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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.
| 1. |
A bone containing `200 g` carbon-14 has `beta`-decay rate of `375` decay/min. Calculate the time that has elapsed since the death of the living one. Given the rate of decay for the living organism is equal to `15` decay per min per gram of carbon and half-life of carbon-14 is `5730 years`.A. 22920 yrB. 11460 yrC. 17190 yrD. None of these |
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Answer» Correct Answer - C `R_(1)=(15xx200)=300` decay/min R = 375 decay/min `=(R_(1))/(8)` `thereforeTime,t=3"(half-liver)"=3xx5730=17190yr` |
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| 2. |
Fusion processes, like combining two deuterons to form a He nucleus are impossible at ordinary temperatures and pressure. The reasons for this can be traced to the fact: (a) nuclear forces have short range. (b) nuclei are positively charged. (c) the original nuclei must be completely ionized before fusion can take place. (d) the original nuclei must first break up before combining with each other. |
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Answer» (a), (b) (a) nuclear forces have short range. (b) nuclei are positively charged. |
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| 3. |
A deuteron strikes `""._(7) N^(14)` nucles with the subsequent emission of an `alpha `-particle ,Find the atomic number , Mass number and chemical name of the element so produced |
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Answer» the nuclear reaction as per the given data is given data is given as , `""_(7)N^(14)+""_(1)h^(2) to ""_(2)Z^(A)+""_(2)He^(4) " " (alpha - " particle ")` According to laws of conservation (i) for change, `sum Z_("intital ")=sum Z_("final ")` `implies 7+1=z+2implies z=6` (ii) for mass number , `sum A_("initial ")=sum A_("final")` `implies 14+2=A+4implies A=12` `implies ""_(6)Z^(12)to ` the element is carbon ""_(6)C^(12)`. |
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| 4. |
A deutron strikes `._8 O^16` nucleus with subsequent emission of an alpha particle. Idenify the nucleus so produced. |
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Answer» The nuclear reaction may be expressed as under `._8O^16 + ._1H^2 to ._zX^A + ._2He^4 (alpha-"particle")` Law of conservation of charge 8+1=Z+2 `rArr` Z=7 The atomic number of the element is 7 `rArr` element is nitrogen Law of conservation of mass number 16+2 =A+4 A=14 `._zX^A=._7N^14` . |
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| 5. |
Fast neutrons can easily be slowed down byA. Passing them through waterB. Applying a strong electric fieldC. The use of lead shieldingD. Elastic collisions with heavy nude |
| Answer» Correct Answer - A | |
| 6. |
Assertion:Nuclear sources will give a million times larger energy than conventional sources. Reason:Nuclear energy sources are massive than conventional energy sources.A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false |
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Answer» Correct Answer - c For same quantity of matter, nuclear sources will give a million times larger energy than conventional sources. One kilogram of coal on burning gives `10^7` J of energy, whereas 1 kg of uranium , which undergoes fission, will generate `10^14` J of energy. |
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| 7. |
Assertion:Fusion of hydrogen nuclei into helium nuclei is the source of energy of all stars. Reason:In fusion heavier nuclei split to form lighter nuclei.A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false |
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Answer» Correct Answer - c In fusion lighter nuclei combine to form a larger nucleus . Fusion of hydrogen nuclei into helium nuclei is the source of energy of all stars including our sun. |
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| 8. |
Assertion:Neutrons penetrate matter more readily as compared to protons. Reason:A neutron has no charge .A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false |
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Answer» Correct Answer - a Protons suffer more collisions when they enter matter than neutrons. In each collision they lose energy. Hence, neutrons penetrate into matter more than protons.The reason why protons suffer more collisions is because of their charge. |
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| 9. |
Assertion:There occurs a chain reaction when uranium is bombarded with slow neutrons. Reason:When uranium is bombarded with slow neutrons more neutrons are producedA. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false |
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Answer» Correct Answer - a When uranium is bombarded by slow neutrons the reaction is represented as `._92^235U + ._0^1n to ._56^144Ba + ._36^89Kr + 3_0^1n`+Energy As more neutrons are produced , the reason is correct. There are additional neutrons strike uranium nuclei to produce even more neutrons. Thus a chain reaction is established. |
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| 10. |
The three stable isotopes of neon : `._(10)^(20)Ne, ._(10)^(21)Ne` and `._(10)^(22)Ne` have respective abundance of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of Neon. |
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Answer» The masses of three isotopes are 19.99 u, 20.99 u, 21.99 u Their relative abundances are 90.51 %, 10.27% and 9.22% `therefore` Average atomic mass of Neon is `m = (90.51xx19.99+0.27xx20.99+9.22xx21.99)/((90.51+0.27+9.22))` `=(1809.29+5.67+202.75)/(100)=(2017.7)/(100)=20.17 u` |
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| 11. |
In pair annihilation, an electron and a positron destroy each other to produce gamma radiation. How is the momentum conserved? |
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Answer» 2γ photons are produced which move in opposite directions to conserve momentum. |
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| 12. |
Consider `alpha-, beta`- particles and `gamma-` rays, each having an energy fo `0.5Mev` in increasing order f penertation power, the radiations are:A. `alpha, beta, gamma`B. `alpha, gamma , beta`C. `beta, gamma , alpha `D. `gamma, beta, alpha` |
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Answer» Correct Answer - a For given energy , the radiations in the increasing order of penetrating power are `alpha lt beta lt gamma ` |
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| 13. |
The binding energy of deuteron `._1^2 H` is `1.112 MeV` per nucleon and an `alpha-`particle `._2^4 He` has a binding energy of `7.047 MeV` per nucleon. Then in the fusion reaction `._1^2H + ._1^2h rarr ._2^4 He + Q`, the energy `Q` released is.A. 23.74 MeVB. 32.82 MeVC. 11.9 MeVD. 4.94 MeV |
| Answer» Correct Answer - A | |
| 14. |
If one microgram of `._(92)^(235)U` is completely destroyed in an atom bomb, how much energy will be released ? |
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Answer» `m = 1 mu g = 1xx10^(-6)g = 1xx10^(-6)xx10^(-3)kg` `= 10^(-9)kg` `c=3xx10^(8)m//s` `E=mc^(2)=1xx10^(-9)xx9xx10^(6)=9xx10^(7)J` |
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| 15. |
An energy of `24.6 eV` is required to remove one of that electrons from a neutal helium atom. The enegy (in `eV`)required to remove both the electrons from a netural helium atom isA. 38.2B. 49.2C. 51.8D. `79.0` |
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Answer» Correct Answer - D After remaining one electron from helium atom it will become hydrogen like atom. `therefore" "E=24.6+(13.6)^(2)=79eV` |
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| 16. |
A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio `8 : 1`. The ratio of radii of the fragments is.A. `1 : 2`B. `1:4`C. `4:1`D. `2:1` |
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Answer» Correct Answer - A By conservation of momentum, `m_(1)v_(1)=m_(2)v_(2)` `implies " "(v_(1))/(v_(2))=8/1=(m_(2))/(m_(1))" "...(i)` Also, from `r propA^(1//3),` `therefore" "(r_(1))/(r_(2))=((A_(1))/(A_(2)))^(1//3)=((1)/(8))^(1//3)=1/2` |
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| 17. |
The activity of a sample of radioactive material is `R_(1)` at time `t_(1)andR_(2)"at time"t_(2)(t_(2)gtt_(1))`. Its mean life is T. Then,A. `R_(1)t_(1)=R_(2)t_(2)`B. `(R_(1)-R_(2))/(t_(1)-t_(2))`=constantC. `R_(2)=R_(1)e^((t_(1)-t_(2))//T`D. `R_(2)=R_(1)e^((t_(1)//t_(2))//T` |
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Answer» Correct Answer - C Activity, `R=R_(0)e^(-lamdat),thereforeR_(1)=R_(0)e^(-t//T)` `implies"Aganin",R_(2)=R_(0)e^(-t2//T)` `therefore R_(2)=R_(1)e^(-t2//T)impliesR_(2)=R_(1)e^(((t_(1)-t_(2)))/(T))` |
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| 18. |
Which of the following cannot be emitted by radioactive substances during their decay ?A. NeutrinosB. ProtonsC. ElectronsD. Helium nuclei |
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Answer» Correct Answer - b Protons are not emitted by radioactive substances during their decay. |
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| 19. |
Which of the following cannot be emitted by radioactive substance during their decay?(A) Protons(B) Neutrinos(C) Helium nuclei(D) Electrons |
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Answer» The answer is (A) Protons. |
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| 20. |
In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundnce on earth. The three isotopes and their masses are `._(12)^(24)Mg (23.98504 u), ._(12)^(25)Mg (24.98584 u)` and `._(12)^(26)Mg(25.98259 u)`. The natural abundance of `._(12)^(24)Mg` is `78.99 %` by mass. Calculate the abundances of other two isotopes. |
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Answer» Let the abundance of `._(12)Mg^(25)` by mass be x% therefore, abundance of `._(12)Mg^(26)` by mass = (100 - 78.99) - x% = (21.01) - x% Now average atomic mass of magnesium is `24.312 = (23.98504xx78.99+24.98584+25.98529 (21.01 - x))/(100)` on solving we get x = 9.303 % for `._(12)Mg^(25)` and for `._(12)Mg^(26), (21.01 -x) = 11.71 %` |
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| 21. |
Show that the density of a nucleus does not depend upon its mass number (density is independent of mass) |
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Answer» Density of nucleus matter `= ("mass of nucleus")/("volume of nucleus")` But mass of nucleus = No. of nucleons (A) `xx` mass of nucleons (m) Volume of nucleus `V=(4)/(3)pi R^(3)` But `R=R_(0)A^(1//3)` `therefore V=(4)/(3)pi R_(0)^(3)A` `therefore` Density of nuclear matter = `(Am)/((4)/(3)pi R_(0)^(3)A=(3m)/(4pi R_(0)^(3))` `therefore` Density of nucleus is independent of mass. |
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| 22. |
The plot of the number (N) of decayed atoms versus activity (R) of a radioactive substance isA. B. C. D. |
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Answer» Correct Answer - D We have, `N=N_(0)e^(-lamdat)and R=R_(0)e^(-lamdat)=lamdaN_(0)e^(-lamdat)` `thereforeN_("decayed")=N_(0)-N=N_(0)e^(-lamdat)=N_(0)-(R)/(lamda)` This is equation of straight line with negative slope. |
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| 23. |
Ceratain radioactive substance reduces to `25%` of its value is `16 days`. Its half-life isA. 32 daysB. 8 daysC. 64 daysD. 28 days |
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Answer» Correct Answer - B We have, `N=No((1)/(2))^(t//T)implies(N_(0))/(4)=N_(0)((1)/(2))^(16//T)impliesT=` 8 days |
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| 24. |
Half-life of radioactive element depend uponA. amount of element presentB. temperatureC. pressureD. nature of element |
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Answer» Correct Answer - D Half-life of a substance does not depends upon amount, temperature and pressure. It depends upon the nature of the substance. |
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| 25. |
The activity of a radioactive sample is measures as `N_0` counts per minute at `t = 0` and `N_0//e` counts per minute at `t = 5 min`. The time (in minute) at which the activity reduces to half its value is.A. `log_e2/5`B. `5/(log_e2)`C. `5log_10 2`D. `5log_e 2` |
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Answer» Correct Answer - d According to activity law `R=R_0e^(-lambdat)` …(i) where, `R_0`=initial activity at t=0 R=activity at time t `lambda`=decay constant According to given problem, `R_0=N_0` counts per minute `R=N_0/e` counts per minute t=5 minutes Substituting these values in equation (i), we get `N_0/e=N_0e^(-5lambda)` `e^(-1)=e^(-5lambda)` `5lambda=1` or `lambda=1/5` per minute At `t=T_(1//2)` , the activity R reduces to `R_0/2`. where `T_(1//2)` = half life of a radioactive sample From equation (i), we get `R_0/2=R_0 e^(-lambdaT_(1//2))` `e^(lambdaT_(1//2))=2` Taking natural logarithms on both sides of above equation, we get `lambdaT_(1//2) = log_e2` or `T_(1//2)=(log_e 2)/lambda=(log_e 2) /((1/5))=5 log_e 2` minutes |
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| 26. |
The half-life period of a radioactive substance is `5 min`. The amount of substance decayed in `20 min` will beA. `25%`B. `6.25%`C. `93.75%`D. `75%` |
| Answer» Correct Answer - C | |
| 27. |
The half-life of pononium is `140` days. After how many days. `16 gm` polonium will be reduced to `1 gm` (or `15 gm` will decay) ?A. 700 daysB. 280 daysC. 560 daysD. 420 days |
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Answer» Correct Answer - C We have `(N)/(N_(0))=((1)/(2))^(t//140)[thereforeN=No((1)/(2))^(n)andt=nt_(1//2)]` `1/16=((1)/(2))^(t//140)impliest/140=4(t_(1//2)=140days)` `impliest=560 days` |
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| 28. |
Mean life of a radioactive sample is 100s . Then ,its half-life (in min) isA. 0.693B. 1C. `10^(-4)`D. 1.155 |
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Answer» Correct Answer - D Mean life, `T=1//lamda=100s` Hilf life `=(0.693)/(lamda)=(0.693xx100)/(60)=1.155` min |
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| 29. |
The half-life of radioactive Polonium `(Po)` is `138.6` days. For ten lakh Polonium atoms, the number of disintegrations in `24` hours is.A. 2000B. 3000C. 4000D. 5000 |
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Answer» Correct Answer - D We, have, `n=(24)/(24xx138.6)=(1)/(138.6)` Now, `(N)/(N_(0))=((1)/(2))^(n)=((1)/(2))^(1//138.6)` `impliesN=1000000((1)/(2))^(1//1.38.6)=995011` So, number of disintengration `=1000000-995011=4989~~5000` |
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| 30. |
If the half-life of a radioactive sample is 10 hours its mean life isA. 14.4 hB. 7.2 hC. 20 hD. 6.93 h |
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Answer» Correct Answer - A Mean life `=("Hilf-life")/(0.6931)=(10)/(0.6931)=14.4h` |
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| 31. |
A radioactive substance has an average life of `5` hours. In a time of `5` hoursA. Less than half of the active nuclei decayB. More than half of the active nuclei decayC. Half of the active nuclei decayD. All active nuclei decay |
| Answer» Correct Answer - B | |
| 32. |
A sample of radioactive element has a mass of `10 g` at an instant `t=0`. The approximate mass of this element in the sample after two mean lives isA. 2.50 gmB. 1.35 gmC. 6.30 gmD. 3.70 gm |
| Answer» Correct Answer - B | |
| 33. |
A sample of a radioactive element has a mass of 10 g at an instant t=0. The approximate mass of this element in the sample left after two mean lives isA. 1.35 gB. 2.50 gC. 3.70 gD. 6.30 g |
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Answer» Correct Answer - a Here, t=2 `tau=2xx1.44 T_(1//2)=2.88 T_(1//2)` As `N/N_0=m/m_0=(1/2)^(t//T_(1//2))=(1/2)^(2.88)` `therefore m=m_0(1/2)^(2.88)=10(1/2)^2.88 ` g =10 x 0.1358 g = 1.358 g |
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| 34. |
Two samples X and Y contain equal amount of radioactive substances. If `1^(th)/16` of the sample Y , remain after 8 hours, then the ratio of half life periods of X and Y isA. `2:1`B. `1:2`C. `1:4`D. `4:1` |
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Answer» Correct Answer - a As `N/N_0=(1/2)^n` where, number of half lives , `n=t/T` T is the half life period For sample X `1/16=(1/2)^(8//T_X)` or `(1/2)^4 =(1/2)^(8//T_X)` `rArr 4=8/T_X` ….(i) For sample Y `(1/256)=(1/2)^(8//T_Y)` or `(1/2)^8=(1/2)^(8//T_Y)` `8=8/T_Y` ...(ii) Dividing (i) by (ii) , we get `4/8=8/T_X xx T_Y/8` `1/2=T_Y/T_X` or `T_X/T_Y=2/1` |
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| 35. |
A radioactive element decays by `beta-`emission. A detector racords `n`-beta particles in `2` sec and in next `2` sec it records `0.65n`-beta particles. Find mean life.A. 4 sB. 2 sC. `(2)/((In 2))s`D. 2 (In 2)s |
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Answer» Correct Answer - C The half-life of radioactive substance is `T_(v2)=2s` Mean life, `lamda=(T_(1//2))/("in" 2)=2/("in"2)s` |
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| 36. |
A radio active isotope has a half-life of T years. How long will it take the activity to reduce to 1% of its original value ? |
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Answer» Here `(N)/(N_(0))=(1)/(100)` From `(N)/(N_(0))=e^(-lambda t)=(1)/(100)` `- lambda t=log 1- log_(e )^(100)` `= 0-2.303 log_(10)^(100)= - 2.203xx2` `= - 4.606` `t = (4.606)/(lambda)=(4.606)/(0.696//T)=6.65 T` |
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| 37. |
Starting with a sample of pure `.^66 Cu, 7//8` of it decays into `Zn` in `15 min`. The corresponding half-life is.A. 5 minB. `7(1)/(2)min`C. 10 minD. 15 min |
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Answer» Correct Answer - A According to the question, (7/8) part decays i.e., remaining part is (1/8). We have, `N =N_(0)((1)/(2))^((t)/(T_(1//2)))implies1/8((1)/(2))^((15)/(T_(1//2)))` `implies"Half-life"T_(1//2)=5min` |
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| 38. |
Obtain the amount of `.^60Co` necessary to provide a radioactive source of `8.0 Ci` strength. The half-life of `.^60Co` is `5.3` years? |
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Answer» The strength of the radioactive source is given as: `(dN)/(dt)=8.0mCi` `8xx10^(-3)xx3.7xx10^(10)` `=29.6xx10^(7)decay//s` Where, N = Required number of atoms Half-life of `_(27)^(60)Co,T_(1/2)=5.3" "years` `=5.3xx365xx24xx60xx60` `=1.67xx10^(8)s` For decay constant `lambda`, we have the rate of decay as: `(dN)/(dh)=lambdaN` Where,`lambda=0.693/T_(1/2)=(0.693)/(1.67xx10)s^(-1)` `therefore N=1/lambda (dn)/(dt)` `=(29.6xx10^(7))/((0.693)/(1.67xx10^(8)))=7.133xx10^(16) " atoms"` `"For " _(27)Co^(60):` `"For " _(27)Co^(60):` Mass of `6.023xx10^(23)` (Avogadro’s number) atoms `= 60 g` `therefore "Mass of "7.133xx10^(16) " atoms "=(60xx7.133xx10^(16))/(6.023xx10^(23))=7.106xx10^(-6)g` Hence, the amount of `_(27)Co^(60)` necessary for the purpose is `7.106xx10−6 g`. |
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| 39. |
The half-life a radioacitve substance is `40` yeard. How long will it take to reduce to one fourth of its original amount and what is the value of decay constant ?A. `40 yr,0.9173//yr`B. `90yr,9.017//yr`C. `80yr,0.0173//yr`D. None of these |
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Answer» Correct Answer - C To reduce one-fourth it takes time, `t=2(T_(1//2))=2xx40=80yr` So, decay constant, `lamda(0.693)/(T_(1//2))=(0.963)/(40)=0.0173yr` |
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| 40. |
Two radioactive nuclei `P` and `Q`, in a given sample decay into a stable nucleus `R`. At time `t = 0`, number of `P` species are `4 N_0` and that of `Q` are `N_0`. Half-life of `P` (for conversation to `R`) is `1mm` whereas that of `Q` is `2 min`. Initially there are no nuclei of `R` present in the sample. When number of nuclei of `P` and `Q` are equal, the number of nuclei of `R` present in the sample would be :A. `(5N_(0))/(2)`B. `2N_(0)`C. `3N_(0)`D. `(9N_(0))/(2)` |
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Answer» Correct Answer - D Initeally `Pto4N_(0),QtoN_(0)` Half-life, `T_(P)=1min,T_(Q)=2min` Let after time t, number of nuclei of P and Q equal `i.e.,(4N_(0))/(2^(t//1))=(N_(0))/(2^(t//2))" "(becauseN=(N_(0))/(2^(t//T_(1)//2)))` `or(4)/(2^(t//2))=1or t=4min` So, at t= 4 min `N_(P)=((4N_(0)))/(2^(4//1))=(N_(0))/(4)` and at `t=4 min,N_(Q)=(N_(0))/(2^(4//2))=(N_(0))/(4)` So, number of nuclei of R in teh sample `=(4N_(0)-(N_(0))/(4))+(N_(0)-(N_(0))/(4))=(9N_(0))/(2)` |
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| 41. |
At a given instant, there are 25% undecayed radioactive nuclei in a sample. After 10 seconds the number of undecayed nuclei reduces to 12.5%, the mean life of the nuclei isA. 10.21 sB. 14.43 sC. 5.31 sD. 7.43 s |
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Answer» Correct Answer - b As the number of undecayed nuclei decreases from 25% to 12.5% in 10 s, it shows that the half life of the sample is 10 s, i.e. `T_(1//2)` =10 s Decay constant ,`lambda=0.06931/T_(1//2)=0.6931/"10 s"` Mean life , `tau = 1/lambda= "10 s"/0.6931 =14.43 s` |
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| 42. |
In a sample of radioactive material , what fraction of the initial number of active nuclei will remain undisintegrated after half of the half life of the sample ?A. `1/4`B. `1/(2sqrt2)`C. `1/sqrt2`D. `sqrt2-1` |
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Answer» Correct Answer - c According to radioactive decay `N/N_0=(1/2)^(t//T_(1//2))` Here, `t=(T_(1//2))/2 , therefore N/N_0=(1/2)^(1//2) = 1/sqrt2` |
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| 43. |
Sn, C, Si and Ge are all group XIV elements . Yet , Sn is a conductor , C is an insulator while Si and Ge are semiconductors . Why? |
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Answer» A marterial is a conductor if in its energy band diagram, there is no energy gap between conduction band and valence band. For insulator, the energy gap is large and semiconductor the energy gap is moderate. The energy gap for Sn is `0 eV`, for C is `5.4 eV`, for Si is `1.1 eV` and for Ge is `0.7 eV`, related their atomic size. Therefore Sn is a conductor, C is an insulator and Ge and Si semiconductors. |
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| 44. |
Name and define, the SI units for the 'activity', of a given sample of radioactive nuclei. |
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Answer» Becquerel On becquerel activity corresponds to one decay disintegration per second. |
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| 45. |
Mention the SI unit of activity. |
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Answer» Becquerel (Bq) |
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| 46. |
Which type of radioactive emission produces a daughter nucleus which is an isobar of the parent? |
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Answer» Beta particle |
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| 47. |
select the point of isotones from the following nuclei `""._(12)Mg^(24),""._(1)H^(3),""._(2)He^(4),""._(11)Na^(23)`. |
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Answer» (i)`._1H^3` and `._2He^4` Number of neutrons =3-1 or 4-2 =2 (ii)`._12Mg^24` and `._11Na^23` Number of neutrons=24-12 or 23-11 =12 |
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| 48. |
1g of a radioactive substance disintegrates at the rate of `3.7xx10^(10)` disintegrations per second. The atomic massof the substance is 226. Calculate its mean life. |
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Answer» We known , number of nuclei present ,`N=(m)/(M)N_(A)` where , m=given mass ,M= atomic mass `N_(A) `= Avogadro number Activity , `R=lamda N =(1)/(tau)(mN_(A))/(M)` where , `tau` = mean life . Maean life of substance , `tau =(mN_(A))/(MR)=(1xx6.023xx10^(23))/(226xx3.7xx10^(10))s` `=7.2xx10^(10)s` |
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| 49. |
During mean life of a radioactive element, the fraction that disintegrates isA. eB. `(e-1)/e`C. `1/e`D. `e/(e-1)` |
| Answer» Correct Answer - B | |
| 50. |
A radioactive substance disintegrates 1/64 of initial value in 60 s. The half-life of this substance isA. 5 sB. 10 sC. 30 sD. 20 s |
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Answer» Correct Answer - B We have , `1/64=((1)/(2))^(n)n=6` 6 half-lives are equal to 60 s. `therefore` 1 half-life = 10 s |
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