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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.
| 401. |
For a reaction, the rate constant is `2.34 s^(-1)`. The half-life period for the reaction isA. 0.30 sB. 0.60 sC. 3.3 sD. Date is insufficient |
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Answer» Correct Answer - A `t_(1//2) = (0.693)/(K) = (0.693)/(2.34) = 0.296s` |
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| 402. |
The half-life of a radionuclide is 69.3 minutes. What is its average life (in minutes)A. 100B. `10^(-2)`C. `(69.3)^(-1)`D. `0.693 xx 69.3` |
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Answer» Correct Answer - A Average life `(tau) 1.44 t_(1//2) = 1.44 xx 69.3 = 99.7 ~~ 100` minutes |
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| 403. |
1.0 g radioactive sodium on decay becomes 0.25 g in 16 hours. How much tie 48 g of same radioactive sodium will need to become 3.0 gA. 48hoursB. 32 hoursC. 20 hoursD. 16 hours |
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Answer» Correct Answer - B 48g of radioactive sodium will need 32 hours to become 3.0 g |
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| 404. |
The half-life of 1 g of radioactive sample is 9 hour. The radioactive decay obeys first order kinetics. The time required for the original sample to reduce to 0.2 g isA. 15.6 hoursB. 156 hoursC. 20.9 hoursD. 2.09 hours |
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Answer» Correct Answer - C `(0.693)/(9) = (2.303)/(t) "log" (1)/(1 - 0.2)` |
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| 405. |
`._(11)Na^(24)` half-life is 15 hours. On heating it willA. ReduceB. Remain unchangedC. Depend on temperatureD. Become double |
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Answer» Correct Answer - B `t_(1//2)` is independent of all external factors and is constant for a given species |
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| 406. |
A radioactive isotope has a half-life of 20 days. If 100 g of the substance is taken, the weight of the isotope remaining after 40 days isA. 25 gB. 2.5 gC. 60 gD. 40 g |
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Answer» Correct Answer - A `n = (40)/(20) = 2` `:.` Amount left `(N_(0))/(2^(n)) = (100)/(2^(2)) = 25g` |
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| 407. |
The decay of a radioactive element follows first order kinetics, as a resultA. Half-life period = constant/k, where k is the decay constantB. Rate of decay is independent of temperatureC. Rate can be changed by changing chemical conditionsD. The element will be completely transformed into a new element after expiry of two half-life period |
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Answer» Correct Answer - A::B For `1^(st)` order `t_(1//2) = 0.693 K^(-1)`. Rate of decay is independent of temperature. |
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| 408. |
Whe a radioactive substance is kept in vacuum, the rate of its disintegration per secondA. Increases considerablyB. Is not affectedC. Suffers a slight decreaseD. Increases only if the products are gaseous |
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Answer» Correct Answer - B As there is no physical change involved in case of radioactive disintegration, so it will not be affected by external changes |
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| 409. |
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)` Amount of radioactive elements (activity) decreases with passage of time asA. LinearlyB. ExponentiallyC. ParabolicallyD. All of these |
| Answer» Correct Answer - A | |
| 410. |
Which is more unstable of each of the following pairs, and in each case what type of process could the unstable nucleus undergo? a. `._(6)C^(16)` b. `._(9)F^(18)`, `._(10)Ne^(18)`A. a. `._(6)C^(16)` ,`._(7)N^(16)`B. b. `._(9)F^(18)` `._(10)Ne^(18)`C.D. |
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Answer» In this case although `._(6)C^(16)` has even number of `p` and `n`, still its `n//p` ratio is higher than `n//p` ratio of `._(7)N^(16)`, `(n)/(p)` of `._(6)C^(16) = (10)/(6) = 1.7` `({:("For" Z = 20 n//p),("should be" = 1),("Forstability"):})` `n//p` ratio is higher than 1. IT would emit `beta^(ɵ)` to get `n//p` ratio back to stable range. `._(6)C^(16) rarr ._(7)N^(14) + ._(-1)e^(0)` `(n)/(p) = (10)/(6) = 1.7, (n)/(p) = (7)/(7) = 1` In case of `._(7)N^(16), (n)/(p) = (9)/(7) = 1.2` `n//p` of `._(6)C^(16) gt n//p` of `._(7)N^(16)`. So `_(6)C^(16)` more unstalbe than `._(7)N^(16)`. b. In this case although `._(9)F^(18)` has odd number `p` and `n`, still its `n//p` ratio is equal 1, has `n//p` of `._(10)Ne^(18)` is less than 1. `(n)/(p)` of `._(9)F^(18) = (9)/(9) = 1` (For `Z = 20`, stability `n//p = 1)` `(n)/(p)` of `._(10)Ne^(18) = (8)/(10) = 0.8` (Its value is less than 1) So `._(10)Ne^(18)` would be unstable than `._(9)F^(18)`. So, it will emitt `beta^(o+)` or K-capture to get the ratio into the range of stability. `._(10)Ne^(18) rarr ._(9)F^(18) + ._(+1)e^(0)` `(n)/(p) = (8)/(10) = 0.8` or `(n)/(p) = (9)/(9) = 1 (0.8 rarr 1)` `._(10)Ne^(18) + ._(+1)F^(18)` |
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| 411. |
Which particle can be used to change `._(13)Al^(27) " into " ._(16)P^(30)`A. NeutronB. `alpha`-particleC. ProtonD. Deuteron |
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Answer» Correct Answer - B `._(13)Al^(27) + ._(2)He^(4) rarr ._(15)P^(30) + ._(0)n^(1)` |
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| 412. |
n atomic reacotrs, graphite is used as aA. LubricantB. Moderator to show down neutronsC. FuelD. Liner of the reactor |
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Answer» Correct Answer - B Graphite is used as moderator to slow down the speed of neutrons in atomic reactors |
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| 413. |
Unit for radioactive constant isA. `"Time"^(-1)`B. TimeC. Mole - `"time"^(-1)`D. Time - `"mole"^(-1)` |
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Answer» Correct Answer - A The rate of disintegrations is expressed in terms of the number of disintegrations per seconds |
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| 414. |
Radioactive idoine is being used to diagnose the disease ofA. BonesB. KidneysC. Blood cancerD. Thyroid |
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Answer» Correct Answer - D `I^(131)` is used for goitre therapy, i.e., idoine deficiency |
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| 415. |
Which of the following is electronic configuration of `Cu^(2+) (Z = 29)` ?A. `[Ar]4s^(1) 3d^(8)`B. `[Ar]4s^(2) 3d^(10) 4p^(1)`C. `[Ar]4s^(1) 3d^(10)`D. `[Ar] 3d^(9)` |
| Answer» Correct Answer - D | |
| 416. |
Explain with reason the nature of emitted particle by:A. `._(7)^(14)N (n, p)`B. `._(18)^(35)Ar`C. `._(32)^(80)Ge`D. `._(20)^(40)Ca` |
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Answer» Correct Answer - (a) `._(20)^(38)Ca` : It has n/p=18/20=0.9, Which lies below the belt of stability and thus positron emitter `._(20)^(38)Ca rarr ._(19)^(38)K+_(+1)^(0)e` (b) `._(18)^(35)Ar` : It has `n/p=17/18=0.994`, which lies below the belt of stability and thus, positron emitter `._(18)^(35)Ar rarr ._(17)^(35)Cl+._(+1)^(0)e` If `n//p lt 1` and nuclear charge is high the nuclide may show K-electron capture. (c) `._(32)^(80)Ge` : It has `n/p=48/32=1.5`, which lies above the belt of stability and thus `beta`-emitter `._(32)^(80)Ge rarr._(33)^(80)As +_(-1)^(0)e` (d) `._(20)^(40)Ca` : It has both magic number `p=20, m=20` and thus, stable. |
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| 417. |
The mass of cathode ray particle is:A. Same for different gasesB. Different for different gasesC. Minimum for `H_(2)` gasD. Different for same gases |
| Answer» Correct Answer - A | |
| 418. |
The element having no neutron in the nucleus of its atom is-(a). hydrogen(b). nitrogen(c). helium(d). boronA. HydrogenB. NitrogenC. HeliumD. Boron |
| Answer» Correct Answer - A | |
| 419. |
Which of the following does not occur?A. `_(20)^(40)Ca+_(0)^(1)n rarr _(19)^(40)K+_(1)^(1)H`B. `_(12)^(24)Mg+_(2)^(4)He rarr _(14)^(27)Si +_(0)^(1)n`C. `_(48)^(113)Cd+_(0)^(1)n rarr_(48)^(112)Cd+_(-1)^(0)e`D. `_(20)^(43)Ca+_(2)^(4)He rarr_(21)^(46)Sc+_(1)^(1)H` |
| Answer» Correct Answer - C | |
| 420. |
if element `,_(25)X+y` has spain magnetric moment 1.732 B.M then:A. number of unpaired electron `=1`B. number of unparired electron `=2`C. `Y=4`D. `Y=6` |
| Answer» Correct Answer - A, D | |
| 421. |
Which consists of charged particles of matter?A. `alpha`-particleB. `beta`-particleC. `gamma`-taysD. Anode rays |
| Answer» Correct Answer - A, B, D | |
| 422. |
the configuration `[Ar]3d^(10)4s^(2)4p^(4)` is similar to that of :A. boronB. oxygenC. sulphurD. aluminium |
| Answer» Correct Answer - B, C | |
| 423. |
The isotope of carbon used in radiocarbon dating isA. `._(6)C^(12)`B. `._(6)C^(13)`C. `._(6)C^(14)`D. `._(6)C^(15)` |
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Answer» Correct Answer - c |
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| 424. |
`._(13)Al^(28)` when radiated by suitable projectile gives `._(15)P^(31)` and neutron. The projectile used isA. ProtonB. NeutronC. Alpha particleD. Deuteron |
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Answer» Correct Answer - C `._(13)Al^(28) + ._(2)He^(4) rarr ._(15)P^(31) + ._(0)n^(1)` |
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| 425. |
In the nuclear reactions the speed of the neutrons is slowed down byA. Heavy waterB. Ordinary waterC. Zinc rodsD. Molten caustic soda |
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Answer» Correct Answer - A Heavy water is `D_(2)O` |
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| 426. |
A sample of hydrogen (in the form of atoms), is made to absorb white light. `52%` of the hydrogen atoms got ionised. In order to calculate the ionisation energy of hydrogen from its absorption spectrum `("assuming the electrons that got ejected have" KE=0)`, it is possible by measuring the frequency of the:A. line of shortest wavelengthB. line of longest wavelengthC. line of greatest intensityD. line of smallest intensity |
| Answer» Correct Answer - 1 | |
| 427. |
Write the four quantum numbers for V and VI electrons of carbon atom. |
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Answer» `_(6)C: 1s^(2), 2s^(2) 2p^(2)` fifth electron : `n=2 l=1 m=-1 or +1 s=+1/2 or -1/2` sixth electron : `n=2 l=1 m=0 s=+1/2 or -1/2` |
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| 428. |
Consider Xenon `(Z=54)`. The maximum number of electrons in this atom that have the values for their quantum numbers as `n=4, l=3` and `s=1/2` in its ground state is:A. ZeroB. `7`C. `9`D. `14` |
| Answer» Correct Answer - 1 | |
| 429. |
Loss of a beta particles is equivalent toA. Increase of one neutron onlyB. Decrease of one neutron onlyC. Both (a) and(b)D. None of these |
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Answer» Correct Answer - B Loss of beta particle is equivalent to decrease of one neutron only `n rarr p + e^(-) + bar(v)` |
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| 430. |
Loss of `beta`-particles is equivalent toA. decrease of one neutron onlyB. increase of one proton onlyC. Both (a) and (b)D. None of the above |
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Answer» Correct Answer - c `._(0)n^(1) to ._(1)H^(1) + ._(-1)e^(0)` So, it is equivalent to increase of one proton and decrease of one neutron. |
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| 431. |
In an atomic orbital , the sign of inhes indicates theA. sign of the probability distributionB. sign of chargeC. sign of the wave functionD. presence or absence of electron |
| Answer» Correct Answer - C | |
| 432. |
which of the given statement(s) is /are false ? (P) orbital angular momentum of the azimuthal quantum number as lowest for any principle quantum number is `(h)/(pi)`, (Q) if n=3 ,l=0 ,m=0 for the last valence shell electron ,them the possible atomic number may be 12 or 13 . (R ) total spin of electrons for the atom `_(25)` Mn is `+-(7)/(2)` . (S ) spin magnetic moment of inert gas is zero .A. I, II and IIIB. II and III onlyC. I and IV onlyD. none of these |
| Answer» Correct Answer - A | |
| 433. |
The conversion of one element into another by aritificial means is termed `………………………..`. |
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Answer» Correct Answer - Artificial transmutation |
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| 434. |
The orbit having Bohr radius equal to 1st Bohr orbit of H-atom is :A. `n=2` of `He^(+)`B. `n=2` of `B^(+4)`C. `n=3` of `Li^(2+)`D. `n=2` of `Be^(+3)` |
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Answer» Correct Answer - D `r_(n)=0.529 n^(2)/Z ` For hydrogen, `n=1` and `Z=1, :. r_(H)=0.529` For `Be^(3+), n=2` and `Z=4, :. r_(Be^(3+))=(0.529xx2^(2))/(4)=0.529` there, (D) is correct option. |
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| 435. |
Carbon 14 dating method is based on the fact thatA. Carbon -14 fraction is the same in all objectsB. Carbon-14 is highly insolubleC. Ratio of carbon -14 and carbon -12 is constantD. All of these |
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Answer» Correct Answer - C A basis for the `C -14` dating technique |
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| 436. |
Carbon 14 dating method is based on the fact thatA. carbon 14 fraction is same in all objectsB. carbon 14 is hightly insolubleC. ratio of carbon 14 and carbon 12 constantD. All of the above |
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Answer» Correct Answer - C By carbon dating method Age of wood =`(2.303)/(0.693)xxT_(0.5) log(N_(0))/(N)` Where , `(N_(0))/(N)=[("Ratio of" C^(14)//C^(12)"in living wood")/("Ratio of" C^(14)//C^(12) "in dead wood")]` Hence it based upon the ratio of `C^(14)` and `C^(12)` |
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| 437. |
`C - 14` is used in carbon dating of dead objects becauseA. Its half-life is `10^(3)` yearsB. Its half-life is `10^(4)` yearsC. It is found in nature abundantly and in definite ratioD. It is found in dead animals abundantly |
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Answer» Correct Answer - C `C -14` is found in nature abundantly and in definite ratio |
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| 438. |
Which of the following nuclei is unstable?A. `._(5)B^(10)`B. `._(4)Be^(10)`C. `._(7)N^(14)`D. `._(8)O^(16)` |
| Answer» Correct Answer - B | |
| 439. |
Unstable nuclei attain stability through disintegration. The nuclear stability is related to neutron proton ratio `(n//p)`. For stable nuclei `n//p` ratio lies close to unity for elements with low atmoic numbers (20 or less) but it is more than 1 for nuclei having higher atomic numbers. Nuclei having `n//p` ratio either very high or low undergo nuclear transformation. When `n//p` ratio is higher than required for stability, the nuclei have the tendency to emit `beta`-rays. while when `n//p` ratio is lower than required for stability, the nuclei either emits `alpha`-particles or a positron or capture `K`-electron.A. `beta`-emissionB. `alpha`-emissionC. `gamma`-emissionD. Positron emission |
| Answer» Correct Answer - A | |
| 440. |
When passing through a magnetic field the largest degflection is experienced byA. `alpha`-raysB. `beta`-raysC. `gamma`-raysD. All equal |
| Answer» Correct Answer - B | |
| 441. |
Which of the following is not correctA. positron emission : increase n/p ratioB. k electron capture : decrease n/p ratioC. `beta`-decay decrease n/p ratioD. `alpha`-decay increases n/p ratio |
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Answer» Correct Answer - b `""_(11)Na""^(23)to""_(10)Ne""^(23)+""_(+1)e""^(0)` |
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| 442. |
The radioactive decay produces the species with fastest speed isA. `alpha`B. `beta`C. `gamma`D. Positron |
| Answer» Correct Answer - c | |
| 443. |
`._(6)^(11) C` on decay produces:A. positronB. `beta`-particleC. `alpha`-particleD. None of these |
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Answer» Correct Answer - a N/P below the stability zone |
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| 444. |
A nucleus with an excess of neutrons may decay with the emission ofA. a neutronB. a protonC. an electronD. a positron |
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Answer» Correct Answer - c `""_(0)n""^(1)to""_(1)H""^(1)+""_(-1)e""^(0)` |
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| 445. |
`._(13)Al^(27)` is a stable isotope. `._(13)Al^(29)` is expected to disintegrate byA. `alpha`-emissionB. `beta`-emissionC. positron emissionD. proton emission |
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Answer» Correct Answer - b N/P is above stability zone |
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| 446. |
`._(13)^(27)Al` is a stable isotope. `._(13)^(29)Al` is expected to disintegrate byA. `alpha-` emissionB. `beta-emissionC. Positron emissionD. Proton emission |
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Answer» Correct Answer - B The `(n)/(p)` ratio of `._(13)Al^(29)` places it above the belt of stability and thus it emits `beta-` particles. |
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| 447. |
A follow parallel path of first-order reactions giving `B` and `C` as If the initial concentration of `A` is `0.25 M`, calculate the concentration of `C` after 5 hr of reaction. |
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Answer» `lambda_(A) = lambda_(1) + lambda_(2) = .15 xx 10^(-5) + 5 xx 10^(-6) = 20 xx 10^(-6) s^(-1)` Also, `2.303 log ([A]_(0))/([A]_(t)) = lambdas xx t` `:. 2.303 log (0.25)/([A]_(t)) = 20 xx 10^(-6) xx 5 xx 60 xx 60` `:. [A]_(t) = 0.1744 M` `:. [A]_("decompound") = [A]_(0) - [A]_(t) = 0/25 - 0.1744` `= 0.-756 = M` Fraction of `C` formed `= [(lambda_(2))/(lambda_(1) + lambda_(2))] xx [A]_("decomposed") xx (2)/(5)` `= 0.0756 xx (5 xx 10^(-6))/(20 xx 10^(-6)) xx (2)/(5)` `= 7.56 xx 10^(-3) M` |
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| 448. |
Which of the following is not deflected by magnetic fieldA. DeutronsB. PositronC. ProtonD. Photon |
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Answer» Correct Answer - D Photons do not carry any charge |
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| 449. |
In a given electric field , `beta-` particles are deflected more than `alpha-` particles, inspite of `alpha-` particles having larger charge. |
| Answer» Correct Answer - 1 | |
| 450. |
Energy required to separate neutron and proton from the nucleus is calledA. Bond energyB. Nuclear energyC. Chemical energyD. Radiation energy |
| Answer» Correct Answer - B | |