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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.
| 351. |
Some ratioisotopes are used for medical imaging . A radioisotope is injected into th blood of a person. Malignant tissue absorb and retian this isotope much more efficiently than other healthy tissues. A sample of radio isotope having half life of 5 hour and activity of 22mCi is injected into the blood aof a patient. the isotope enters into thyroid glnads through the blood streams. The isotope radiates gamma photons each having energy of `2xx10^(-8)muJ`. (a) Qualitatively sketch the amount of radio isotope in the thyroid glands as a function of time. (b) Sometime after the injection a scan of the glands is taken over a period of 20 minute. The total energy emitted by the glands was found to be `6xx10^(-4)J`. Assume that the thyroid glands are quickly saturated (with isotopes) after the injection and almost 50% of the injected isotopes are absorbed by the glands. Make an estimate of the time after the injection when the scan was performed. |
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Answer» Correct Answer - (b) 20 hours |
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| 352. |
`N` atoms of a radioactive element emit `n` alpha particles per second. The half-life of tge element is.A. `n/N` secB. 1.44 `n/N` secC. 0.69 `n/N` secD. 0.69 `N/n` sec |
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Answer» Correct Answer - D |
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| 353. |
A radioactiev nuleus `X` deays to a stable nuleus `Y`. Then, time graph of rate of formation of `Y` against time `t` will be:A. B. C. D. |
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Answer» Correct Answer - c `N=N_(0) e^(- lmabda t),N_(Y) =N_(0)(1-e^(-lambda t))` Rate of formation of `Y` is `(dN)/(dt) =+ lambda N_(0)e^(-lambda t)` Which decreases exponentially with time. |
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| 354. |
A heavy nuleus having mass number `200` gets disintegrated into two small fragmnets of mass numbers `80` and `120`. If binding energy per nulceon for parent atom is `6.5 MeV` and for daughter nuceli is `7 MeV` and `8 MeV`, respectivley , then the energy released in the decay will be.A. `200 Me V`B. `-220 M eV`C. `220 M eV`D. `180 M eV` |
| Answer» Correct Answer - C | |
| 355. |
A radioactiev nuleus `X` deays to a stable nuleus `Y`. Then, time graph of rate of formation of `Y` against time `t` will be:A. B. C. D. |
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Answer» Correct Answer - C |
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| 356. |
In the fission of `._(94)^(239) Pu` by a thermal neutron, two fission fragmnets of equal masses and sizes are produced and four neutrosn are emitted. Find the force between the two fission fragments at the moment they are produced. Given : `R_(0) =1.`1 fermi. |
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Answer» Correct Answer - `4.37 xx 10^(3)N` Total mass number of `._(94)^(239)Pu` + neutron (thermal) `=239 +1=240`. Since `4` neutrons are produced, the mass number of each fragment `(A)=(240-4)/(2) =118`. The atomic number of each fragment `=94//2=47`. Therefore charege of each fragment is `q=47 xx 1.6 xx 10^(-19)=7.52 xx 10^(-18) C` The radius of each nucleus of the fragment is `R=R._(0)(A)^(1//3)` `=1.1 xx 10^(-15) xx (118)^(1//3)` `=5.395 xx 10^(-15)m` Distance between the centres of the two fragments at the moment they are ptoduced is `r =2 xx 5.395 xx 10^(-15) =10.79 xx 10^(-15)m ` The electrostatic force between them is `F=(1)/(4 pi epsilon_(0)) (q^(2))/(r^(2)) =9 xx 10^(9) (7.5 xx 10^(-18))^(2)/((10.79 xx 10^(-15)).^(2))=4.37 xx 10^(3)N` . |
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| 357. |
The chemical behaviour of a atom depends upon:A. The number of electrons orbiting aroung its nucleus.B. The number of protons in its nucleusC. The number of neutrons in its nucleusD. The number of nucleons in its nucleus |
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Answer» Correct Answer - A |
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| 358. |
A factory produces a radioactive substance A at a constant rate R which decays with a decay constant ` lmabda` to form a stable substance. Find (a) the number of nuclei of A and (b) number of nuclei of B, at any time t assuming the production of A starts at `t=0`. (c ) Also, find out the maximum number of nuclei of `A` present at any time during its formation. |
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Answer» Factory `underset(const.rate)overset(R) rarr A underset(decay)overset(lambda) rarrB` Let `N` be the number of nuclei of A any time t. `because (dN)/(dt)=R-lambdaN" "underset(0)overset(N)(int)(dN)/(R-lambdaN)=underset(0)overset(t)(int)dt` On solving, we will get `N = R//lambda (1-e^(-lambda t))` (b) Number of nuclei of B at any tiem t, `N_(B) =Rt -N_(A)` `=R t-R//lambda(1-e^(-lambda t) )=R//lambda(lambda t -1 +e^(- lambda t))` Maximum number of nuclei of 'A' present at any time during its formation ` =R//lambda` . |
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| 359. |
In nuclear power station energy of uranium is used for producing-A. Electric energyB. Mechanical energyC. Heat energyD. magnetic energy |
| Answer» Correct Answer - A | |
| 360. |
Order of ionization power of `alpha, beta` and `gamma` rays isA. `alpha gt gamma gt beta`B. `alpha gt beta gt gamma`C. `alpha lt beta lt gamma`D. `alpha gt beta lt gamma` |
| Answer» Correct Answer - B | |
| 361. |
The half -life of cobalt-60 is 5.25 years. How long after a new sample is delivered, will the activity have decreased to about one third `(1//3)` of its original value-A. 5.25 yearsB. 8.3 yearsC. 10.50 yearsD. 15.75 years |
| Answer» Correct Answer - B | |
| 362. |
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 a long time bombardment, number `.^(56)Mn` nuclei present in the target depends upon.A. All (i), (ii), and (iii) are correct.B. Only (i) and (ii) are correct.C. Only (ii) and (iii) are correct.D. Only (i) and (iii) are correct. |
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Answer» Correct Answer - c As `N =(Pt_(1//2))/(In2)` It is dependent upon `P` and `t_(1//2)`. Initial number of `.^(56)Mn` nuclei will make no differnce as in equilibrum, rate of production equals rate of decay. Large intial number will only make equilibrium come sooner. |
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| 363. |
The half-life of a radioactive decay is x times its mean life. The value of zx isA. `0.3010`B. `0.6930`C. `0.6020`D. `(1)/(0.6930)` |
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Answer» Correct Answer - b `T_(1//2) =(0.693)/(lambda)`or `T_(1//2)=0.693[(1)/(lambda)]` or `T_(1//2)=0.693 tau` Clearly, `x=0.693`. |
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| 364. |
Neutron decay in the free space is given follows: `._0n^1 rarr. _1H^1 +._(-e)^0 +[]` Then, the parenthesis representsA. photonB. gravitonC. neutrinoD. antineutrino |
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Answer» Correct Answer - d The emission of antineutron is a must for the vallidity of different laws. |
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| 365. |
Fusion reaction takes place at high temperature becauseA. atoms are ionised at high temperatureB. molecules break-up at high temperatureC. nuclei break-up at high temperatureD. kinetic energy is high enough to overcome repulsion between nuclei |
| Answer» Correct Answer - D | |
| 366. |
An `alpha` particle is bombarded on `^14N`. As a result, a `^17O` nucleus is formed and a particle is emitted. This particle is aA. neutronB. protonC. electronD. positron |
| Answer» Correct Answer - B | |
| 367. |
A gamma ray photon of energy `1896 MeV` annihilates to produce a photon-antiproton pair. If the rest mass of each of the particles involved be `1.007276 a.m.u` aapproximately, find how much `K.E` these will carry? |
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Answer» Working on the same lines as an electron-position pair production, we notice that the reaction. `gamma rarr` Proton + antiproton, has the energy balance `E=m_(0("Proton"))C^(2)+K.E_(("Proton"))+` `m_(0("antiproton"))C^(2)+K.E_(("antiproton"))` But `m_(0)C^(2)=` energy equivalent of `1.007276 a.m.u` `~~ 938 MeV.[ because 1.007276xx931~~938 MeV]` Thus `K.E`. of each particle `=(1)/(2)[1896MeV-2xx938MeV]=10 MeV`. |
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| 368. |
During a beta decayA. An atomic electron is ejectedB. An electron which is already present within the nucleus is ejected.C. A neutron in the nucleus decays emitting an electronD. A part of the binding energy is converted into electron. |
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Answer» Correct Answer - C ( c) Negative `beta-decay` is expressed by the equation `n = p^+ + p^- + v^-` `gamma-rays` are highly penetrating. |
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| 369. |
A carbon nucleus emits a particles x and changes into nitrogen according to the equation: `._(6)^(14)Crarr._(7)^(14)N+x` What is x?A. An electonB. A protonC. An alpha particleD. A photon |
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Answer» Correct Answer - A |
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| 370. |
A nucleus moving with velocity `bar(v)` emits an `alpha`-particle. Let the velocities of the `alpha`-particle and the remaining nucleus be `bar(v)_(1)` and `bar(v)_(2)` and their masses be `m_(1)` and `(m_2)` then,A. `underset (v) rarr , underset (v_(1))rarr and underset (v_(2))rarr` must be parallel to each otherB. none of the two of `underset (v) rarr , underset (v_(1))rarr and underset (v_(2))rarr` should be paralle to each otherC. ` underset (v_(1))rarr + underset (v_(2))rarr` must be parallel to `underset (v) rarr `.D. ` m_(1) underset (v_(1)) rarr + m_(2)underset (v_(2)) rarr ` must be parallel to `underset (v) rarr` |
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Answer» Correct Answer - d Since no external force is present, so momnetum conservation principle is completely appllicable `:. M because m vecv =vec m_(1)vecv_(1)+m_(2) vecv_(2)` or `(m_(1) +m_(2) vecv)=m_(1)vecv_(1)+m_(2)vecv_(2)`. |
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| 371. |
Which of the following statements is incorrect for nuclear forces?A. These are strongest in magnitude.B. They are charge dependent.C. They are effective only for short ranges.D. They are from interaction of every nucleon with the nearest limited number of nucleons. |
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Answer» Correct Answer - d Nulcear forces are charge independent. |
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| 372. |
A certain radioactive material can undergo three constant `lambda,2lambda` and `3 lambda`. Then, the effective decay constant `lambda_(eff)` is equal to `n lambda`. What is the value of n?A. ` 6 lambda`B. ` 4 lambda`C. `2 lambda`D. ` 3 lambda` |
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Answer» Correct Answer - a `(dN_(A))/(dt)=(- lambda N_(A))+(- 2 lambda N_(A)) +(-2 lambdaN_(A))=-6 lambda N_(A)`. |
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| 373. |
A certain radioactive material can undergo three constant `lambda,2lambda` and `3 lambda`. Then, the effective decay constant `lambda_(eff)` is equal to `n lambda`. What is the value of n? |
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Answer» Correct Answer - 6 Effective decay constant will be sum of all different decay constant So, `lambda_(eff)=lambda+2lambda+3lambda=6lambda,`hence `n=6`. |
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| 374. |
The radioactive of a sample is `R_(1)` at a time `T_(1)` and `R_(2)` at a time `T_(2)`. If the half-life of the specimen is `T`, the number of atoms that have disintegrated in the time `(T_(2)-T_(1))` is equal to`(n(R_(1)-R_(2))T)/(ln4) `. Here n is some integral number. What is the value of n? |
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Answer» Correct Answer - B `R_(1)=lambdaN_(1) , R_(2)=lambda N_(2)` No. of atoms decayed in `T_(1)-T_(2)=N_(1)-N_(2)=(R_(1)-R_(2))/(lambda)` `=((R_(1)-R_(2)))/(log 2)=(2(R_(1)-T_(2)))/(log 4)rArr n=2` |
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| 375. |
If a radioactive substance reduces to `(1)/(16)` of its original mass in `40` days, what is its half-life ?A. 10B. 5C. 2.5D. 20 |
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Answer» Correct Answer - A (a) From Rutherford and Soddy law for radioactive decay `N = N_0 e^(-lamda t)` where `N_0` and `N` are number of atom in a radioactive substance at time` t = 0` and `t = t` and `lamda` is decay constant. Also, half-life `T_(1//2) = (0.693)/(lamda)` `:. (N)/(N_0) = e^((0.683)/(-T^(1//2)) t)` Given, `t = 40 days, (N)/(N_0) = (1)/(16)` `:. (1)/(16) = e^((-(0.693 xx 40))/(T^(1//2)))` `rArr (0.693 xx 40)/(T^(1//2)) = log 16` `T_(1//2) = (0.693 xx 40)/(log 16)` =`9.99 ~~ 10 days`. |
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| 376. |
The dependence of binding energy per nucleon, `B_N` on the mass number, `A` is represented by. (a) , (b) , ( c) , (d) .A. .B. .C. .D. . |
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Answer» Correct Answer - B (b) From the concept of binding energy, the binding energy per nucleon increases on an average and reaches a maximum of about `8.7 MeV` per nucleon for `A = 50 - 80` and generally a peak is found for `._26 Fe^56`, which is an iron isotope. This is also the most stable nucleus form all them because it has even number of protons (26) and even number of neutrons (30) i.e., even-even pair. Stability of iron can also be judged from the fact that, maximum energy is needed to pull a nucleus away from it. Also it has even number of protons and neutrons, hence it is most stable. |
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| 377. |
A neutron of energy 1 MeV and mass `1.6 xx 10^(-27)` kg passes a proton at such a distance that the angular momentum of the neutron relative to the proton approximately equals `10^(-33) J s`. The distance of closest approach neglecting the interaction between particles siA. `0.44nm`B. `0.44mm`C. `0.44Å`D. `0.44fm` |
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Answer» Correct Answer - d If `d` is the distance of closet approach given, then the angular momentum `=mvd=10^(-33) J s` `E=(1)/(2) mv^(2) =1 MeV =1.6 xx 10^(-33) J` Momentum, `P=sqrt(2m_nE)=sqrt(2 xx 1.6xx10^(27) xx1.6 xx 10^(-13))` `=1.6 sqrt2 xx 10 ^(20) kg m s^(1)` Distance of clossest approach, `d=(10^(33))/(1.6sqrt2xx10^(-20))` `=(1)/(1.6sqrt2) xx 10^(-13)=(100)/(1.6sqrt2)fm =0.44 fm` |
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| 378. |
In problem` 43`, number of atoms decayed between time interval `t_(1)` and `t_(2)` areA. `(ln(2))/(lambda) (R_(1) R_(2))`B. `R_(1) e^(-lambda t_(2)) -R_(2) e^(-lambda t_(2))`C. `lambda(R_(1)-R_(2))`D. `(R_(1)-R_(2))/(lambda))` |
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Answer» Correct Answer - d `R_(1) =lambdaN_(1) rArr N_(1) =(R_(1))/(lambda)` and `R_(2) =lambdaN_(2)rArr N_(2) =(R_(2))/(lambda)` Therefore, number of atoms decayed `=N_(1)-N_(2) =((R_(1)-R_(2))/(lambda))`. |
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| 379. |
The ratio of molecular mass of two radioactive substances is `3//2` and the ratio of their decay cosntatnt is `4//3`. Then. The ratio of their initial activity per mole will beA. `2`B. `(8)/(9)`C. `(4)/(3)`D. `(9)/(8)` |
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Answer» Correct Answer - c Activity, `R=lambdaN`. Number of nuclei `N` per mole are equal for both the substance. `:. R prop lambda` or `(R_(1))/(R_(2)) =(lambda_(1))/(lambda_(2)) =(4)/(3)`. |
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| 380. |
The half life of `.^(218)Pa` is 3 minutes. What mass of a 16 g sample of `.^(218)Pa` will remain after 15 minutes ?A. 3.2 gB. 2.0 gC. 1.6 gD. 0.5 g |
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Answer» Correct Answer - D |
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| 381. |
Find the energy required for separation of `a_(10)Ne^(20)` nucleus into two `alpha`-particles and `a_(6)C^(12)` nucleus if it is known that the binding energies per nucleon in `._(10)Ne^(20), ._(2)He^(4)` and `._(6)C^(12)` nuclei are equal to `8.03, 7.07` and `7.68 MeV` respectively. |
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Answer» Correct Answer - [`E = 20 ._(Ne) - 2.4 E_(alpha) - 12 ._(c) = 11.9 MeV`, where E is the B.E per nucleon in the corresponding nucleus] |
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| 382. |
In pair annihiliation two `gamma`-ray photons are produced it is due toA. Law of conservation of energyB. Law of conservation of massC. Law of conservation of momentumD. Law of conservation of angular momentum |
| Answer» Correct Answer - C | |
| 383. |
Positronium is converted intoA. `2` Photons each of energy `0.51 MeV`B. Photon of energy `1.02 MeV`C. `2` Photons each of energy `1.02 MeV`D. One photon of energy `0.51 MeV` |
| Answer» Correct Answer - A | |
| 384. |
The phenomenon of pair production is theA. formation of an electron and a position from `gamma`-raysB. ejection of an electron from a metal surface when exposed to ultra-violet lightC. ejection of an electron form a nucleusD. ionization of a neutral atom |
| Answer» Correct Answer - A | |
| 385. |
A positron and an electron come close together to give a neutral one calledA. ElectroniumB. PositroniumC. `gamma`-photonD. `beta`-particle |
| Answer» Correct Answer - C | |
| 386. |
Following process is known as `hv rarr e^(+) +e^(-)`A. Pair productionB. Photoelectric effectC. Compton effectD. Zeeman effect |
| Answer» Correct Answer - A | |
| 387. |
The rest mass energy of electron or positron is (in `MeV`)A. `0.51`B. `1`C. `1.02`D. `1.5` |
| Answer» Correct Answer - A | |
| 388. |
To produce an electron-position pair, the minimum energy of `gamma`-ray photon must beA. `0.15`B. `1`C. `1.02`D. `1.5` |
| Answer» Correct Answer - C | |
| 389. |
To produce an electron-position pair, the minimum energy of `gamma`-ray photon must beA. `1.02keV`B. `1.02MeV`C. `1.02BeV`D. `1.02eV` |
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Answer» Correct Answer - B Rest energy of electron `= (9.1xx10^(-31)xx(3xx10^(8))^(2))/(1.6xx10^(-13))MeV` `= 0.510 MeV` `gamma rarr e^(+) +e^(-)` Energy of photon `= 2xx0.51 = 1.02MeV` |
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| 390. |
When radioactive substance emits an `alpha-`particle, then its position in the periodic table is lowered by.A. three placesB. one placeC. five placesD. two places |
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Answer» Correct Answer - D (d) When an `alpha-`particle is emitted from the nucleus of a radioactive atom, the atomic number is reduced by `2` and mass number is reduced by `4`. `._Z X^A rarr ._(Z-2)Y^(A - 4) + ._2He^4 (alpha-`particle) Elements are placed in the periodic table according to their atomic number. The emission of an `alpha-`particle from the atom of an element reduces its atomic number by `2`, so the element is converted into a new element which occurs `2` steps earlier in the periodic table. |
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| 391. |
The fraction of heavy water in a nuclear reactor is to.A. increase the neutronsB. slow down the neutronsC. stop the electronsD. None of the above |
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Answer» Correct Answer - B (b) A nuclear reactor is a device in which a self-sustaining controlled chain reaction is produced in a fissionable material. A nuclear reactor consists of a moderator which is used to slow down the neutrons. Heavy water is the best moderator. |
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| 392. |
Which one of the following has the highest neutrons ratio ?A. `._8 O^16`B. `._2 He^4`C. `._92 U^235`D. `._26 Fe^56` |
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Answer» Correct Answer - C ( c) The representation `._92 U^235` shows `235` is mass number and `92` is atomic number. Hence, number of neutrons =`235 - 92 = 143` `._8 O^16` has `16` as mass number and `8` as atomic number. Number of neutorn `= 16 - 8 = 8` `._2 He^4` has `4` as mass number and `2` as atomic number. Number of neutrons `= 4 - 2 = 2` `._26 Fe^56` has `56` as mass number and `26` as atomic number, Number of neutrons `= 56 - 26 = 30` Hence, `._92 U^235` has highest neutrons ratio. |
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| 393. |
Which of the following is a correct statement?A. Beta rays are same as cathode raysB. Gamma rays are high energy neutronsC. Alpha particle are singly ionized helium atomsD. Protons and neutrons have exactly the same mass. |
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Answer» Correct Answer - A (a) Both the `beta-`rays and the cathode rays are made up of electrons. `gamma-`rays are `EM` waves. `alpha-`particles are doubly ionized helium atoms and protons and neutrons have approximately the same mass. |
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| 394. |
Some radioactive nucleus may emitA. only one `(alpha, gamma)(beta, gamma)` at a timeB. all the three `alpha, beta` and `gamma` one after anotherC. all the three `alpha, beta` and `gamma` sinultaneouslyD. only `alpha` and `beta` simultaneously |
| Answer» Correct Answer - A | |
| 395. |
Some radioactive nucleus may emit.A. Only one `alpha, beta` or `gamma` at a time.B. All the three `alpha, beta` and `gamma` one after another.C. All the three `alpha, beta` and `gamma` simultaneouslyD. Only `alpha` and `beta` simultaneously |
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Answer» Correct Answer - A (a) No radioactive substance emit both `alpha` and `beta` particles simultaneously. Some substance emit `alpha-`particles and some other emits `beta-`particles. `gamma-`rays are emitted along with both `alpha` and `beta`-particles. |
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| 396. |
A solution contains a mixture of two isotopes A (`half-life=10days`) and B (`half-life=5days`). Total activity of the mixture is `10^10` disintegration per second at time `t=0`. The activity reduces to `20%` in `20 days`. Find (a) the initial activities of A and B, (b) the ratio of initial number of their nuclei. |
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Answer» Correct Answer - `[(a) 0.73xx10^(10) dps 0.27xx10^(10) dps, (b) 5.4]` |
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| 397. |
The energy spectrum of `beta` - particle [number `N€` as a function of `beta` - energy E] emitted from a radioactive source isA. B. C. D. |
| Answer» Correct Answer - D | |
| 398. |
A radiaoactive nucleus a series of decays according to the scheme `A underset rarr (alpha) A_(1) underset rarr (beta) A_(2) underset rarr (alpha)A_(3) underset rarr(gamma) A_(4)` If the mass number and atomic number of `A `are `180` and `72`, respectively, then what are these number for `A_(4)` ?A. `172 and 69`B. `174 and 70`C. `176 and 69`D. `176 and 70` |
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Answer» Correct Answer - a Two `alpha`-particles reduce mass number by `8 `. Therefore, new mass number `=180 -8=172` Emission of two `alpha`-particles reduces charge number by `4`. Emission of `beta`-particles increases charge number by `1`. Therefore the new charge number `=72-4+1=69`. |
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| 399. |
The percentage of quantity of a radioactive material that remains after 5 half-lives will be .A. `0.3%`B. 0.01C. 0.31D. `3.125%` |
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Answer» Correct Answer - D (d) Fraction of atoms remains after five half lives `(N)/(N_0) = ((1)/(2))^(t//T) = ((1)/(2))^(5T//T) = (1)/(32)` rArr Percentage atom remains `= (1)/(32) xx 100 = 3.125 %`. |
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| 400. |
The percentage of quantity of a radioactive material that remains after 5 half-lives will be .A. `1%`B. `3%`C. `5%`D. `20%` |
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Answer» Correct Answer - B `N = N_(0) ((1)/(2))^(n) = N_(0)((1)/(2))^(5) = (N_(0))/(32)` `(N)/(N_(0)) xx 100 = (100)/(32) ~~ 3%` |
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