InterviewSolution
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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.
| 101. |
A small quantity of solution containing `Na^24` radio nuclide `(half-life=15 h)` of activity `1.0 microcurie` is injected into the blood of a person. A sample of the blood of volume `1cm^3` taken after `5h` shows an activity of `296` disintegrations per minute. Determine the total volume of the blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of person. (1 curie `=3.7 xx 10^10` disintegrations per second) |
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Answer» `t_(1//2) = 15h, A_(0) = 1muCi = 10^(-6) xx 3.7 xx 10^(10)` `= 3.7 xx 10^(4)dps` `t = 5h, A = (296)/(60)dps`, volume of blood `V = 1 cm^(3)` Let initial volume of blood = `V_(0)cm^(3)` `(A)/(V) = (A_(0))/(V_(0)) e^(-lambdat) = (A_(0))/(V_(0))e^(-(0.693)/(t_(1//2))t)` `(296//60)/(1) = (3.7 xx 10^(4))/(V_(0)) e^(-(0.693)/(15)xx5)` `= (3.7xx10^(4))/(V_(0)xxe^(0.231))` `V_(0) = 5953.17cm^(3)` `= 5.953` litre |
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| 102. |
`.^22 Ne` nucleus after absorbing energy decays into two `alpha-`particles and an unknown nucleus. The unknown nucleus is.A. NitrogenB. CarbonC. BoronD. Oxygen |
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Answer» Correct Answer - B (b) `._10^22 Ne rarr ._2^4 He + ._2^4 He + ._6^14 X` , hence `X` is carbon. |
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| 103. |
Ne nucleus , the after absorbing energy , decays into two `alpha` particle and an unknown nucleus . The unknown nucleus isA. nitrogenB. carbonC. boronD. oxygen |
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Answer» Correct Answer - b `._10^(22)Ne rarr ._2^4He + ._2^4He +._6^(14)X` The new elemen `X` has a atomic number `6`. Therefore, the element is carbon. |
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| 104. |
The half - life of` `I ^ (131)` is 8 days. Given a sample of `I^(131)` at time `t = 0` , we can assert thatA. No nucleus will decay before `t = 4` daysB. No nucleus will decay before `t = 8` daysC. All nuclei will decay before `t = 16` daysD. A given nucleus may decay at any time after `t = 0` |
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Answer» Correct Answer - D (d) Number of nuclei decreases exponentially `N = N_0 e^(-lamda t)` and Rate of decay `(-(dN)/(dt)) = lamda N` Therefore, decay process lost up to `t = oo` Therefore, a given nucleus may decay at any time after `t = 0`. |
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| 105. |
Staements I: The fission of a heavy nucleus is always accompanied with the neutrons along with two product nuclei. Staements II: For a lighter stble nuclide, the `N/Z` ratio has to be slightly greater than 1. |
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Answer» Correct Answer - a When fission of heavy nucleus takes place,it splits itself into two lighhter nuclei which are having too many neutrons and are highly unstable. To attain stablility, they decay neutrons and hence try to achieve `N//Z` ratio somewhat greater than 1. |
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| 106. |
Staements I: In alpha decay of different radioactive nuclides, the energy of alpha particles has been compared. It is found that as the energy of alpha particle increases the half-life of the decay goes on decreasing. Staements II: More is the energy in any decay process, more is the probability of decaying the nuclide which leads to faster rate of decay. |
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Answer» Correct Answer - a Staement II is correctly explaining Staement I. More than probability of decay means faster decay process and hence shorter half-life. |
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| 107. |
`alpha, beta` and `gamma` radiations come out of a radioactive substanceA. spontaneouslyB. when it is put in a reactorC. when it is heatedD. under pressure |
| Answer» Correct Answer - A | |
| 108. |
In which of the following decays the element does not change?A. `alpha`-decayB. `beta^(+)`-decayC. `beta^(-)`-decayD. `gamma-`decay |
| Answer» Correct Answer - D | |
| 109. |
Two radioactive materials have decay constant `5lambda&lambda`. If initially they have same no. of nuclei. Find time when ratio of nuclei become `((1)/(e))^(2)` : |
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Answer» Correct Answer - D `N = N_(0)e^(-lambdat)` `N_(A) = N_(0)e^(-5lambdat)` `N_(B) = N_(0)e^(-lambdat)` `(N_(A))/(N_(B)) = e^(-4lambdat) rArr ((1)/(e))^(2) = ((1)/(e))^(4lambdat)` `4lambdat = 2 rArr t = (1)/(2)lambda` |
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| 110. |
Choose the correct statements from the following:A. Like other light nuclei, the .`_2He^4` nuclei also have a low value of the binding energy per nucleon .B. The binding energy per nucleon decreases for nuclei with small as well as large atomic number.C. The energy requuired to remove one neutron from `._3Li^7` to transform it into the isotpes `._3Li^6` is` 5.6 MeV`, which is the same as the binding energy per nucleon of `._3Li^6`D. When two deuterium ncueli fuse together, they give rise to a tritrium nucleus accompained by a release of energy. |
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Answer» Correct Answer - d,c Statement (A) is in incorrect. The `_(2)He^(4)` nucleus (or the `alpha`-particles) is exceptionally stable and has a much higher value of BE per nucleon than that for most other light nuclei. Statement (b) is correct but the reason of decrease in binding energy is different for the caese of smaller and larger values of A. The reason for the decreases in the BE per nucleon for nuclei with large A is that with an increases in the number of protons, the Coulomb repulsion increase. On the other hand, the decrease in the BE per nucleon for nuclei with small A is due to a surface effect: the nucleons at the surface being less strongly bound than those in the interior. Statement (c ) is also correct. The energy required to remove one neutron (i.e., one nucleon) is the same as the binding energy per nucleon for a given isotopes. Statement (d) is incorrect. To ensure both charge and mass number conservaion, a proton must be produced as a by-product of the reaction: `._1D^2+._1D^2 rarr ._1T^3+._1P^1+Q`. |
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| 111. |
Statement I:Heavy nuclides tend to have more number of neutrons than protons. Staements II: In hevay nuclei, as there is coloumbic repulsion between protons, so excess of neutrons are preferable: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 falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - A (a) Assertion is true, reason is true and reason is a correct explanation for assertion. |
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| 112. |
In a radioactive decay, neither the atomic number nor the mass number changes. Which of the following particles is emitted in the decay?A. protonB. neutronC. electornD. photon |
| Answer» Correct Answer - D | |
| 113. |
It has been found that nuclides with `2,8,20,50,82,` and `126` protons or neutrons are exceptionally stable. These numbers are refferd to as the magic numbers and their existance has led to.A. the idea of periodictiy in nuclear properties similar to the periodicity of chemical elements in periodic tableB. the so-called 'liquid drop model of the nucleus'C. the so-called 'shell model of the nuclus'D. have a conveninet exploanation of 'nuclear fission' |
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Answer» Correct Answer - a,c The idea of magic number has led to the shell model and the nuclides with these number of protons or neutrons have been compared with the inert gases vis-à-vis-à-vis stabitity in terms of clossed shells. |
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| 114. |
When a nucleus with atomic number `Z` and mass number `A` undergoes a radioactive decay process, (i) Both `Z` and `A` will decrease, if the process is `alpha` decay (ii) `Z` will decrease but `A` will not change, if the process is `beta^(+)-`decay (iii) `Z` will increase but `A` will not change, if the process is `beta-`decay (iv) `Z` and `a` will remain uncharged, if the prices is `gamma` decayA. `(i),(ii),(iii)`B. `(ii),(iii),(iv)`C. `(i),(ii),(iv)`D. all |
| Answer» Correct Answer - D | |
| 115. |
Binding energy per nucleon of fixed nucleus `X^(A)` is `6 MeV`. It absorbs a neutron moving with `KE= 2 MeV` and converts into Y at ground state, emitting a photon of energy `1 MeV`. The Binding energy per nucleon of Y (in MeV) is -A. `((6A+1))/((A+1))`B. `((6A-1))/((A+1))`C. `7`D. `(7)/(6)` |
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Answer» Correct Answer - B `BE` of `X=6A` `BE` of `Y=6A-2+1=6A-1` [Because absorption of energy decreases `BE` and releas of energy increases `BE`] `:. (BE)/("nucleon")=(6A-1)/(A+1)` |
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| 116. |
The mass defect for the nucleus of helium is `0.0303` a.m.u. What is the binding energy per nucleon for helium in `MeV` ?A. 28B. 7C. 4D. 1 |
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Answer» Correct Answer - B (b) `("Binding energy")/("Nucleon") = (0.0303 xx 931)/(4) ~~ 7`. |
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| 117. |
The sequence of decay of radioactive nucleus is `D overset alphato D_1oversetbeta toD_2 oversetalpha toD_3`. If nucleon number and atomic number of `D_2` are 176 and 71 respectively, what are their values for `D` and `D_3`? |
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Answer» As mass number of each `alpha` particles is `4` units and its charge number is `2` units, therefore, for `D_(4)A=176-8=168 Z=71-4=67` Now charge no. of `beta` is `-1` and its mass number is zero, therefore, for `D` `A=176+0+4=180 Z=71-1+2=72` |
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| 118. |
One microgram of matter converted into energy will give.A. `90 J`B. `9 xx 10^3 J`C. `9 xx 10^10 J`D. `9 xx 10^5 J` |
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Answer» Correct Answer - C ( c) `E = Delta mc^2 = 10^-6 xx (3 xx 10^8) = 9 xx 10^10 J`. |
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| 119. |
An antomic Power station has a generating capacity of `200 MW`. The energy generated in a day by this station is.A. 200 M JB. 200 JC. `4800 xx 10^6 J`D. `1728 xx 10^10 J` |
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Answer» Correct Answer - D (d) Energy/day `= 200 xx 10^6 xx 24 xx 3600` =`2 xx 2.4 xx 3.6 xx 10^12 = 1728 xx 10^10 J`. |
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| 120. |
In the reaction `._1^2 H + ._1^3 H rarr ._2^4 He + ._0^1 n`, if the binding energies of `._1^2 H, ._1^3 H` and `._2^4 He` are respectively `a,b` and `c` (in MeV), then the energy (in MeV) released in this reaction is.A. `c + a - b`B. `c - a - b`C. `a + b + c`D. `a + b - c` |
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Answer» Correct Answer - B (b) During fusion binding energy of daughter nucleus is always greater than the total energy of the parent nuclei so energy released `= c - (a + b) = c - a - b`. |
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| 121. |
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 |
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| 122. |
The activity of a sample of radioactive material `A_(1)` at time `t_(1)` and `A_(2)` at time `t_(2)(t_(2)gtt_(1))`. Its mean life is `T`.A. `A_(1)t_(1)=A_(2)t_(2)`B. `(A_(t)-A_(2))/(t_(2)-t_(1))="constant"`C. `A_(2)=A_(1)e^(((t_(1)-t_(2)))/(T))`D. `A_(2)=A_(1)e^((t_(1)//t_(2)//T)` |
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Answer» Correct Answer - C `A_(1)=A_(0)e^(-lambdat_(1))` and `A_(2)=A_(0)e^(-lambdat)` Dividing both, we get `lambda` and mean life is `(1)/(lambda)` |
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| 123. |
A radioactive sample has initial concentration `N_(0)` of nuclei. Then,A. The number of undecayed nuclei present in the sample decays exponentially with timeB. The activity (R) of the sample at any instant is directly proportional to the number of undecayed nuclei present in the sample at that timeC. The number of decayed nuclei grows linearly with timeD. The number of decayed nuclei grows exponentially with time |
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Answer» Correct Answer - A, B, D |
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| 124. |
Consider the situation of the previous problem.suppose the production of the radioactive isotope starts at t=0. Find the number of active nuclei at time t. |
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Answer» Correct Answer - `[(R)/(lambda) (1-e^(-lambda t))]` |
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| 125. |
The activity of a sample of radioactive material `A_(1)` at time `t_(1)` and `A_(2)` at time `t_(2)(t_(2)gtt_(1))`. Its mean life is `T`.A. `A_(1)t_(1)-A_(2)t_(2)`B. `(A_(1)-A_(2))/tau`C. `(A_(1)-A_(2))(t_(2)-t_(1))`D. `(A_(1)-A_(2))tau` |
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Answer» Correct Answer - D |
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| 126. |
The half life of a radioactive material is 12.7 hr. What fraction of the original active material would become inactive in 63.5 hr.A. `1//32`B. `1//23`C. `31//32`D. `23//32` |
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Answer» Correct Answer - C |
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| 127. |
For a certain radioactive substance, it is observed that after `4h`, only `6.25%` of the original sample is left undeacyed. It follows that.A. the half-life of the sample is `1h`B. the mean life of the sample is `(1)/(1n2)h`C. the decay constant of the sample is `1n(2) h^(-1)`D. after a further `4h`, the amount of the substance left over would by only 0.39% of the original amount |
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Answer» Correct Answer - a,b,c,d We have, `6.25%=(6.25)/(100)=(1)/(16)` The given time of `4h` thus equals `4 half-life is `1 h. Since half-life `=(1n 2)/(decay constant) and mean life =(1)/(decay constant)`, after further `4 h`, the amount left over would be `(1)/(2^(4)) xx 1/2^(4)`, i.e., `(1)/(256)` or `(100)/(256)` or `39%` of original amount., |
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| 128. |
The half life of a radioactive substance is `T_(0)`. At `t=0`,the number of active nuclei are `N_(0)`. Select the correct alternative.A. The number of nuclei decayed in time internal `0-t` is `N_(0)e^(-lambda t)`B. The number of nuclei decays in time interval `0-t` is `N_(0) (1-e^(-lambda t))`C. The probability that a radioactive nuclei does not decay in interval `0-t` is `e^(-lambdat)`D. The probability that a radioactive nuclei does not decay in interval `1-e^(-lambdat)` |
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Answer» Correct Answer - B, C |
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| 129. |
The activity of a radioactive sample decreases to one tenth of the original activity `A_(0)` in a period of one year further After further 9 years, its activity will be :A. `A_(0)/100`B. `A_(0)/90`C. `A_(0)/10^(10)`D. None of these |
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Answer» Correct Answer - C |
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| 130. |
Choose the correct alternativeA. `K_(alpha)` wavelength emitted by an atom of atomic number `Z=21` is `lambda` then `K_(alpha)` wavelength emitted by an atom of atomic number `Z=31` is `(4 lambda)/(9)`B. Half life of radioactive substance is `5` year, Probability that a nucleus decays in `10` years is `3//4`C. Mass number of a nucleus is always greater than its atomic numberD. Gamma rays are emitted due to nuclear deexciation |
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Answer» Correct Answer - A::B::D `(lambda_(1))/(lambda_(2))=(E_(2))/(E_(1))=((Z_(2)-a)^(2))/((Z_(1)-1)^(2))` |
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| 131. |
For a certain radioactive substance, it is observed that after `4h`, only `6.25%` of the original sample is left undeacyed. It follows that.A. the half life of the sample is `1` hourB. the mean life to the sample is `1//ln 2` hourC. the decay constant of the sample is `ln2 hour^(-1)`D. after a further `4` hours, the amount of the substance left over would be only `0.39%` of the original amount |
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Answer» Correct Answer - A::B::C::D 6.25% means, `N=(N_(0))/(16),` that means, `t=4T_(1//2)` |
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| 132. |
A nucleus `._Z^AX ` emits an `alpha`-particel. The resultant nucleus emits a `beta^(+)` particle. The respective atomic and mass numbers of the final nucleus will beA. `Z - 3, A - 4`B. `Z - 1, A - 4`C. `Z - 2, A - 4`D. `Z, A - 2` |
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Answer» Correct Answer - A (a) `._Z^A X rarr ._(Z-2)^(A-4)Y + ._2^4 He rarr ._(Z-3)^(A-4) Z + ._1^0 B`. |
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| 133. |
A nucleus of an element `._84 X^202` emits an `alpha-`particle first a `beta-`particle next and then a gamma photon. The final nucleus formed has an atomic numberA. 200B. 199C. 83D. 198 |
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Answer» Correct Answer - C ( c) `._84 X^202 overset (alpha -decay)rarr ._82Y^198 + ._2 He^4` and `._82Y^198 overset (beta-decay)rarr ._83Z^198 + ._-1 beta^0`. |
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| 134. |
If `M` is the atomic mass and `A` is the mass number, packing fraction is given by.A. `(A)/(M - A)`B. `(A - M)/(A)`C. `(M)/(M - A)`D. `(M - A)/(A)` |
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Answer» Correct Answer - D (d) Paking fraction `(M - A)/(A)`. |
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| 135. |
Assertion : Though light of a single frequency (monochromatic) is incident on a metal , the energies of emitted photoelectrons are different. Reason : The energy of electrons emitted from inside the metal surface is lost in collision with the other atoms in the metal.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 falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - A (a) When a light of single frequency falls on the `e^-` of inner layer of metal, then this `e^-` comes out of the metal surface after a large number of collisions with atoms of its upper layer. |
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| 136. |
The most common kind of iron nucleus has a mass number of `56`. Find the radius, approximate mass, and approxiamte density of the nucleus. |
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Answer» We use two key ideas. The radius and mass of a nucleus depend on the mass number A and density is mass divided by volume. We use Eq.(i) to determine the radius of the nucleus. The mass of the nucleus in aatomic mass units is approximately equal to the mass number. The radius is `R = R_(0) A^(1//3)` ` =(1.2 xx 1.0^(-15) m) (56)^(1//3)` `=4.6 xx 10^(-15) m =4.6 fm` Since `A =56`, the mass of the nucleus is approximately 56 u, or ` m~~(56)(1.66 xx 10^(-27) kg) =9.3 xx 10^(-26) kg`. The volume is `nu =(4)/(3) pi R^(3) =(4)/(3) xx 3.14 xx(4.6 xx10^(-15))^(3)` `~~ 4.1 xx 10^(-13) m^(3)` And the density `rho` is approxiimately `rho=(m)/(V)~~(9.3 xx 10^(-26) kg)/(4.1 xx 10^(-43) m^(3)) =2.3 xx 10^(27) kg m^(-3)` The density of solid iron `700 kg m^(3)`, So we see that the iron nucleus is more than `10^(13)` times as dense as the bulk material. Densities of this magnitude are also found in neutron stars, which are similar to gigantic nuclei made almost entirely of neutrons. A `1 cm` cube of material with this density would have a mass of `2.3 xx 10^(11) kg` or `230` million metric tons. |
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| 137. |
A sample of a radioactive element has a mass of `10 g` at an instant `t=0`. The approxiamte mass of this element in the sample after two mean lives is .A. `1.35 g`B. `2.50 g`C. `3.70 g`D. `6.30 g` |
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Answer» Correct Answer - a Let `t=0, M_(0) =10g` `t=2 tau 2((1)/(lambda))` Then, `M=M_(0) e^(- lambdat) =10e^(-lambda)((2)/(lambda)) =10((1)/(e))^(2)1.35g `. |
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| 138. |
The half life of a radioactive substance is `20` minutes . The approximate time interval `(t_(2) - t_(1))` between the time `t_(2)` when `(2)/(3)` of it had decayed and time `t_(1)` when `(1)/(3)` of it had decay isA. `7 min`B. `14 min`C. `20 min`D. `28 min` |
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Answer» Correct Answer - C `(2)/(3)N_(0)=N_(0)e^(-lambdat_(1))` `(1)/(3)N_(0)=N_(0)e^(-lambdat_(2))` `2=e^(lambda(t_(2)-t_(1)))` `lambda(t_(2)-t_(1))= l n 2` `(t_(2)-t_(1))= ( l n 2)/(lambda)=20 "min"` |
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| 139. |
The half life of a radioactive substance is `20` minutes . The approximate time interval `(t_(2) - t_(1))` between the time `t_(2)` when `(2)/(3)` of it had decayed and time `t_(1)` when `(1)/(3)` of it had decay isA. `14 min`B. `20 min`C. `28 min`D. `7 min` |
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Answer» Correct Answer - B At `t_(1) (2)/(3)=(1)/(2^(t_(1)//20))` At `t_(2)=(1)/(3)=(1)/(2^(t_(2)//20))` `t_(2)-t_(1)=20 "mins"` |
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| 140. |
A radioactive element decays by `beta-emission`. A detector records `n` beta particles in `2 s` and in next `2 s` it records `0.75 n` beta particles. Find mean life correct to nearest whole number. Given ln `|2| = 0.6931`, ln `|3|=1.0986`. |
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Answer» Correct Answer - `6.954 sec` Let `n_(0)` be the number of radioactive nuclei at time `t=0`. Number of nuclei decayed in time `t` are given `n_(0)(1-e^(-2lambda))`, which is also equal to the number of beta particles emitted the same interval of time. For the given condition, `n=n_(0)(1-e^(-2lambda))`......(i) `(n+0.75n)=n_(0)(1-e^(-4lambda))`.......(ii) Dividing (ii) by (i) we get `1.75=(1-e^(-4lambda))/(1-e^(-2lambda))` or `1.75-1.75e^(-2lambda)=1-e^(-4lambda)` `:. 1.75e^(-2lambda)=(3)/(4)`.......(iii) Let us take `e^(-2lambda)=x` Then the above equation is, `x^(2)-1.75x+0.75=0` or `x=(1//75sqrt((1.75)^(2)-(4)(0.75)))/(2)` or `x=1` and `(3)/(4)` `:.` From equation (iii) either `e^(-2lambda)=1 or e^(-2lambda)=(3)/(4)` but `e^(-2lambda)=1` is not accepted because which means `lambda=0`. Hence `e^(-2lambda)=(3)/(4) or " " -2lambda ln (e )= l n (3) - l n(4) = l n (3)-2 l n(2):. " "lambda= l n (2)-(1)/(2) l n(3)` Substituting the given values, `lambda=0.6931-(1)/(2)xx(1.0986)=0.14395s^(-)` `:.` Mean life `t_("means")=(1)/(lambda)=6.947 sec` |
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| 141. |
If the nuclear radius of `.^27 A1` is `3.6` Fermi, the approximate nuclear radius of `64 Cu` in Fermi is :A. 3.6B. 2.4C. 1.2D. 4.8 |
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Answer» Correct Answer - D (d) Nuclear radius `r prop A^(1//3)`, where `A` is mass number `r = r_0(27)^(1//3) = 3 r_0 rArr r_0 = (3.6)/(3) = 1.2 fm` For `.^64 Cu, r = r_0 A^(1//3) = 1.2 fm (64)^(1//3) = 4.8 fm`. |
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| 142. |
At a given instant, there are `25%` undecayed radio-active nuclei in a sample. After `8 sec`, the number of undecayed nuclei reduced to `12.5%`. The time after which the number of undecayed unclei will further reduce to `6.25%` of the reduced number is ______sec |
| Answer» To reduced by half number the time required`=` one half life`=8 sec` | |
| 143. |
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. `3 N_0`B. `(9 N_0)/(2)`C. `(5 N_0)/(2)`D. `2 N_0` |
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Answer» Correct Answer - B (b) Initially `p rarr 4 N_0` `Q rarr N_0` Half-life `T_p rarr 1 min` `T_Q rarr 2 min` Let after time `t` number of nuclei of `P` and `Q` are equal, i.e., `(4 N_0)/(2^(t//1)) = (N_0)/(2^(t//2))` `4 = 2^(t//2)` `2^2 = 2^(t//2) rArr (t)/(2) = 2` `t = 4 min` Deactivate nucleus or nuclei of `R` =`(4 N_0 - (4 N_0)/(2^4)) + (N_0 - (N_0)/(2^2))` =`4 N_0 (N_0)/(4) + N_0 - (N_0)/(4) = 5 N_0 - (N_0)/(2) = (9)/(2) N_0`. |
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| 144. |
A `5 xx 10^(-4) Å` photon produces an electron-positron pair in the vincinity of a heavy nucleus. Rest energy of electron is `0.511 MeV`. sIf they have the same kinetic energies, the energy of each paricles is nearlyA. `1.2 MeV`B. `12 MeV`C. `120MeV`D. `1200MeV` |
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Answer» Correct Answer - b If the kinetic energy of each particle is k, then `2+2(0.511 MeV)=(hc)/(lambda)=(12.4xx10^(-3) MeV Å)/(5 xx 10^(-4)Å) =24.8 MeV` `rArr k=(24.8 -1.022)/(2)=11.9 MeV` . |
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| 145. |
A freshly prepared radioactive source of half-life `2 h` emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source isA. `6 h`B. `12 h`C. `24 h`D. `128 h` |
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Answer» Correct Answer - B `64=2^(6)` After `6` half lives activity will become `=(1)/(64)` Hence required time `=6xx2h=12h` |
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| 146. |
A freshly prepared radioactive source of half-life `2 h` emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source isA. 6 hoursB. 12 hoursC. 24 hoursD. 128 hours |
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Answer» Correct Answer - B (b) `(N)/(N_0) = ((1)/(2))^n rArr (1)/(64) = ((1)/(2))^6` =`((1)/(2))^n rArr n = 6` after `6` half lives intensity emitted will be safe. `:.` Total time taken `= 6 xx 2 = 12 hrs`. |
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| 147. |
A freshly prepared radioactive source of half-life `2 h` emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source isA. `6h`B. `12 h`C. `24 h`D. `128 h` |
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Answer» Correct Answer - b `(A)/(A_(0)) =(N)/(N_(0))` Let safe level avtivity be `A`. Initial activity =`64 A`. Hence, `(N)/(N_(0)) =(A)/(A_(0))=(A)/(64A)=(1)/(64)` or `((1)/(2))^(n)=(1)/(64)` or `n=6` Hence, `(t)/(T) =n=6` `becauseT=2h` `:. t= 12 h`. |
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| 148. |
A freshly prepared radioactive source of half-life `2 h` emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source isA. `6 h`B. `12 h`C. `24 h`D. `28 h` |
| Answer» Correct Answer - b | |
| 149. |
An explosion of atomic bomb releases an energy of `7.6xx10^(13)J`. If `200 MeV` energy is released on fission of one `.^(235)U` atom calculate (i) the number of uranium atoms undergoing fission. (ii) the mass of uranium used in the atom bomb |
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Answer» `E=7.6xx10^(13)J` , Energy released per fission`=200 MeV` `=200xx10^(6)xx1.6xx10^(-19)=3.2xx10^(-11)J` `"Number of uranium atoms"(n)=("Total energy")/("Energy per fission")` `n=(7.6xx10^(13))/(3.2xx10^(-11))=2.375xx10^(24)"atoms"` Avagadro number `(N)=6.023xx10^(23)"atoms"` Mass of uranium `=` `(nxx235)/(N)=(2.375xx10^(24)xx235)/(6.023xx10^(23))=92.66g` |
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| 150. |
The percentage of mass lost during nuclear fusion isA. `0.1%`B. `0.4%`C. `0.5%`D. `0.65%` |
| Answer» Correct Answer - D | |