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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.
| 251. |
Radioactivity of neptunium stops when it is converted toA. BiB. RnC. ThD. Pb |
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Answer» Correct Answer - A Bi is a stable end product of Neptunium series |
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| 252. |
The last member of `4n+1` series is an isotope of lead. |
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Answer» Correct Answer - F |
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| 253. |
The last member of `4n+1` series is an isotope of `………………………………….` . |
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Answer» Correct Answer - Bimuth |
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| 254. |
Rate constant for a reaction is `lamda`. Average life is representant byA. `1//lamda`B. `ln 2//lamda`C. `(lamda)/(sqrt2)`D. `(0.693)/(lamda)` |
| Answer» Correct Answer - D | |
| 255. |
Positronim is the name given to an atom -like combination formed betweenA. A positron and a protonB. A positron and a neutronC. A positron and `alpha`-particleD. A positron and an electron |
| Answer» Correct Answer - D | |
| 256. |
Electromagnetic radioation with maximum wavelength isA. Ultraviolet rayB. RadiowaveC. X-rayD. Infrared |
| Answer» Correct Answer - BIts (B) Radiowave | |
| 257. |
Calculate difference between number of a `beta` particles emmited in the nuclear reaction `""_(92)U""^(23)to""_(82)Pb""^(206)` |
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Answer» Correct Answer - 2 No. of `alpha=(Ap-Ad)/4`,No. of `beta=2xxalpha-(Zp-Zd)` |
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| 258. |
Match column I with column II |
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Answer» Correct Answer - A-q;B-p;C-r;D-s `(A)""_(11)Na""^(23)+""_(1)H""^(2)to""_(11)Na""^(24)+""_(1)H""^(1)` (B) `2""_(1)H""^(3)to2He""^(4)+2""_(0)n""^(1)` `(C ) ""_(92)U""^(238)to""_(90)Th""^(234)+""_(2)He""^(4)` (D)`""_(29)Cu""^(63)to""_(28)Ni""^(63)+""_(1)e""^(0)` |
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| 259. |
Match the column I and column II and pick the correct matching from the codes given below. The correct matching isA. `{:(A,B,C,D,E),(5,4,1,2,3):}`B. `{:(3,1,2,5,4):}`C. `{:(2,1,4,5,3):}`D. `{:(2,5,1,4,3):}` |
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Answer» Correct Answer - b |
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| 260. |
Which of the following is used as neutron absorber in the nuclear reactor?A. WaterB. DeuteriumC. Some compound of uraniumD. Cadmium |
| Answer» Correct Answer - D | |
| 261. |
In a nuclear reactor, heavy water is used toA. Increase the speed of neutornsB. Decreases the speed of neutronsC. Transfer the heat from the reactorD. None of above |
| Answer» Correct Answer - B::C | |
| 262. |
Radioactive substance of 1 curie is the amount that can produce `……………………………` disintegrations per second. |
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Answer» Correct Answer - `3.7xx10^(10)` |
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| 263. |
The value of one mincrocurie = .... disintegrations/secondA. `3.7 xx 10^(5)`B. `3.7 xx 10^(7)`C. `3.7 xx 10^(4)`D. `3.7 xx 10^(10)` |
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Answer» Correct Answer - C `1 Ci = 3.7 xx 10^(10)dps or 3.7 xx 10^(10)`Bq. `1m Ci = 3.7 xx 10^(4) dps` |
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| 264. |
A radioactive element has a half-life of 20 minutes. How much time should elapse before the element is reduced to `(1)/(8)th` of the original massA. 40 minutesB. 60 minutesC. 80 minutesD. 160 minutes |
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Answer» Correct Answer - B `t_(1//2) = 20` minute, `N = (1)/(9)N_(0)` Use, `t = (2.303)/(0.693) xx t_(1//2) "log"(N_(0))/(N)` |
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| 265. |
`1 g` atom of an `alpha`-emitting `._(z)X^(4)` (half life = 10 hr) was placed in sealed containers, `4.52 xx 10^(25)`. Helium atoms will accumulate in the container afterA. 4.52 hrB. 10.00 hrC. 9.40 hrD. 20.00 hr |
| Answer» Correct Answer - D | |
| 266. |
The activity of `1 g` radium is found to be 0.5. Calculate the half-life period of radium and the time required for the decay of `2 g` of radium to give `0.25 g` of radium (atomic mass off radium = 226). |
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Answer» Number of atoms in `1 g` of `Ra^(226) = (6.02 xx 10^(23))/(226)` Activity, i.e., `(dN)/(dt) = 0.5` curie `= 0.5 xx 3.7 xx 10^(10) dps` But `(dN)/(dt) = K//N`, i.e., `1.85 xx 10^(10) = K xx (6.02 xx 10^(23))/(226)` or `K = 6.945 xx 10^(-12) s^(-1)` `:. t_(1//2) = (0.693)/(K) = (0.693)/(6.945 xx 10^(-12)) s` `= 9.978 xx 10^(10) s` `= (9.978 xx 10^(10))/(3600 xx 24 xx 365)` years = 3164 years Time requried for decay of `2 g` of `Ra` to `0.25 g` = three half lives `= 3 xx 3164 = 9492` years |
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| 267. |
S : `""^(60)Co` and `""^(60m)Co` are nuclear isomers E : The two different nuclear energy stateA. Both S and E are correct and E is cporrect explaination of SB. Both S and E are correct and E not correct explaination of SC. S is correct but E is wrongD. S is wrong but the E is correct |
| Answer» Correct Answer - a | |
| 268. |
The amount of `._(53)I^(128)` (`t_(1//2) = 25` minutes) left after 50 minutes will beA. One-halfB. One-thirdC. One-fourthD. Nothing |
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Answer» Correct Answer - C The amount of `._(53)I^(128)` left after 50 minutes will be = 25 minutes `= (100)/(25) = (1)/(4)` |
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| 269. |
The half-life of the ratio element `._(83)Bi^(210)` is 5 days. Starting with 20 g of this isotope, the amount remaining after 15 days isA. 10 gB. 5 gC. 2.5 gD. 6.66 g |
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Answer» Correct Answer - C `n = (15)/(5) = 3, N = (N_(0))/(2^(n)) = (20)/(2^(3)) = (20)/(8) = 2.5 g` |
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| 270. |
The half life of C-14 is 5600 years. A sample of freshely cut wood from a tree contains 10 mg of C-14. The amopunt left in the sample after 50000 years is (a-x)`xx100`. The value of `(a-x)xx100` is |
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Answer» Correct Answer - 2 `a-x=N""_(0)/2""^(n)=10/2""^(9)=10/512xx100=1.953=2,n=t/t""_(1//2)` |
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| 271. |
Identify the incorrect statements is/areA. `""_(z)X""^(A)overset(-alpha)tooverset(-beta)tooverset(-beta)to` in the above nuclear disintegration the no. of nuclides having isotopic number are 2.B. Activity of 1 gm `RaCl""_(2)` is more than that of 1gm `RaSO""_(4)`C. `""_(6)C""^(14)` emits `alpha`-particleD. `""_(15)CP""^(30)` decays by emitting positron |
| Answer» Correct Answer - c | |
| 272. |
The half-life of a radioactive element is 10 hours. How much will be left after 4 hours in 1 g atom sampleA. `45.6 xx 10^(23)` atomsB. `4.56 xx 10^(23)` atomsC. `4.56 xx 10^(24)` atomsD. `4.56 xx 10^(25)` atoms |
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Answer» Correct Answer - B `N_(t) = N_(0) ((1)/(2))^(n) n = 4//10` `:. N_(t) = 6.022 xx 10^(23) xx ((1)/(2))^(-.4) = 4.56 xx 10^(23)` |
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| 273. |
`._(90)Th^(234)` disintegrates to give `._(82)Pb^(206)Pb` as the final product. How many alpha and beta particles are emitted during this process ? |
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Answer» Correct Answer - `13` |
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| 274. |
Which one of the following statement is not correctA. `._(6)^(14)C` is a non-radioactive isotope of carbonB. `._(27)^(6)Co` is an unstable radioisotope of cobaltC. `BF_(3)` is a Lewis acidD. `CN^(-)` is a very strong ligand |
| Answer» Correct Answer - A | |
| 275. |
Elements having different nuclear charge but the same mass number are calledA. IsotopesB. IsobarsC. IsomersD. Isotones |
| Answer» Correct Answer - B | |
| 276. |
Werner Heisenberg considered the limits of how precisely we can measure the properties of an electron or other microscopic particle. He determined that there is a fundamental limit to how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse also true. This is summed up in what we now call the Heisenberg uncertainty principle. The equation si `deltax.delta (mv)ge(h)/(4pi)` The uncertainty in the position or in the momentum of a marcroscopic object like a baseball is too small to observe. However, the mass of microscopic object such as an electon is small enough for the uncertainty to be relatively large and significant. What would be the minimum uncetaintty in de-Broglie wavelength of a moving electron accelerated by potential difference of 6 volt and whose uncetainty in position is `(7)/(22)` nm?A. `6.25 Å`B. `6 Å`C. `0.625 Å`D. `0.1325 Å` |
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Answer» Correct Answer - C `lambda_(D.B.)=sqrt(150/6) Å=5 Å` and `Deltax.Deltap ge h/(4pi), p=h/lambda` or `Deltap=h/lambda^(2) Deltalambda` `rArr Deltax. h/lambda^(2)xxDeltalambda ge h/(4pi)` `rArr 1/pixx10^(-9)/lambda^(2)xxDeltalambda ge 1/(4pi)rArr Deltalambda ge 2.5/4xx10^(-10)` `Deltalambda ge 0.625 Å` |
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| 277. |
The weight of `1` curie `._(82)Pb^(214) (t_(1//2)=26.8 min)` in grams isA. `3.1xx10^(-8) g`B. `1.55xx10^(-8) g`C. `6.2xx10^(-8)g`D. `3.1xx10^(-10)g` |
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Answer» Correct Answer - a We know that, `(-dN)/(dt) = lambda(6.023xx10^(23)xxw)/(M)` Here, `(-dN)/(dt)` = 1 curie `= 3.7xx10^(10)dps` `lambda = (0.693)/(t_(1//2)) = (0.693)/(26.8xx60)s^(-1)` `:. 3.7xx10^(10) = (0.693)/(26.8xx60)xx(6.023xx10^(23)xxw)/(214)` or `w = 3.1xx10^(-8)g` |
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| 278. |
The amount of substance that give `3.7xx10^(7)` dps isA. one becquerelB. one curieC. one millicurieD. one rutherford |
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Answer» Correct Answer - c The amount of substance, which give `3.7xx10^(10)` disintegration per second, is known as 1 curie. Hence, `3.7xx10^(7) dps =` 1 millicuire |
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| 279. |
Natural radioactivity was discovere byA. SchmidtB. CurieC. BecquerelD. Rutherford |
| Answer» Correct Answer - C | |
| 280. |
In the decay process: `A overset(- alpha)rarr B overset(-beta)rarr C overset(-beta)rarr D`A. `A` and `B` are isodiaphersB. `A` and `C` are isotonesC. `A` and `C` are isotopesD. `B, C` and `D` are isobars |
| Answer» Correct Answer - A::C::D | |
| 281. |
What percentage of decay takes place in the average life of a substance ?A. `63.21%`B. `36.79%`C. `90%`D. `99%` |
| Answer» Correct Answer - A | |
| 282. |
In a chain reaction uranium atom gets fissioned forming two different material. The total weight of these put together isA. More than the weight of parent uranium atomB. Less than the weight of parent uranium atomsC. More of less depends upon experimental conditionsD. Neither more nor less |
| Answer» Correct Answer - B | |
| 283. |
In the case of a radioisotope, the value of `t_(1//2)` and `lambda` are identical in magnitude. The value of `t_(1//2)` isA. `1//0.693`B. `(0.693)^(1//2)`C. `(0.693)/(2)`D. `0.693` |
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Answer» Correct Answer - b We know that, `t_(1//2) = (0.693)/(k) or (t_(1//2))^(2) = 0.693` (as `t_(1//2) = K`) `t_(1//2) = (0.693)^(1//2)` |
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| 284. |
The `alpha`-particle is identical withA. Helium nucleusB. Hydrogen nucleusC. ElectronD. Proton |
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Answer» Correct Answer - A `alpha`-particle is identical with `._(2)He^(4)`, helium nucleus |
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| 285. |
Uncertainty in position of a hypothetical subatomic particle is `1 "Å"` and uncertainty in velocity is `(3.3)/(4 pi) xx 10^(5)` m/s then the mass of the particle is approximately ( h = `6.6 xx 10^(-34)` Js) :A. `2xx10^(-28) kg`B. `2xx10^(-27) kg`C. `2xx10^(-29) kg`D. `4xx10^(-29) kg` |
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Answer» Correct Answer - C `Deltax xx m xx Deltav ge h//4pi` `1 xx10^(-10)xxmxx3.3/(4pi)xx10^(5) ge (6.6xx10^(-34))/(4xxpi) m=2xx10^(-29) kg` |
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| 286. |
An electron moving near an atomic nucleus has a speed of `6xx10^(6) +- 1% m//s`. What is the uncertainty in its position? |
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Answer» Correct Answer - `ge 1xx10^(-9) m` Findding the uncertainty in speed `Deltau, Deltau =(6xx10^(6) m//s)(0.01)=6xx10^(4) m//s` Calculating the uncertainty in position, `Deltax: Deltax ge (h)/(4pimDeltau) ge (6.626xx10^(-34) kg. m^(2)//s)/(4pi(9.11xx10^(-31) kg)(6xx10^(4) m//s)) ge 1xx10^(-9) m` |
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| 287. |
An electrons in a hydrogen atom finds itself in the fourth energy level. (i) Write down a list of the orbits that it might be in. (ii) Can it be in all of these orbitals at once? (iii) Can you tell which orbital it is in ? |
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Answer» Correct Answer - (i) `4s, 4P, 4d, 4f` (ii) No, it will only be in one of them. (iii) No. For the hydrogen atom all orbitals with the same principal quantum number have the same energy (they are degenerate). (i) `4s, 4P, 4d, 4f` (ii) No, it will only be in one of them. (iii) No. For the hydrogen atom, all orbitals with the same principal quantum number have the same energy (they are degenerate). (i) `4s, 4P, 4d, 4f` |
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| 288. |
An isotope of `._(90)Th^(231)` can be converted into `._(90)Th^(227)` by the emission ofA. one `alpha`-particleB. one `beta`-particleC. two `alpha` and one `beta`-particleD. one `alpha` and `2 beta`- particles |
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Answer» Correct Answer - d Isotopes of an element can be obtained by emitting one `alpha`- and `2beta`- particles from its nucleus `._(90)Th^(231) overset(-alpha)(to) ._(88)Ra^(227) overset(-beta)(to) ._(89)Ac^(227) overset(-beta)(to) ._(90)Th^(227)` |
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| 289. |
An isobar of `._(20)Ca^(40)` isA. `._(18)Ar^(40)`B. `._(20)Ca^(38)`C. `._(20)Ca^(42)`D. `._(18)Ar^(38)` |
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Answer» Correct Answer - a |
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| 290. |
A chemist prepares `1.00 g` of pure `._(6)C^(11)`. This isotopes has half life of 21 min, decaying by the equation: a. What is the rate of disintegration per second (dps) at starts ? b. What is the activity and specific activity of `._(6)C^(11)` at start? c. How much of this isotope `(._(6)C^(11))` is left after 24 hr its preparation? |
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Answer» Applying `- (dN)/(dt) = lambda N_(0)` `= (0.693)/(21 xx 60) xx (1 xx 6.02 xx 10^(23))/(11)` `= 3 xx 10^(19) dps` b. Acitivity `= (3 xx 10^(19))/(3.7 xx 10^(10))` (1 curie `= 3.7 xx 10^(10) dps)` `= 8.108 xx 10^(8)` curie Specifics activity `= 3 xx 10^(19) xx 10^(3) xx 10^(22)` dis `(kg s)^(-1)` `= 8.108 xx 10^(11)` curie c. Applying `N = N_(0) ((1)/(2))^(n) [n = (t)/(t_(1//2)) = (24 xx 60)/(21) = 68.57]` `N = 1 xx ((1)/(2))^(68.57) = 2.29 xx 10^(-21) g` |
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| 291. |
Particles having energy of several hundred `MeV` are known asA. ElectronsB. NucleonsC. Fast particlesD. Super fast particles |
| Answer» Correct Answer - D | |
| 292. |
An isotope of `._(32)Ge^(76)` isA. `._(32)Ge^(77)`B. `._(33)As^(77)`C. `._(34)Se^(77)`D. `._(34)Se^(78)` |
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Answer» Correct Answer - A Atoms having same number of protons are called isotopes. |
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| 293. |
Wave length associated with electron motionA. increases with increase in the speed of electronB. remains same irrespective of the speed of electronC. decreases with increase of the speed of electronD. changes with the atomic number of the atom to which it belond |
| Answer» Correct Answer - C | |
| 294. |
The correct set of quantum numbers for the unpaired electron of chlorine atom isA. `n=2, l=1, m=0`B. `n=2, l=1, m=1`C. `n=3, l=1, m=1`D. `n=3, l=0, m=0` |
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Answer» Correct Answer - 3 `Cl_(17): [Ne] 3s^(2) 3p^(5)` Unpaired electron is in `3p` orbital. `:. N=3, l=1, m=1, 0, -1`. |
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| 295. |
The metal that shows photoelectric emission at lowest frequency radiation isA. berylliumB. lithiumC. sodiumD. magnesium |
| Answer» Correct Answer - C | |
| 296. |
Which of the following has the maximum number of unpaired electrons?A. `Mn`B. `Ti`C. `V`D. `Al` |
| Answer» Correct Answer - 1 | |
| 297. |
The number of unpaired electrons in the scandium atom isA. `1`B. `2`C. `0`D. `3` |
| Answer» Correct Answer - A | |
| 298. |
Hydrogen atom: The electronic ground state of hydrogen atom contains one electron in the first orbit. If sufficient energy is provided, this electron can be promoted to higher energy levels. The electronic energy of a hydrogen-like species (any atom//ions with nuclear charge Z and one electron) can be given as `E_(n)=-(R_(H)Z^(2))/(n^(2))` where `R_(H)= "Rydberg constant," n= "principal quantum number"` The ratio of energy of an electron in the ground state `Be^(3-)` ion to that of ground state H atom is: The kinetic and potential energies of an electron in the H atoms are given as `K.E. =e^(2)/(4 pi epsilon_(0)2r)` and `P.E.=-1/(4pi epsilon_(0)) e^(2)/r`A. `16`B. `4`C. `1`D. `8` |
| Answer» Correct Answer - `16` | |
| 299. |
Hydrogen atom: The electronic ground state of hydrogen atom contains one electron in the first orbit. If sufficient energy is provided, this electron can be promoted to higher energy levels. The electronic energy of a hydrogen-like species (any atom//ions with nuclear charge Z and one electron) can be given as `E_(n)=-(R_(H)Z^(2))/(n^(2))` where `R_(H)= "Rydberg constant," n= "principal quantum number"` A gaseous excited hydrogen-like species with nuclear charge Z can emit radiations of six different photon energies. (a) The principal quantum number of the excited state is : (b) It was observed that when this excited species emits photons of energy`=2.64 eV` when it comes to next lower energy state. Calculate the nuclear charge of the species. The least energy required to remove an electron from a species is know as the ionization energy (I.E.) of the species. The experimental I.E. of He atom is `24.58 eV`.A. `6`B. `5`C. `4`D. `3` |
| Answer» Correct Answer - (a) `4` (b) `2` | |
| 300. |
In `1894`, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables `1` and `2`. Latter, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were mearsured by using a glass vessel with a known volume under atmospheric pressure `(1.013xx10^(5)Pa)`. `|{:("From nitric oxide",,2.3001 g),("From nitrous oxide",,2.2990 g),("From amonium nitrite purified at a red heat",,2.2987 g),("From urea",,2.2985 g),("From ammonium nitrite purified in the cold",,2.2987 g),("Mean",,2.2990 g):}|` `|{:(O_(2) "was removed by hot copper" (1892),,2.3103 g),(O_(2) "was removed by hot iron" (1893),,2.3100 g),(O_(2) "was removed by ferrous hydrate (1894),2.3102 g,),(Mean,,2.3102 g):}|` Ramsay and cleve discovered helium in cleveite (a mineral consiting of uranium oxide and oxides of lead, thorium, and rare earths, an impure variety of uraninite) independently and virtually simultaneously in `1895`. The gas extracted from the rock showed a unique spectroscopic line at around `588` nm (indicated by `D3` in Figure `1`), which was fist observed in the spectrum of solar prominence during a total eclipse in `1868`, near the well-known `D_(1)` and `D_(2)` lines of sodium. `ul("Which")` equation explains the occurrence of helium in cleveite among `[A]` to `[D]` below? Mark one.A. `.^(238)U rarr .^(234)Th+alpha`B. `Uhe_(2) rarr U+2 He`C. `.^(240)U rarr .^(240)Np+beta^(-)`D. `.^(235)U+n rarr .^(95)Y+^(139)l+2 n` |
| Answer» Correct Answer - Considering that the a particle is the nucleus of helium a decay `[A]` is the relevent source of helium in such rocks. No compounds of He such as `UHe2` in `[B]` is known to be stable at ambient temperature. `[C]` is a radioactive decay of `240 U` in the thorium series. `[D]` is a nuclear fission reaction of `235U` occurring in nuclear reactors. Thus, the correct answer is `[A]` | |