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.
| 501. |
Nitrogen has several isotopes where atomic masses ranging from 10 to 25. Out of which `._(7)N^(17)` is radioactive and is converted into `._(8)O^(17)` by emission ofA. Alpha particleB. PositronC. Beta particleD. Neutron |
|
Answer» Correct Answer - C `._(7)N^(17) rarr ._(8)O^(17) + ._(-1)e^(0)` (`beta` emission) |
|
| 502. |
The disintegration of an isotope of sodium `._(11)Na^(24) rarr ._(12)Mg^(24) + ._(-1)e^(0)` shown is due toA. The emission of `beta-`radiationB. The formation of a stable nuclideC. The fall in the neutron : proton ratioD. None of these |
|
Answer» Correct Answer - A::B::C An emission of `beta-`particles that atomic number increases by 1 but mass number remain unaffected and neutron -protons ratio decreases. |
|
| 503. |
The positron was discovered byA. PaulingB. AndersonC. YukawaD. Segar |
|
Answer» Correct Answer - B Anderson discovered positron in 1932 |
|
| 504. |
Which of the following elements belongs to `4n`- series?A. `Pb-207`B. `Bi-209`C. `Pb-208`D. `Pb-206` |
|
Answer» Correct Answer - C `Pb-208` belong to `4n` series. |
|
| 505. |
To determine the masses of the isotopes of an element which of the following techniques is usefulA. The acceleration of charged atoms by an electric field and their subsequent deflection by a variable magnetic fieldB. The spectroscopic examination of the light emitted by vaporised elements subjected to electric dischargeC. The photographing of the diffraction patterns which arise when X-rays are passed through crystalsD. The bombardment of metal foil with alpha particles |
|
Answer» Correct Answer - A It is the required technique |
|
| 506. |
Which of the following properties are different for neutral atoms of isotopes of the same elementA. MassB. Atomic numberC. General chemical reactionsD. Number of electrons |
|
Answer» Correct Answer - A Isotopes have same atomic number but different mass number |
|
| 507. |
Which of the radioactive isotopes is used for temperature control in blood disease?A. `P^(32)`B. `H^(3)`C. `Rn^(233)`D. `I^(131)` |
| Answer» Correct Answer - A | |
| 508. |
The symbol x in the following equation is `""_(11)Na^(23)+""_(1)H^(1) rarr ""_(12)Mg^(23)+x`A. NeutronB. DeutronC. `alpha-`particlesD. Positron |
|
Answer» Correct Answer - A `._(11)Na^(23) + ._(1)H^(1) rarr ._(12)Mg^(23) + ._(0)n^(1)` |
|
| 509. |
The instability of a nucleus is due toA. High proton: electron ratioB. High proton : neutron ratioC. Low proton : electron ratioD. Low proton : neutron ratio |
|
Answer» Correct Answer - D The nuclei of those atoms are stable whose ratio of neutrons to proton (n/p ratio) lies in the range 1 to 1.5 |
|
| 510. |
Which is different in isotopes of an element?A. Atomic numberB. Mass numberC. Number of protonsD. Number of electrons |
|
Answer» Correct Answer - B Isotopes have same atomic number but different mass number. |
|
| 511. |
Atoms with the same mass number but having different nuclear charges are calledA. IsotopesB. IsobarsC. IsohoresD. Isotones |
|
Answer» Correct Answer - B Isobars have different mass number. |
|
| 512. |
`._(6)C^(14)` in upper atmosphere is generated by the nuclear reactionA. `._(7)N^(14) + ._(1)H^(1) rarr ._(6)C^(14) + ._(+1)e^(0) + ._(1)H^(1)`B. `._(7)N^(14) rarr ._(6)C^(14) + ._(+1)e^(0)`C. `._(7)N^(14) + ._(0)n^(1) rarr ._(6)C^(14) + ._(1)H^(1)`D. `._(7)N^(14) + ._(1)H^(3) + ._(0)n^(1) rarr ._(6)C^(14) + ._(2)He^(4)` |
|
Answer» Correct Answer - C `._(7)N^(14) + ._(0)n^(1) rarr ._(6)C^(14) + ._(1)H^(1)` |
|
| 513. |
`._(6)C^(14)` in the upper atmosphere is formed by the action of neutron onA. `._(7)N^(14)`B. `._(8)O^(17)`C. `._(6)C^(12)`D. `._(8)O^(18)` |
|
Answer» Correct Answer - A `._(7)N^(14) rarr ._(1)n^(0) + ._(6)C^(14)` |
|
| 514. |
`._(6)C^(14)` is formed from `._(7)N^(14)` in the upper atmosphere by the action of the fundamental particleA. PositronB. NeutronC. ElectronD. Proton |
|
Answer» Correct Answer - B The radioactive isotope `._(6)C^(14)` is produced in the atmosphere by the action of cosmic ray neutrons on `._(7)N^(14)` `._(7)N^(14) + ._(0)n^(1) rarr ._(6)C^(14) + ._(1)H^(1)` |
|
| 515. |
Which of the following nuclei is not double magic?A. `._(2)He^(4)`B. `._(8)O^(16)`C. `._(82)Pb^(208)`D. `._(92)U^(238)` |
|
Answer» Correct Answer - d The magic number for protens are 2, 8, 20, 28, 50, 82, 114 and for neutrons are 2, 8, 20, 28, 50, 82, 126, 184, 196. |
|
| 516. |
The nuclear reaction `._(1)^(2)H + ._(1)^(2)H to ._(2)^(4)He` is calledA. fission reactionB. fusion reactionC. chain reactionD. thermal reaction |
|
Answer» Correct Answer - b |
|
| 517. |
The function of the cadmium rod in a nuclear reactor isA. to produce raysB. to produce neutronsC. to absorb neutronsD. to speed up neutrons |
|
Answer» Correct Answer - c Cadmium is used to absorb neutron. |
|
| 518. |
If halfe-life of a substance is 5 yrs, then the total amount of substance left after 15 years, when initial amount is 64 grams isA. 16 gramsB. 2 gramsC. 32 gramsD. 8 grams |
|
Answer» Correct Answer - D `t_(1//2) = 5 yrs, t = 15 yrs` `:. n = (t)/(t_(1//2)) = (15)/(5) = 3` Now `N = (N_(0))/(2^(n)) = (N_(0))/(2^(3)) = (1)/(8) N_(0) = (1)/(8) xx 64 = 8` grams, |
|
| 519. |
Initial mass of a radioactive element is 40 g. How many grams of it would be left after 24 years, if its half-life period is 8 yearsA. 2B. 5C. 10D. 20 |
|
Answer» Correct Answer - B `N = (N_(0))/(2^(n)) and n = (24)/(8) = 3` `N = (40)/(2^(3)) = (40)/(8) = 5` |
|
| 520. |
The half-life period of a radioactive substance is 8 years. After 16 years, the mass of the substance will reduce from starting 16.0 g toA. 8.0 gB. 6.0 gC. 4.0 gD. 2.0 g |
|
Answer» Correct Answer - C `n = (16)/(8) = 2, N = (N_(0))/(2^(n)) = (16.0)/(2^(2)) = (16.0)/(4) = 4.0 g` |
|
| 521. |
Calculate the average life of a radioactive substance whose half-life period is 100 years. |
|
Answer» Given `t_(1//2) = 100` years We know that average life `= 1.44 xx t_(1//2)` `= 1.44 xx 100` `= 144` years |
|
| 522. |
The half -life period of `._(84)Po^(210)` is 140 days. In how many days `1 g` of this isotope is reduced to `0.25 g`? |
|
Answer» Original quantity of the isotope `(N_(0)) = 1 g` Final quantity of the isotope `N = 0.25 g` We know that, `N = ((1)/(2))^(n) N_(0)` So, `(1)/(4) = ((1)/(2))^(n) xx 1` or `((1)/(2))^(2) = ((1)/(2))^(2)` or `n = 2` Time taken `T = n xx t_(1//2) = 2 xx 140 = 280` days |
|
| 523. |
According to Schrodinger model nature of electron in an atom is as :-A. Particle onlyB. Wave onlyC. Particle and wave nature simultaneousD. Sometimes waves and sometimes particle |
| Answer» Correct Answer - C | |
| 524. |
Which of the following d-orbitals has dough-out shape ?A. `d_(xy)`B. `d_(yz)`C. `d_(x^(2)-y^(2))`D. `d_(z^(2))` |
| Answer» Correct Answer - D | |
| 525. |
Which of the `d` orbitals not lies in the `xy`-plane.A. `d_(x^(2)-y^(2))`B. `d_(xy)`C. `d_(xz)`D. `d_(xy)` and `d_(x^(2)-y^(2))` |
| Answer» Correct Answer - C | |
| 526. |
The `SI` unit of radioactivity is curie. |
|
Answer» Correct Answer - F |
|
| 527. |
If wavelength is equal to the distance travelled by the electron in one second , then :A. `lambda=h/p`B. `lambda=h/m`C. `lambda=sqrt(h/p)`D. `lambda=sqrt(h/m)` |
| Answer» Correct Answer - D | |
| 528. |
`A` particle X moving with a certain velocity has a debroglie wave length of `1A^(@)`. If particle Y has a mass of `25%` that of X and velocity `75%` that of X, debroglies wave length of Y will be :-(a). `3A^(@)`(b). `5.33A^(@)`(c). `6.88A^(@)`(d). `48A^(@)`A. `3 Å`B. `5.33 Å`C. `6.88 Å`D. `48 Å` |
| Answer» Correct Answer - B | |
| 529. |
What will be the product, if `._(92)U^(235)` emits two `alpha`-and one `beta`-particle?A. `._(87)Ac^(211)`B. `._(89)Ac^(325)`C. `._(89)Ac^(225)`D. `._(89)Ac^(227)` |
|
Answer» Correct Answer - d `._(92)U^(235) overset(-2alpha)(to) ._(88)X^(227) overset(-beta)(to) ._(89)Ac^(227)` |
|
| 530. |
Two radioactive elements `X` and `Y` half-live pf `50` and `100` minute respectiv ely. Intial sample of both the elements have same number of atoms. The ratio of the reamaining number of atoms of `X` and `Y` after 20 minute is:A. 2B. `1//2`C. 4D. `1//4` |
|
Answer» Correct Answer - d Amount left of `X=N""_(0)/2` `because` n=4(half-lives) |
|
| 531. |
If the binding energy of `2^(nd)` excited state of a hydrogen like sample os `24 eV` approximately, then the ionisation energy of the sample is approximatelyA. `54.4 eV`B. `24 eV`C. `122.4 eV`D. `216 eV` |
|
Answer» Correct Answer - D `(13.6(Z)^(2))/((3)^(2))=24` I.E. `=13.6(Z)^(2)=(24xx9)=216 ev` |
|
| 532. |
The beta activity of `1 g` of carbon made from green wood is 15.3 counts per minute. If the activity of `1 g` of carbon derived from the wood of an Egyptian mummy case is 9.4 counts per minute under the same conditions, how old is the wood of the mummy case? |
|
Answer» ` K = 0.693//t_(1//2) = 0/693//5770` `= 1.20 xx 10^(4) "year"^(-1)` log `N_(0)//N_(1) = Kt//2.303` `1.20 xx 10^(-4) xx t//2.303 = "log" N_(0)//N_(t)` `"log" 15.3//9.4` Hence `t = 2.303//1/20 xx 10^(-4) "log" 15.3//9/4` `= 3920` years |
|
| 533. |
If `R""_(1),R""_(2)` are activity of radioactive element at time `t""_(1)t""_(2)(t""_(2)gtt""_(1))`. The correct expression for mean life of nuclide isA. `(t""_(2)-t""_(1))/("in"(R""_(1)//R""_(2)))`B. `(t""_(2)-t""_(1))"in"(R""_(1)//R""_(2))`C. `((t""_(2)-t""_(1)))/("in"(R""_(2)//R""_(2))`D. `("in"(R""_(2)//R""_(2)))/((t""_(2)-t""_(1)))` |
|
Answer» Correct Answer - a `t""_(2)="in"/lamdaxx1/R""_(2),t""_(1)="in"/lamdaxx1/R""_(1)` |
|
| 534. |
During `beta-` decayA. An atomic electron is ejectedB. An electron which is already present with in the nucleus is ejectedC. A neutron in the nucleus decays emitting an electronD. A part of binding of the nucleus is converted into an electron |
| Answer» Correct Answer - C | |
| 535. |
Which of the following statement is/are correct ?A. The nuclear composition of atoms of the same element is always the same but that of atoms of different elements is different.B. Nuclei with a very high or very low neutron - to -proton ratio are radioactiveC. The shorter the half-life, the more intense is the radioactive decay of a given sampleD. Nuclei containing the magic number of neutrons and protons are the most stable |
| Answer» Correct Answer - B::C::D | |
| 536. |
Which of the following statement is/are correctA. Isotopes are always radioactive speciesB. Beta rays are always negatively charged particlesC. Alpha rays are always negatively charged particlesD. Radioactivity is due to unstable electronic configuration |
| Answer» Correct Answer - A::C::D | |
| 537. |
If `""_(92)U""^(235)` nucleus absorbes neutron and disintegrates `""_(54)X""^(139).""_(38)Sr""^(95)` and X. Then what will be the product XA. `alpha`-particlesB. `beta`-particleC. 2-neutronD. 3-neutron |
|
Answer» Correct Answer - c `""_(92)U""^(235)+""_(0)n""^(1)to""_(54)Xe""^(139)+""_(38)Sr""^(95)+2 ""_(0)n""^(1)` |
|
| 538. |
The difference between the wave number of 1st line of Balmer series and last line of Paschen series for `Li^(2+)` ion is :A. `R/36`B. `(5R)/(36)`C. `4R`D. `R/4` |
| Answer» Correct Answer - D | |
| 539. |
Light of wavelength `lambda` falls on metal having work functions `hc//lambda_(0)` . Photoelectric effect will take place only if :A. `lambda ge lambda_(0)`B. `lambda ge 2lambda_(0)`C. `lambda le lambda_(0)`D. `lambda le lambda_(0)//2` |
| Answer» Correct Answer - C | |
| 540. |
Which one of the following processes `alpha, beta^(o+), beta^(o+)`, or K-capture cause a. An increase in atomic number b. A decrease in atomic number c. Emission of X-raysA. a. An increase in atomic numberB. b. A decrease in atomic numberC. c. Emission of X-raysD. |
| Answer» a.`beta^(ɵ)` b. `alpha, beta^(o+)`, and K-capture c. K-caputre. | |
| 541. |
To what stable isotope `._(103)Lr^(257)` decay? |
| Answer» `._(103)Lr^(257)` is a member of `(4n + 1)` series and would decay to `._(83)Bi^(209)`. | |
| 542. |
Several short-lived radioactive species have been used to determine the age of wood or animal fossils. One of the most intresting substance is `""_(6)C""^(14)`(half-life fossils, etc.). Carbon-14 is produced by the bombardment of nitrogen atoms present in the upper atmosphere with neutrons (from cosmic rays). `""_(7)N""^(14)+""_(0)n""^(1)to""_(6)C""^(14)+""_(1)H""_(1)` Thus carbon-14 is oxidised, to `CO""_(2)` and eventually lingested by land animals. The death of plants or animals put an end to the intake of `C""_(14)` from the atmosphere. After this the amount of `C""_(14)` in the dead tissues starts decreasing due to its disintegration as per the following reaction : `""_(6)C""^(14)to""_(7)N""^(14)+""_(-1)beta""^(0)` The `C""^(14)` isotope enters the biosphjere when carbon dioxide is taken up in plant photosynthesis. Plants are eaten by animals, which exhale `C""^(14)` as `CO""_(2)`. Eventually, `C""^(14)` participates in many aspects of carbon cycle. The `C""^(14)` lost by radioactive decay-replenishment process, a dynamic rquillibrium is established whereby the ratio of `C""^(14)` to `C""^(12)` remains constant in living matter. But when an individual plant or an animal dies, the `C""^(14)` isotpe inn it is no longer replenished, so the ratio decreases as `C""^(14)` decays. So, ther number `C""^(14)` nuclei after time t (after the death of living matter) would be less than in a livingthe following formula, `t""_(1//2)=0.693`/? The intensity of the cosmic rays have remain the same for 30,000 years. But since some years changes in this are observed due to excessive burning of fossil fuel and nuclear test? A nuclear explosion has taken place leading to increase in concentration of `C""^(14)` in nearby areas. `C""^(14)` concentration is `C""_(1)` in nearly areas and `C""_(2)` in areas far away. If the age of the fossil is determined to be `T""_(1)` and `T""_(2)` at the respective places thenA. The age of the fossil will inrease at the where explosion has taken place and `T""_(1)-T""_(2)=1/lamda"in"C""_(1)/(C""_(2))`B. The age of the fossil will decrease at the place where explosion has taken place and `T""_(1)-T""_(2)=1/lamda"in"C""_(1)/(C""_(2))`C. The age of the fossil will determined to be sameD. `T""_(1)/T""_(2)=C""_(1)/C""_(2)` |
| Answer» Correct Answer - a | |
| 543. |
Several short-lived radioactive species have been used to determine the age of wood or animal fossils. One of the most intresting substance is `""_(6)C""^(14)`(half-life fossils, etc.). Carbon-14 is produced by the bombardment of nitrogen atoms present in the upper atmosphere with neutrons (from cosmic rays). `""_(7)N""^(14)+""_(0)n""^(1)to""_(6)C""^(14)+""_(1)H""_(1)` Thus carbon-14 is oxidised, to `CO""_(2)` and eventually lingested by land animals. The death of plants or animals put an end to the intake of `C""_(14)` from the atmosphere. After this the amount of `C""_(14)` in the dead tissues starts decreasing due to its disintegration as per the following reaction : `""_(6)C""^(14)to""_(7)N""^(14)+""_(-1)beta""^(0)` The `C""^(14)` isotope enters the biosphjere when carbon dioxide is taken up in plant photosynthesis. Plants are eaten by animals, which exhale `C""^(14)` as `CO""_(2)`. Eventually, `C""^(14)` participates in many aspects of carbon cycle. The `C""^(14)` lost by radioactive decay-replenishment process, a dynamic rquillibrium is established whereby the ratio of `C""^(14)` to `C""^(12)` remains constant in living matter. But when an individual plant or an animal dies, the `C""^(14)` isotpe inn it is no longer replenished, so the ratio decreases as `C""^(14)` decays. So, ther number `C""^(14)` nuclei after time t (after the death of living matter) would be less than in a livingthe following formula, `t""_(1//2)=0.693`/? The intensity of the cosmic rays have remain the same for 30,000 years. But since some years changes in this are observed due to excessive burning of fossil fuel and nuclear test? Why do we use the carbon dating to calculate the age of the fossil?A. Rate of exchange of carbon beween atmosphere and living is slower than decay of `C""^(14)`B. It is not appropriate to use `C""^(14)` dating to determine imageC. Rate of exchange of `C""^(14)` between atmosphere and living organism is so fast that an equilibrium is set up between the intake of `C""^(14)` by organisms and its exponential decayD. None of these |
| Answer» Correct Answer - c | |
| 544. |
Assertion `(A) : ` Nuclear isomers have same atomic number and same mass number but with different radioactive properties. Reason `(R) : U_((A))` and `U_((Z))` are nuclear isomers.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 - A | |
| 545. |
Assertion `(A) : K-` shell electron capture is detected by analyzing the wavelength of `X-` ray emitted. Reason `(R):` The wavelength of the `X-` ray is characteristic of the daughter element and not the parent element.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. If `(A)` is incorrect, but `(R)` is correct. |
| Answer» Correct Answer - B | |
| 546. |
If 75% of a frist-order reaction is completed in 32 min, than 50% of the reaction would complete in? a. 24 min b. 16 min c. 8 min d. 4 minA. a. 24 minB. b. 16 minC. c. 8 minD. d. 4 min |
|
Answer» Correct Answer - b. 16 min Use direct formula `(t_(x%))/(t_(y%)) = ("log" (a)/(a - x))/("log" (a)/(a - y))` `:. (t_(50%))/(t_(75%)) = ("log" (100)/(100 - 50))/("log" (100)/(100 - 75)) implies t_(50%) = (t_(75%))/(2)` `:.t_(32)/(2) = 16 "min"` |
|
| 547. |
The half life of `C^(14)(lambda=2.31xx10^(-4)` per year) isA. `2xx10^(2)` yrB. `3xx10^(3)` yrC. `3.3xx10^(4)` yrD. `4xx10^(3)` yr |
|
Answer» Correct Answer - B Radio active decay is first order reaction So `lambda =(0.69)/(t_(1//2)` `t_(1//2)=(0.693)/(2.31xx10^(-4))yr` `=0.3xx10^(4)` `=3xx10^(3)yr` |
|
| 548. |
In the case of a radio isotope the value of `T_(1//2) and lamda` are identical in magnitude. The value isA. 0.693B. `(0.693)^(1//2)`C. `1//0.693`D. `(0.693)^(2)` |
|
Answer» Correct Answer - B `t_(1//2) = (0.693)/(k or lamda) " i.e., " 0.693 = lamda^(2) rArr lamda = (0.693)^(1//2)` |
|
| 549. |
Only `1//8th` of the original amount of a radioactive element remains after 96 min. The value of `t_(1//2)` of this element isA. 12.0 minB. 32.0 minC. 24.0 minD. 48.0 min |
|
Answer» Correct Answer - b Given that, `N_(0) = 1, N = (1)/(8), t = 96` min We know that, `N = N_(0)((1)/(2))^((t)/(t(1//2))) or (1)/(8) = ((1)/(2))^((96)/(t_(1//2)))` or `((1)/(2))^(3) = ((1)/(2))^((96)/(t_(1//2))) or (96)/(t_(1//2)) = (96)/(3) = 32` min |
|
| 550. |
If the disintegration constant is `6.93 xx 10^(-6) yr^(-1)`, then half-life of `._(6)C^(14)` will beA. `10^(2) yr`B. `10^(3) yr`C. `10^(4) yr`D. `10^(5)yr` |
|
Answer» Correct Answer - D `t_(1//2) = (0.693)/(k) = (0.693)/(6.93 xx 10^(-6)) = 0.1 xx 10^(6) = 10^(5) yr` |
|