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| 539. |
When `._(15)P^(30)` decays to become `._(14)Si^(30)`, the particle released isA. electronB. `alpha`-particleC. neutronD. positron |
| Answer» Correct Answer - D | |
| 540. |
The enegry released per fission of `.^(235)u` is nearlyA. `200 eV`B. `20eV`C. `2000 eV`D. `200MeV` |
| Answer» Correct Answer - D | |
| 541. |
A slow neutron can cause fission inA. `.^(235)U`B. `.^(238)U`C. `.^(235)Pu`D. `.^(232)Th` |
| Answer» Correct Answer - A | |
| 542. |
A slow neutron can cause fission inA. `U^(238)`B. `U^(235)`C. `Pb^(206)`D. `Sr^(90)` |
| Answer» Correct Answer - B | |
| 543. |
How many atoms decay in one mean life time of a radioactive sample-A. `37%`B. `63%`C. `50%`D. `100%` |
| Answer» Correct Answer - B | |
| 544. |
The count rate of `10 g` of radioactive material was measured at different times and times has been shown in the figure. The half-life of material and the total counts (approximately) in the first half life period, respectively are. .A. `4h, 9000`B. `3h, 14000`C. `3h,235`D. `3h, 50` |
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Answer» Correct Answer - B (b) Read time for `50` count rate, it gives half-life period of `3 hrs`, one small square gives `600` counts `(10 xx 60)`. The number of small squares between graph and time axis are approximately `24` Hence count rate `= 24 xx 600 = 14400`. |
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| 545. |
The count rate for `10 g` of radioactive material was measured at different times and this has been shown in the above graph with scale given. The half-life of the material and the total count in the first half-value period, respectively are.A. 4 hours and 9000 (approximately)B. 3 hours and 14100 (approximately)C. 3 hours and 235 (approximately)D. 10 hours and 150 (approximately) |
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Answer» Correct Answer - B (b) Half-value period `= 3 hours` `= 3 xx 60 minutes = 180 minutes` Approximate mean disintegration rate =`(100 + 50)/(2) = 75` Total count in first half-valur period =`75 xx 180 ~~ 13500` Total number of disintegrations is slightly greater than `13500`. |
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| 546. |
Which of the following changes in the artificial transmutation of elements?A. number of neutronsB. number of electronsC. atomic weightD. nucleus |
| Answer» Correct Answer - D | |
| 547. |
During the fission process of Uranium, the amount of energy liberated per fission is nearlyA. `100 Me V`B. `200 Me V`C. `150 Me V`D. `300 Me V` |
| Answer» Correct Answer - B | |
| 548. |
A chain reaction in fission of Uranium is possible becauseA. Large amount of energy is releasedB. Two intermediate size nuclear fragments are formedC. More than one neutron is given out in each fissionD. Fragments in fission are radioactive |
| Answer» Correct Answer - C | |
| 549. |
The count rate of `10 g` of radioactive material was measured at different times and times has been shown in the figure. The half-life of material and the total counts (approximately) in the first half life period, respectively are. .A. 4h,9000B. 3h,14000C. 3h,235D. 3h,50 |
| Answer» Correct Answer - B | |
| 550. |
The SI unit of activity is-A. CurieB. rutherfordC. becquerelD. sievert |
| Answer» Correct Answer - C | |