InterviewSolution
Saved Bookmarks
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.
| 601. |
Calculate the number of `alpha`- and `beta`-particles emitted when `._(92)U^(238)` into radioactive `._(82)Pb^(206)`.A. `8alpha, 6beta`B. `6alpha, 6 beta`C. `6alpha, 8beta`D. `4alpha, 4beta` |
| Answer» Correct Answer - A | |
| 602. |
Calculate the number of `alpha`- and `beta`-particles emitted when `._(92)U^(238)` into radioactive `._(82)Pb^(206)`. |
|
Answer» Parent element =`._(92)U^(238)`, daughter element =`._(82)Pb^(206)`. Let `x alpha-` and `y beta`- Particles be given out during the change `._(92)U^(238) rarr ._(82)Pb^(206) + x_(2)He^(4) + y ._(-1)e^(0)` Equating mass number on both sides `238 = 206 + 4x y xx 0` or `x = 8` Equating atomic number on both sides `92 = 82 + 2x + y (-1)` `= 82 + 2 xx 8 + y (-1)` or `y = 6` `:.` Number of `alpha`-particles = 8, Number of `beta- particles = 6 |
|
| 603. |
In the decay series `._(92)U^(238)` to `._(82)Pb^(206)`, how many `alpha`-paritcles and how many `beta^(ɵ)`-particles are emitted?A. 6,8B. 9,6C. 8,8D. 8,6 |
|
Answer» Correct Answer - D `._(92)U^(238) rarr ._(82)Pb^(206) + x ._(2)He^(4) + y ._(-1)e^(0)` Equating mass number on both sides `238 = 206 + x xx 4 + y xx 0` `:. X = (238 - 206)/(4) = 8` Equating atomic number on both sides `92 = 82 + 2 xx 8 + y (-1)` `:. y = 98 - 92 = 6` |
|
| 604. |
In the decay series `._(92)U^(238)` to `._(82)Pb^(206)`, how many `alpha`-paritcles and how many `beta^(ɵ)`-particles are emitted? |
| Answer» The change in mass is 238 - 206 = 32 unit. It means that `32//4 = 8 alpha`-particles are emitted. With the emission of `8 alpha`-particles, the change in atomic number will be `8 xx 2 = 16`, i.e., the new element would have atomic number `92 - 16 = 76`. But the final product Pb has atomic number 82. It means there would have been an emission of 82 - 46 = 6 `beta`-particles. | |
| 605. |
Which of the following nuclides are `beta^(ɵ)` and `beta^(o+)` emitter, and stable nuclides? `._(20)Ca^(49)` b. `._(80)Hg^(195)` c. `._(82)Pb^(208)` d. `._(5)B^(8)` e. `._(67)Ho^(150)` f. `._(13)Al^(30)` e. `._(50)Sn^(120)` g. `._(36)Kr^(94)` |
|
Answer» `beta^(ɵ)`- emitter `implies` a. `._(20)Ca^(49)` f. `._(13)Al^(30)` h. `._(36)Kr^(94)` All of them have `Z` less than 82. a. `._(20)Ca^(49) implies (n)/(p) = (29)/(20) = 1.45` `[{:(n//p "higher and is"),("lowered by"),(beta^(ɵ)-"emission"),((atomic weight" lt 82)):}]` f. `._(13)Al^(30) implies (n)/(p) = (17)/(13) = 1.3` h. `._(36)Kr^(94) implies (n)/(p) = (58)/(36) = 1.3` `beta^(o+)`-emitter `implies` b. `._(80)Hg^(195)` d. `._(5)B^(8)` e. `._(67)Ho^(150)` All of them are `Z` less than 82. b. `._(80)Hg^(195) , (n)/(p) = (115)/(80) = 1.43` `[{:(n//p "is lower than the"),("stability belt for Hg"),(n//p "should be equal"),("to" 153):}]` `n//p` is lower than 1.53, so it will emit `beta^(o+)` to incrase `n//p` ratio. d. `._(5)B^(8)`,(n)/(p) = (3)/(5) = 0.6`, (for stability `n//p` ratio - 1) `n//p` is lower than 1. `(Z "up" 20, (n)/(p) = 1)` It is `beta^(o+)` emitter e. `._(67)Ho^(150), (n)/(p) = (33)/(67) = 1.2` `({:("For stability with" Z gt 60),(n//p "should be" 1.50):})` `n//p` is lower than 1.5. So `beta^(o+)` emitter. Stable nuclides `._(92)Pb^(208)`, `._(50)Sn^(120)` Both have even number `p` and even number of `n`. These are stable nuclides. |
|
| 606. |
The nuclide `X` undergoes `alpha`-decay and another nuclides `Y` undergoes `beta^(ɵ)`-decay, which of the following statement `"is"//"are"` correct?A. The `beta^(ɵ)`-particles emitted by `Y` may have widely different speeds.B. The `alpha`-particles emitted by `X` may have widely different speeds.C. The `alpha`-particles emitted by `X` will have almost same speed.D. The `beta`-particles emitted by `Y` will have the same speed. |
| Answer» Correct Answer - B::C | |
| 607. |
The wave number of the first line in the Balmer series of hydrogen is `15200 cm^(-1)`. What would be the wavenumber of the first line in the Lyman series of the `Be^(3+)` ion?A. `2.4xx10^(5) cm^(-1)`B. `24.3xx10^(5) cm^(-1)`C. `6.08xx10^(5) cm^(-1)`D. `1.313xx10^(6) cm^(-1)` |
|
Answer» Correct Answer - D Given `15200=R(1)^(2)[1/((2)^(2))-1/((3)^(2))] ...(1)` Then `bar(v)=R(4)^(2)[1/((1)^(2))-1/((2)^(2))] ...(2)` from (1) and (2) equation `bar(v)=1.313xx10^(6) cm^(-1)` |
|
| 608. |
Which of the following statements is/are correct?A. the decay constant is independent of external factor like temperature and pressureB. nuclear isomers have same numbers of protons and neutronsC. the decay constant is indepndent of the amount of substance usedD. the value of decay constant generally decreases with rise in temperature |
| Answer» Correct Answer - b,c,d | |
| 609. |
Which of the following is used as neutron absorber in the nuclear reactor?A. WaterB. `D_(2) O`C. Some compounds of uraniumD. Cadmium |
|
Answer» Correct Answer - `(B,D)` |
|
| 610. |
It is observed that only 0.39% of the original radioactive sample remains ndecayed after eighrt hours. Hence:A. the half-life of that substance is 1 hourB. the mean life of the substance is `1/(log""_(e)2)hour`C. decay constant of the substance is `(log""_(e)2)hour""^(-1)`D. if the number of the radioactive nuclei of this substance at a given instant is `10""^(8)` then the number left after 30 min would be `sqrt2xx10""^(9)` |
|
Answer» Correct Answer - a,b,c `lamda=2.303/tlog"N""_(0)/N`,mean life `=1//lamda` |
|
| 611. |
Which of the following is/are correct for nuclear reactor?A. A typical fission is represented by `""_(92)""^(235)U=""_(0)""^(1)nto""_(54)""^(140)Ba+""_(36)""^(93)Kr+` EnergyB. Heavy water (`D""_(2)O`) is used as moderator in preference to orduinary water (`H""_(2)O`) because hydrogen may capture neutrons, while D would not do thatC. Cadmium rods increse the reactor power when they go in and decrease when they go outwardsD. Slower neutrons are more effective in causing fission than faster neutrons in the case of `""_(235)U` |
| Answer» Correct Answer - b,d | |
| 612. |
In the nuclear transmutation `""_(4)""^(9)Be+Xto""_(4)""^(8)Be+Y`, (X,Y) is(are):A. (`gamma`,n)B. (p,D)C. (n,D)D. (`gamma`,p) |
|
Answer» Correct Answer - a,b `(A)""_(4)""^(9)Be+gammato""_(4)""^(8)Be+""_(0)n""^(1) (B), ""_(4)""^(9)Be+""_(1)p""^(1)to""_(4)""^(8)Be+""_(1)""^(2)D` |
|
| 613. |
In the nuclear transmutation `""_(4)""^(9)Be+Xto""_(4)""^(8)Be+Y`, here (X,Y) is/are:A. (`gamma`,n)B. (p,D)C. (n,D)D. (`gamma`,p) |
| Answer» Correct Answer - a,b | |
| 614. |
Assertion: `._(1)H^(1), ._(1)H^(2) and ._(1)H^(3)` are isotopes of hydrogen. Reason : Nuclides of the same element of different mass numbers are called isotopes of that elementA. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanantion of the assertionC. If assertion is true but reason is falseD. If assertion is false but reason is true |
| Answer» Correct Answer - A | |
| 615. |
The most effective projectile in the transmutation of heavy element is `……………….` . |
|
Answer» Correct Answer - Neutron |
|
| 616. |
Arrange the following particles in the increasing order of their penetration capacity as the projectiles for artificial transmutation of elements: Proton `(._(1)H^(1))`, alpha particle `(._(2)He^(4))`, deuteron `(._(1)H^(2))`, neutron `(._(0)n^(1))`A. `._(2)He^(4) lt ._(1)H^(2) lt ._(1)H^(1) lt ._(0)n^(1)`B. `._(1)H^(1) lt ._(1)H^(2) gt ._(0)n^(1) lt ._(2)He^(4)`C. `._(1)H^(1) lt ._(1)H^(2) lt ._(2)He^(1) lt ._(0)n^(1)`D. `._(0)n^(1) lt ._(1)H^(2) lt ._(2)He^(4) lt ._(1)H^(2)` |
|
Answer» Correct Answer - a The penetrating power increases with increases in velocity of particles. Thus, the correct order of penetrating power of given particles is `._(0)n^(1) gt ._(1)H^(1) gt ._(1)H^(2) gt ._(2)He^(4)` |
|
| 617. |
In which radioation mass number and atomic number will not changeA. `alpha`B. `beta`C. `gamma`D. `alpha and 2 beta` |
| Answer» Correct Answer - C | |
| 618. |
`Nd (Z = 60)` is a member of group 3 in periodic table. An isotope of it is `beta-`active. The daughter nuclei will be a member ofA. Group -3B. Group -4C. Group -1D. Group -2 |
|
Answer» Correct Answer - A Element 57 to 71 are placed in III group |
|
| 619. |
All nuclides exhibit radioactivity when the atomic number exceedsA. 80B. 83C. 90D. 92 |
|
Answer» Correct Answer - B Radioactive element has atomic number `gt 83`. |
|
| 620. |
The first artificial disintegration of an atomic nucleus was achieved byA. GeigerB. WilsonC. Madame curieD. Rutherford |
| Answer» Correct Answer - D | |
| 621. |
On bombardin g`._(8)O^(16)` with deutrons, the nuclei of the product formed will beA. `._(9)F^(18)`B. `._(9)F^(17)`C. `._(8)O^(17)`D. `._(7)N^(14)` |
|
Answer» Correct Answer - A `._(8)O^(16) + ._(1)H^(2) rarr ._(9)F^(18)` |
|
| 622. |
Alpha particles are positively charged. |
|
Answer» Correct Answer - T |
|
| 623. |
An isotope `._(Y)A^(X)` undergoes a series of m alpha and n beta disintegrations to form a stable isotope `._(Y -10)B^(X -32)`. The value of m and n are respectivelyA. 6 and 8B. 8 and 10C. 5 and 8D. 8 and 6 |
|
Answer» Correct Answer - D `._(Y)A^(X) rarr ._(Y - 10)B^(X - 32) + m ._(2)He^(4) + n ._(+1)e^(0)` Value of `m = (X -(X) -32)/(4) = 8` Value of `n = Y - Y - 10 - 2 xx 8 = 6` |
|
| 624. |
Which of the following is artificial radioactive series?A. `4n + 1`B. `4n + 2`C. `4n`D. `4n + 3` |
|
Answer» Correct Answer - A Artificial series is `(4n + 1)`. |
|
| 625. |
When a radioactive element emits an alpha particle, the daughter element is placed in the periodic tableA. Two positions to the left of the parent elementB. Two positions to the right of the parent elementC. One position to the right of the parent elementD. In the same position as the parent element |
|
Answer» Correct Answer - A According to group displacement law |
|
| 626. |
Tritium is `beta-` radioactive in nature. |
|
Answer» Correct Answer - T |
|
| 627. |
The decay of a radioactive element follows first order kinetic. Thus,A. Half-life period `= a` constant`//K`,B. The rate of decay is independent of temperatureC. The rate can be altered by changing chemical conditionsD. The element will be completely transformed into new element after expiry of two half-life period |
|
Answer» Correct Answer - A `t_(1//2) = (0.693)/(K)` |
|
| 628. |
If a radioactive isotope with atomic number A and mass number M emits an `alpha-`particle, the atomic number and mass number of that new isotope will becomeA. `A -2, M -4`B. `A -2, M`C. `A, M -2`D. `A -4, M -2` |
|
Answer» Correct Answer - A `._(A)X^(M) overset(-alpha)rarr ._(A-2)Y^(M-4)` |
|
| 629. |
Tritium undergoes radioactive decay givingA. `alpha-`particleB. `beta-`particleC. NeutronsD. None of these |
|
Answer» Correct Answer - B Tritium `(._(1)H^(3) rarr ._(2)He^(3) + ._(-1)e^(0))` is a `beta`-emitter |
|
| 630. |
A radioactive element belongs to the group 14 of the periodic table, it undergoes `beta-`emission, the product obtained belongs to the following group of the periodic tableA. Group 12B. Group 13C. Group 15D. Group 16 |
|
Answer» Correct Answer - C `._(6)^(14)C overset(beta " decay")rarr ._(7)^(14)N + underset(-1)overset(0)beta` |
|
| 631. |
In the nuclear fission `._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(4)` the masses of `._(1)H^(2) and ._(2)he^(4)` are 2.014 mu and 4.003 mu respectively. The energy released/atom of helium formed is ....MeVA. 16.76B. 26.38C. 13.26D. 23.275 |
|
Answer» Correct Answer - D `Delta m = (2 xx 2.014 - 4.003) m u = 0.023"mu"` 1 amu `~~ 936 MeV` `E = 0.025 xx 936` E per nucleon `= (93.6)/(4) = 23.4MeV` |
|
| 632. |
When `n//p` ratio of an isotope is greater than the stable isotope of that element, it emitsA. `beta`-particlesB. `alpha`-particlesC. NeutronD. Positron |
|
Answer» Correct Answer - A Isotope must have same number of proton, therefore `beta`particle will be emitted. |
|
| 633. |
What is the symbol for the nucleus remaining after `._(20)Ca^(42)` undergoes `beta-`emissionA. `._(21)Ca^(42)`B. `._(20)Sc^(42)`C. `._(21)Sc^(42)`D. `._(21)Sc^(41)` |
|
Answer» Correct Answer - C `._(20)Ca^(42) rarr ._(21)Sc^(42) + ._(-1)e^(0)` |
|
| 634. |
Radioactive disintegration differs from a chemical change in beingsA. An exothermic changeB. A spontaneous processC. A nuclear processD. A unimolecular first order reaction |
|
Answer» Correct Answer - C Chemical change is extra nuclear phenomenon |
|
| 635. |
Identify the nuclear reaction that differs from the restA. Positron emissionB. K-captureC. `beta-`decayD. `gamma-`decay |
|
Answer» Correct Answer - D `beta-`emission occurs in those radioactive elements in which n/p ratio is higher than required for stability. In `beta`-emission n/p ratio decrease, where as `alpha` emission , positron emission and K-capture occus in those elements in which n/p ratio is lower than required for stability. Alpha emission is usually observed in nautral radioactive isotope while emission of positron or K-capture is observed in artificial radioactive isotope. There is no change in n/p ratio due of `gamma`-emission |
|
| 636. |
The atomic mass and atomic number of lead are 208 and 82. The atomic mass and atomic number of bismuth are 209 and 83. The `"neutron"//"proton"` ratio in an atomA. Higher of leadB. Higher of bismuthC. SameD. None of these |
|
Answer» Correct Answer - A `(n)/(p) " of " ._(82)Pb^(208) = (126)/(82) = 1.53` `(n)/(p) " of " ._(83)Bi^(209) = (126)/(83) = 1.51` |
|
| 637. |
The radioactive series whose end product is `._(83)^(209)Bi` isA. Thorium seriesB. Fourier seriesC. Actinium seriesD. Neptunium series |
|
Answer» Correct Answer - D `._(83)^(209)Bi` is the end product of Neptunium series i.e., `4n +1` (Artificial) series |
|
| 638. |
The atomic mass and atomic number of lead are 208 and 82. The atomic mass and atomic number of bismuth are 209 and 83. The `"neutron"//"proton"` ratio in an atomA. Is higher in lead than in bismuthB. Is lower in lead than in bismuthC. Is equal in both lead and bismuthD. None of these |
|
Answer» Correct Answer - D `._(82)Pb^(208), p = 82, n = 208 - 82 = 126` `._(83)Bi^(209), p = 83, n = 209 - 83 = 126` `n//p` for `Pb = (126)/(82) = 1.53` `n//p` for `Bi = (126)/(83) = 1.51` `:. (n)/(p)` for `Pb gt (n)/(P)` for `Bi` |
|
| 639. |
Atomic weight of Th is 232 and its atomic number is 90. The number of `alpha`- and `beta`-particles which be lost so that an isotope of lead (atomic weight 208 and atomic number 82) is produced isA. `4 alpha + 6 beta`B. `6 alpha + 4 beta`C. `8 alpha + 2 beta`D. `10 alpha + 2 beta` |
|
Answer» Correct Answer - B Let the reaction emits `x alpha`-particles and `y beta`-particle `._(90)Th^(232) rarr x ._(2)He^(4) + y ._(-1)e^(0) + ._(82)Pb^(208)` Equating atomic mass number on both sides `232 = 4x + y (0) + 208` `:.x = (232 - 208)/(4) = (24)/(4) = 6` `:. alpha`-particles emitted = 6 Equating atomic number on both sides `90 = 82 + 6 xx 2 + y (-1)` `:. y = 94 - 90 = 4` `:. beta`-particle emitted = 4 |
|
| 640. |
If an isotope of hydrogen has two neutrons in its atom, its atomic number and mass number will respectively beA. 2and 1B. 1 and 1C. 3 and 1D. 1 and 3 |
|
Answer» Correct Answer - D The atomic number of isotope of hydrogen is alasy opne so it have only one protonn in its nuclesus and given that it have two neutrons.It menas that its mass number is 3. |
|
| 641. |
Atomic weight of the isotope of hydrogen which contains 2 neutrons in the nucleus would beA. 2B. 3C. 1D. 4 |
|
Answer» Correct Answer - B `._(1)T^(3)` - Tritium has 2 neutron and 1 proton |
|
| 642. |
The sum of the number of neutrons and proton in the radio isotope of hydrogen isA. 6B. 5C. 4D. 3 |
|
Answer» Correct Answer - D Tritium `(._(1)H^(3))` consist of 1 proton and 2 neutrons. |
|
| 643. |
An electron beam can undergo defraction by crystals .Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to `1.54 Å`A. `54.3 volt`B. `63.3 volt`C. `66.2 volt`D. None of these |
|
Answer» Correct Answer - 2 For an electron `1/2 m u^(2)=eV` " " and `lambda=h/(m u)` Thus, `1/2 m=(h^(2))/(m^(2)lambda^(2))=eV` or `V=1/2h^(2)/(mlambda^(2)e)=(1xx(6.62xx10^(-34))^(2))/(2xx9.108xx10^(-31)(1.54xx10^(-10))^(2)xx1.602xx10^(-19))=63.3 volt.` |
|
| 644. |
Assertion : The activiation energies for fusion reactions are very low. Reaosn : They require very low temperature to overcome electrostatic repulsion between the nuclei.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanantion of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
|
Answer» Correct Answer - D The activation energies for fusion reactions are very high. They require very high temperature `(gt 10^(6))` to over come electrostatic repulsion between the nuclie |
|
| 645. |
Assertion : A beam of electrons deflect more than a beam of `alpha`-particles in an electric field Reaosn : Electrons possess negative charge while `alpha` particles possess positive chargeA. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanantion of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
|
Answer» Correct Answer - D The activity of 1g of pure `U - 235` and that in `U_(3)O_(8)` is same. Activity does not depend upon the state of combination |
|
| 646. |
Which of the folowing is/are correct when a nuclide of mass number (A) and atomic number (Z) undergoes radioactive process?A. Both A and Z decrease, the process is called `alpha`-decayB. A remains unchanged and Z decreases by 1. The process is called `beta""^(oplus)` or positron decay or `K""^(-)` electron capture.C. Both A and Z increases, the process is called nuclear isomerismD. Both A and Z increases, the process is called nuclear isomerism |
| Answer» Correct Answer - a,b,c | |
| 647. |
Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are `tau = 1// lambda` and `t_(1//2) = 0.693//lambda`. Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after `n` half lives of radioactive elements can be calculated using the relation: `N = N_(0) ((1)/(2))^(n)` Select the correct relation.A. `t_(1//2) = (0.693)/(lambda)`B. `tau = (1)/(lambda)`C. `tau = 1.44 xx t_(1//2)`D. `tau = (t_(1//2))/(0.693)` |
|
Answer» Correct Answer - C `tau = 1.44 xx t_(1//2)` |
|
| 648. |
Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are `tau = 1// lambda` and `t_(1//2) = 0.693//lambda`. Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after `n` half lives of radioactive elements can be calculated using the relation: `N = N_(0) ((1)/(2))^(n)` Which `"is"//"are"` true about the decay cosntant?A. Unit of `lambda` is `"time"^(-1)`B. `lambda` is independent of temperatureC. `lambda` depends on the initial amount of element taken.D. `lambda` depends on the nature of radioactive element. |
| Answer» Correct Answer - A | |
| 649. |
`beta`-decay from a radioactive nuclide leads toA. `alpha`-decayB. `beta`-decayC. Positron decayD. K-electron decay |
| Answer» Correct Answer - a,b | |
| 650. |
Radioactive decay is aA. Second order reactionB. First order reactionC. Zero order reactionD. Third order reaction |
|
Answer» Correct Answer - B Radioactive decays is a first order reaction |
|