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.
| 651. |
If 2.0g of a radioactive isotope has a half-life of 20 hr, the half-life of 0.5 g of the same substance isA. 20 hrB. 80 hrC. 5 hrD. 10 hr |
|
Answer» Correct Answer - A Half-life period is a characteristic of radioactive isotope which is independent of initial concentration. |
|
| 652. |
Which of the following particles is emitted in the nuclear reaction: `._(13)Al^(27) + ._(2)He^(4) rarr ._(14)P^(30 +) ….?`A. `._(0)n^(1)`B. `._(-1)e^(0)`C. `._(1)H^(1)`D. `._(1)H^(2)` |
|
Answer» Correct Answer - C `._(13)A^(27) + ._(2)He^(4) rarr ._(14)P^(30) + ._(Z)X^(A)` Equating mass number on both sides `27 + 4 = 30 + A` `:. A = 31 - 30 = 1` Equating atomic number on both sides `13 + 2 = 14 + Z` `:. Z = 1` `:.` Particles is `._(1)H^(1)`. |
|
| 653. |
A radioactive element with half-life 6.5 hr has `48 xx 10^(19)` atoms. Number of atoms left after 26 hrA. `24 xx 10^(19)`B. `12 xx 10^(19)`C. `3 xx 10^(19)`D. `6 xx 10^(19)` |
|
Answer» Correct Answer - C `N_(t) = N_(0) ((1)/(2))^(n) = 48xx10^(19) ((1)/(2))^(26//65)` `= 48 xx 10^(19) ((1)/(2))^(4) = 3xx 10^(19)` |
|
| 654. |
Atoms of the same element possessing identical mass but differing in half`-` life period are called `………………………….` . |
|
Answer» Correct Answer - Nuclear isomers |
|
| 655. |
Nuclear fusion takes place at very high temperature and hence it is called `……………………..` . |
|
Answer» Correct Answer - Thermonuclear |
|
| 656. |
`O_(2)` undergoes photochemical dissociation into one normal oxygen atom and one excited oxygen atom. Excited oxygen atom is `1.967 eV` more energetic than normal . The dissociation of `O_(2)` into two normal atoms of oxygen required `498 kJ mol^(-1)`, what is the maximum wavelength effective for photochemical dissociation of `O_(2)`?A. `1.01 nm`B. `1.64 nm`C. `1.74 nm`D. `2.74 nm` |
|
Answer» Correct Answer - C `O_(2) rarr O +O^(**)`, Dissociation energy `=468 kJ mol^(-1)=(498xx10^(3))/(6.02xx10^(23)) J mol^(-1)=8.27xx10^(-19) J molecules^(-1)` Extra energy of the excited atom `=1.967 eV=1.967xx1.6xx10^(-19) J=3.15xx10^(-19) J atom^(-1)` `lambda=(hc)/(E)=(6.626xx10^(-24)xx3xx10^(8))/(11.42xx10^(-19))=174xx10^(-9) m` Thus, `lambda=174 nm` |
|
| 657. |
A first order nuclear reaction is half completed in 45 minutes. How long does it need 99.9% of the reaction to be completedA. 5 hoursB. 7.5 hoursC. 10 hoursD. 20 hours |
|
Answer» Correct Answer - B `k = (0.693)/(0.75 hr) = (2.303)/(t) "log" (a)/(a - 0.999a)` `= (2.303)/(t) log 10^(3) = 7.5 hr` |
|
| 658. |
`._(92)U^(238)` emits `8 alpha-` particles and `6 beta-` particles. The `n//p` ratio in the product nucleus isA. `(62)/(41)`B. `(60)/(41)`C. `(61)/(42)`D. `(62)/(42)` |
|
Answer» Correct Answer - A `8 alpha-` particle emission decreases the atomic mass number by `8xx4=32`. Thereofre ,a atomic mass number of product nucleus is `238-32=206`. `6 beta-` particles and `8 alpha-` particles emission decreases the atomic number of by `8xx2-6xx1=10`. Therefore, atomic number of product nucleus `=92-10=82`. `:. ` number of neutrons in nucleus `=206-82=124`. `:. (n)/(0)=(cancel(124)^(2))/(cancel(82)^(2))=(62)/(41)` |
|
| 659. |
Consider an `alpha-` particle just in contact with a `._(92)U^(238)` nucleus. Calculate the coulombic repulsion energy (i.e., the height of the coulombic barrier between `U^(238)` and alpha particle) assuming that the distance between them is equal to the sum of their radiiA. `23.851 xx 10^(4) eV`B. `26.147738 xx 10^(4) eV`C. `25.3522 xx 10^(4) eV`D. `20.2254 xx 10^(4) eV` |
|
Answer» Correct Answer - B `r_("nucleus") = 1.3 xx 10^(-13) xx (A)^(1//3)`, where A is mass number `r_(U^(238)) = 1.3 xx 10^(-13) xx (238)^(1//8) = 8.06 xx 10^(-13) cm` `r_(He^(4)) = 1.3 xx 10^(-13) xx (4)^(1//3) = 2.06 xx 10^(-13) cm` `:.` Total distance in between uranium and `alpha` nuclei `= 8.06 xx 10^(-13) + 2.06 xx 10^(-13) = 10.12 xx 10^(-13) cm` Now repulsion energy `= (Q_(1)Q_(2))/(r) = (92 xx 4.8 xx 10^(-10) xx 2 xx 4.8 xx 10^(-10))/(10.12 xx 10^(-13)) erg` `= 418.9 xx 10^(-7) erg = 418.9 xx 10^(-7) xx 6.242 xx 10^(11) eV` `= 26.147738 xx 10^(4) eV` |
|
| 660. |
Assertion `(A):` `._(92)U^(238) (IIIB)overset(-alpha)rarrAoverset(-alpha)rarrB overset(-beta)rarrC` Reason `(R):` Element `B` will be of `II A ` group.A. If both `(A)` and `(R)` are correct , and `(R)` is the correct explanation of `(A)`B. If both `(A)` and `(R)` are correct, but (R) is not the correct explanation of `(A)`C. If `(A)` is correct, but `(R)` is incorrect.D. |
| Answer» Correct Answer - B | |
| 661. |
Assertion (A) : `""_(92)U""^(238)(IIIB)overset(-alpha)toAoverset(-alpha)toBoverset(-beta)toC` Reason (R ) : Element B will be II A groupA. If both (A) and (R ) are correct, and (R ) is the correct explaination of (A)B. If both (A) and (R ) are correct, but (R ) is not the correct explaination of (A)C. If (A) is correct,but (R ) is incorrectD. If both (A) and (R ) are incorrect. |
| Answer» Correct Answer - b | |
| 662. |
Calculate the number of neutrons in the remaining atoms after the emission of an alpha particle from `._(92)U^(238)` atom. |
|
Answer» Correct Answer - 144 Parent element = `._(92)U^(238)` On account of emission of an alpha particle, the atomic mass is decreased by 4 units and atomic number by 2 units. So, atomic mass of daughter element = 234 Atomic number of daughter element = 90 Number of neutrons = Atomic mass - Atomic number `= 234 - 90 = 144` |
|
| 663. |
Consider the following statements: (a) Electron density in the `XY` plane in `3d_(x^(2)-y^(2))` orbital is zero (b) Electron density in the `XY` plane in `3d_(z^(2))` orbital is zero. (c ) `2s` orbital has one nodel surface (d) for `2p_(z)` orbital, `XY` is the nodal plane. Which of these are incorrect statements :A. a & cB. b & cC. Only bD. a, b |
|
Answer» Correct Answer - D (a) Electron density in the `XY` plane in `3d_(x^(2)-y^(2))` orbital is not zero (b) Electron density in the `XY` plane in `3d_(z^(2))` orbital is not zero. (c) `2s` orbital has one nodal surface (d) For `2p_(z)` orbital, `XY` is the nodal plane. |
|
| 664. |
Balance the following nuclear reactions: a. `._(3)Li^(7) + ._(0)n^(1) rarr 2 ._(2)He^(4) + ?` b. `._(42)Mo^(94) + ._(1)H^(2) rarr ._(0)n^(1) + ?` |
|
Answer» On `LHS` of the equation Sum of atomic number `= 3 + 0 = 3` Sum of mass numbers `= 7 + 1 = 8` On `RHS` of the equation, Let the missing particles be `._(z)X^(a)` Sum of atomic number `= Z + 4` Sum of mass numbers `= a + 8` Sum of atomic number of `LHS` = Sum of atomic numbers of `RHS` `3 = Z + 4` or `Z = -1` Sum of numbers of `LHS` = Sum of mass numbers of `RHS` `8 = a + 8` So `a = 0` Thus, the missing particle is `._(1)X^(0)`, i.e., `._(1)e^(0)` The balanced equation is `._(3)Li^(7) + ._(0)n^(1) rarr ._(2)He^(4) + ._(-1)e^(0)` b. Lets the missing particle be `._(z)X^(a)`. On `LHS` of the equation, Sum of the atomic numbers `= (42 + 1) = 43` Sum of mass numbers `= (94 + 2) = 96` On `RHS` of the equation, Sum of atomic numbers `= Z + 0` Sum of mass numbers `= a + 1` |
|
| 665. |
Write the complete nuclear reactions `:` ltbr. `a.` `._(4)Be^(9)+ ._(9)He^(4) rarr ._(6)C^(12)+...................` b. ` ._(1)H^(3)rarr ._(2)He^(3)+..............` `c.` `._(7)N^(14)+._(2)He^(4)rarr ._(8)O^(17)+.........` `d.` `._(92)U^(235)+._(0)n^(1) rarr ._(38)Sr^(92)+....Xe3._(0)n^(1)` `e.` `._(3)Li^(7)+ ._(0)n^(1)rarr2 ._(2)He^(4+)..........` `f.` `._(92)U^(238)+........rarr ._(92)U^(239)rarr ._(93)Np^(239)+....` `g.` `._(7)N^(14)+._(0)n^(1)rarr ._(1)H^(3)+..........` `h.` `._(3)Li^(7)+............. rarr._(4)Be^(8)+gamma-` radiations `i.` ` ._(1)H^(2)+............ rarr ._(2)He ^(4)+._(0)n^(1)` |
|
Answer» Correct Answer - `{:(a. ._(0)n^(1),,,,b. ._(-1)e^(0)),(c. ._(1)H^(1),,,,d. ._(54)Xe^(141)),(e. ._(-1)e^(0),,,,f. ._(0)n^(1) ._(-1)e^(0)),(g. ._(1)H^(3),,,,h. ._(1)H^(1)),(i. ._(0)n^(1),,,,):} ` |
|
| 666. |
S : `""^(14)C` is `beta`-emiter. E : Its electrons are loosely bonded in comparison to `""^(12)C` and `""^(13)C`A. 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 - c | |
| 667. |
A small amount of solution containing a radioactive nucleide `A""^(x)` was administrated into the blood of a patient. The activity of the nucliede id `2xx10""^(3)`dps. Its half life is 15 hours. After 5 hours a sample of the blood drawn out from the patient. It,s activity was 16 dpm per mL. What is the activity of the sample after another 5 hours time?(L og1.59=0.2)A. 11.18 dpm per mLB. 1.118 dpm per mLC. 12.71 dpm per mLD. 1.271 dpm per mL |
|
Answer» Correct Answer - c Activity of sample after another 5 hours I y `0.693/15=2.303/5log.(16)/3,thereforey=12.71` dpm mL |
|
| 668. |
Which of the following has magic number of protons and neutrons? a. `._(82)Pb^(208)` b. `._(2)He^(3)` c. `._(50)Sn^(120)` d. `._(82)Pb^(206)`A. a. `._(82)Pb^(208)`B. b. `._(2)He^(3)`C. c. `._(50)Sn^(120)`D. d. `._(82)Pb^(206)` |
|
Answer» Correct Answer - a. `._(82)Pb^(208)` 208 and 82 both are magic numbers |
|
| 669. |
A small amount of solution containing a radioactive nucleide `A""^(x)` was administrated into the blood of a patient. The activity of the nucliede id `2xx10""^(3)`dps. Its half life is 15 hours. After 5 hours a sample of the blood drawn out from the patient. It,s activity was 16 dpm per mL. In the following graph, binding energy per nucleon is plotted against mass number (a). three elements `A_(1),A_(2)and A_(3)` are located in the graph. Select the false statement about the graph. A. elment `A""_(2)` is more stableB. element `A""_(3)` is less stable than `A""_(2)`C. element `A""_(1)` is more stable than both `A""_(2)` and `A""_(3)`D. `A""_(2)` is metallic element |
| Answer» Correct Answer - c | |
| 670. |
Monazite sample contains 9% `ThO""_(2)` and 0.35% `U""_(3)O""_(8).Pb""^(208)` and `Pb""^(206)` are the stable and products in the radioactive decay series of `Th""^(232)` and `U""^(238)` respectively. All the lead in mozaite is of radiogenic origin. The isotopic ratio of `Pb""^(208)/Th""^(232)` was found to be 0.104. The half lives of Th and U are `1.41xx10""^(16)` years and `4.47xx10""^(9)` years respectively. Select the information incorrect about `Th""^(232)`A. It belonges to third group of actinide deriesB. `Th""^(232)` is a fissile materialC. It is a fissile matrialD. It belonges to 4n series |
| Answer» Correct Answer - b | |
| 671. |
The unit for radioactive constant isA. timeB. time `mol^(-1)`C. `time^(-1)`D. `mol time^(-1)` |
| Answer» Correct Answer - C | |
| 672. |
One curie of activity is equivalent toA. `3.7 xx 10^(17)` disintegrations per secondB. `3.7xx10^(10)` disintegrations per secondC. `3.7 xx 10^(14)` disintegration per secondD. `3.7xx10^(3)` disintegration per second |
| Answer» Correct Answer - B | |
| 673. |
The spectrum of `He^(+)` is expected to be similar to that ofA. `Li^(2+)`B. `He`C. `H`D. `Na` |
| Answer» Correct Answer - A, C | |
| 674. |
When alpha particle are sent through a thin metal foil ,most of them go straight through the foil becauseA. alpha particles are much heavier than electronsB. alpha particles are positively chargedC. most part of the atom is empty spaceD. alpha particles move with high speed |
| Answer» Correct Answer - A, C | |
| 675. |
Which is the correct statementA. Isotopes are always radioactiveB. `beta-`rays are always negatively charged particlesC. `alpha-`rays are always negatively charged particlesD. `gamma`-rays can be deflected in magnetic field |
|
Answer» Correct Answer - B `alpha`-rays positively charged, `beta-`rays are negatively charged, `gamma`-rays carry no charge and thus are not deflected in field |
|
| 676. |
Which of the following does not contain material particlesA. Alpha raysB. Beta raysC. Gamma raysD. Canal rays |
|
Answer» Correct Answer - C `gamma-`rays does not contain material particles |
|
| 677. |
Which of following contains (s) material particles?A. `alpha`-raysB. `beta`-raysC. `gamma`-raysD. Anode rays |
| Answer» Correct Answer - A::B::D | |
| 678. |
Which one of the following statements `"is"//"are"` correct?A. Neutron was discovered by Chadwick.B. Nuclear fission was discovered by Hahn and Strassmann.C. Polonium was discovered by Madam Curie.D. Nuclear was discovered by Fermi. |
| Answer» Correct Answer - A::B::C | |
| 679. |
In a certain electronic transition in the hydrogen atoms from an initial state `(1)` to a final state `(2)`, the difference in the orbit radius `((r_(1)-r_(2))` is 24 times the first Bohr radius. Identify the transition-A. `5 rarr 1`B. `25 rarr 1`C. `8 rarr 3`D. `6 rarr 5` |
| Answer» Correct Answer - A | |
| 680. |
A phot of energy `h upsilon` is absorbed by a free electron of a metal having work function `phi lt h upsilon`A. The electron is sure to come outB. The electron is sure to come out with a kinetic energy `(hv-w)`C. Either the electron does not come out or it comes with a kinetic energy `(hv-w)`D. It may come out with a kinetic energy less than `(hv-w)` |
|
Answer» Correct Answer - D Photoelectric effect is a eandom phenomena. So, electron It may come out with a kinetic energy less than `(hv-w)` as some energy is lost while escaping out. |
|
| 681. |
Energy of an electron is givem by `E = - 2.178 xx 10^-18 J ((Z^2)/(n^2))`. Wavelength of light required to excited an electron in an hydrogen atom from level `n = 1` to `n = 2` will be `(h = 6.62 xx 10^-34 Js` and `c = 3.0 xx 10^8 ms^-1`).A. `1.214xx10^(-7) m`B. `2.816xx10^(-7) m`C. `6.500xx10^(-7) m`D. `8.500xx10^(-7) m` |
|
Answer» Correct Answer - 1 `DeltaE=2.178xx10^(-18)(1/1^(2)-1/2^(2))=(hC)/lambda` `2.178xx10^(-18)(1/1^(2)-1/2^(2))=(6.62xx10^(-34)xx3.0xx10^(8))/(lambda)" " :. lambda~~1.214xx10^(-7)m` |
|
| 682. |
In the radioactive decay `._(92)X^(232) rarr ._(89)Y^(220)`, how many `alpha and beta-`particles are ejected from X to form YA. `3 alpha and 3 beta`B. `5 alpha and 3 beta`C. `3 alpha and 5 beta`D. `5alpha and 5 beta` |
|
Answer» Correct Answer - A `._(92)X^(232) rarr ._(89)Y^(220) + x_(2) he^(4) + y ._(-1)e^(0)` No. of `alpha -`particles `= (232 - 220)/(4) = 3` No. of `beta-`particles `= 2 xx - 92 + 89 = 3` |
|
| 683. |
What is enriched uranium? a. `U-238` b. `U-235` c. `U-235 +` Radium d. `U-235 + U-238`A. a. `U-238`B. b. `U-235`C. c. `U-235 +` RadiumD. d. `U-235 + U-238` |
| Answer» Correct Answer - d. `U-235 + U-238` | |
| 684. |
Radioactive isotopes that have an excessive neutron/proton ratio generally exhibitA. `e^(-)` emissionB. `._(2)He^(4)` emissionC. `e^(+)` emissionD. K-electron capture |
|
Answer» Correct Answer - A The isotopes having an excessive n/p ratio exhibit `e^(-)` emission |
|
| 685. |
Which of the following is iso-electronic with neon-A. `O^(2-)`B. `F^(-)`C. `Mg`D. `Na` |
|
Answer» Correct Answer - A, B, C, D Ne contains `10` electrons `O^(2-)` and `F^(-)` contain `10` electrons |
|
| 686. |
Which is true about an electron-(a). rest mass of electron is `9.1xx10^(-28)g`(b). mass of electron increases with the increase in velocity(c). molar mass of electron is `5.48xx10^(-4)g//"mole"`(d). `e//m` of electron is `1.7xx10^(8) "coulomb"//g`A. Rest mass of electron is `9.1xx10^(-28) g`B. Mass of electron increases with the increase in velocityC. Molar mass of electron is `5.48xx10^(-4) g//mole`D. `e//m` of electron is `1.7xx10^(8) coulomb//g` |
|
Answer» Correct Answer - A, B, C, D `m_(e)=9.1xx10^(-31)kg=9.1xx10^(-28) g` `m=m_(0)/sqrt(1-v^(2)/c^(2)) (m_(0): "rest mass , m : dynamic mass")` `As vuarr, (1-v^(2)/c^(2))darr :. m uarr` Molar mass of `e=9.1xx10^(-28)xx6.023xx10^(23)=5.48xx10^(-4) g//`mole |
|
| 687. |
In Bohr series of lines of hydrogen spectrum, third line from the red end corresponds to which one of the following inner orbit jumps of electron for Bohr orbit in atom in hydrogen :A. `3 rarr 2`B. `5 rarr 2`C. `4 rarr 1`D. `2 rarr 5` |
|
Answer» Correct Answer - 2 The electron has minimum energy in the orbit and its energy increases as `n` increases. Here `n` represents number of orbit, i.e., `1^(st), 2^(nd), 3^(rd)`….. The thired line from the red end corresponds. To yellow region i.e., `5`. In order to obtain less energy electron tends to come `1^(st)` or `2^(nd)` orbit. So jump may be involved either `5 rarr 1` or `5 rarr 2`. Thus option (2) is correct here. |
|
| 688. |
In two individual hydrogen atoms electrons move around the nucleus in circular orbits of radii R and 4R. The ratio of the time taken by them to complete one revolution is:A. `1:4`B. `4:1`C. `1:8`D. `8:1` |
|
Answer» Correct Answer - C Angular momentum `J=mvr` `J^(2)=m^(2)v^(2)r^(2)` or `J^(2)/2=(1/2 mv^(2))mr^(2)` or `K.E.=J^(2)/(2mr^(2))` |
|
| 689. |
A single electron system has ionization energy ` 1. 118 xx 10^7 J mol^(-1)`. Calculate the number of protons in the nucleus of the system . |
|
Answer» `I.E.=Z^(2)/n^(2)xx21.69xx10^(-19) J` `(11180xx10^(3))/(6.023xx10^(23))=Z^(2)/1^(2)xx21.69xx10^(-19) " " Ans. Z=3` |
|
| 690. |
The ratio of the de Broglie wavelength of a proton and alpha particles will be `1:2` if theirA. velocity are in the ratio `1:8`B. velocity are in the ratio `8:1`C. kinetic energy are in the ratio `1:64`D. kinetic energy are in the ratio `1:256` |
|
Answer» Correct Answer - B `lambda=v` then `lambda=h/(mV)` or `lambda^(2)=h/m` So, `lambda=sqrt(h/m)` |
|
| 691. |
The de Broglie wavelength associated with particle isA. inversely proportional to its momentumB. inversely proportional to its energyC. directly proportional to its velocityD. directly proportional to its momentum |
| Answer» Correct Answer - A | |
| 692. |
The equation `E=hv` indicates thatA. photons have both particle and wave natureB. photons are wavesC. photons are stream of particlesD. no such inference can be drawn from the given equation |
| Answer» Correct Answer - A | |
| 693. |
The species which has its fifth ionization potential equal to 340 V is :A. `B^(+)`B. `C^(+)`C. `B`D. `C` |
| Answer» Correct Answer - C | |