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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.
| 51. |
An electron in hydrogen atom is in the n=4 energy level. When it makes a transition to some lower energy level, to which series the wavelength of emitted photon belong? |
| Answer» When transition is form n=4 to n=1, wavelength of photon emitted belongs to Lyman series. When transition is form n=4 to n=2, wavelength of photon emitted belongs to Balmer series. And when transition is form n=4 to n=3, wavelength of photon emitted belongs to Paschen series. | |
| 52. |
Why did Thomson atom model fail? |
| Answer» This model could not explain scattering of `alpha`- particle through large angles. | |
| 53. |
Why is that mass of nucleus does not enter the formula for impact parameter but its charge does? |
| Answer» The scattering of `alpha`-particle occurs due to electrostatic field of the nucleus. That is why charge on nucleus enters the expression for impact parameter and not its mass. | |
| 54. |
Calculate the impact parameter of a 5MeV alpha particle scatterd by `10^@` when it approaches a gold nucleus. Take Z=79 for gold. |
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Answer» Here, `K.E. =1/2 mv^(2)=5MeV` `5xx1.6xx10^(-13)J` , `theta=10^@, Z=79,b=?` As `b=(Ze^(2) cot theta//2)/(4pi in_0(1/2 mv^(2)))` `:. b=(9xx10^(9)xx79(1.6 xx10^(-19))^(2) cot 5^@)/(5xx1.6xx10^(-13))` `(tan 5^@ =0.0875)` `=(9xx79xx1.6xx1.6xx10^(-16))/(8xx0.0875)=2.6 xx 10^(-13) m` |
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| 55. |
What is the value of Rydberg constant? |
| Answer» Correct Answer - `R=1.097xx10^7m^-1` | |
| 56. |
Can a hydrogen atom absorb a photon whose energy exceeds its binding energy? |
| Answer» Correct Answer - Yes, it can absorb. | |
| 57. |
In Rutherford scattering experiment, if a proton is taken instead of an alpha particle, then for same distance of closest approach, how much K.E. in comparison to K.E. of a particle will be required? |
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Answer» At the distance of closest approach `(r_0)`. `KE_(alpha)=((Ze^2)(2e))/(4pi in_0r_0) and KE_(p)=((Ze^2)(e))/(4pi in_0r_0)` Clearly, `KE_P=1/2 KE_(alpha)`. |
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| 58. |
What is the impact parameter for scattering of `alpha`-particle by `180^(@)`? |
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Answer» Zero. This follows form the relation `b=(Ze^2 cot theta//2)/(4piepsilon_0(1/2mv^2))` |
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| 59. |
An alpha particle of energy 4MeV is scattered through an angle of `180^(@)` by a gold foil (Z=79). Calculate the maximum volume in which positive charge of the atom is likely to be concetrated? |
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Answer» Correct Answer - `7.7xx20^(-40)m^(3)` Maximum volume in which positive charge of atom is likely to be concentrated is `V_(0)=4/3pir_(0)^(3)` Calculate `r_(0)` form the usual formula. |
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| 60. |
Explain distance of closest approach and impact parameter with illustrations. |
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Answer» Distance of closest approach is the distance between the centre of nucleus and the point form which the alpha particle appraoching directely to the nucleus returns. Impact parameter is the perpendicular distance of the velocity vector of the alpha particle form the central line of the nuclues, when the particle is far away form the atom. |
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| 61. |
The diameter of first orbit of hydrogen atom is of the order ofA. `0.5Å`B. `1Å`C. `2Å`D. `4Å` |
| Answer» Correct Answer - B | |
| 62. |
Calculate the impact parameter of a 5MeV alpha particle scattered by `10^@` when it approaches a gold nucleus. Take Z=79 for gold. |
| Answer» Correct Answer - `0.95xx10^(-13)m` | |
| 63. |
Write the relation for (i) the distance of closest approach and (ii) impact parameter. |
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Answer» (i) `r_0=(Ze(2e))/(4pi in_0(1/2 mv^2))` (ii) `b=1/(4pi in_0) (Ze^2cot theta//2)/((1/2 mv^(2)))` |
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| 64. |
Calculate the impact parameter of a 5 MeV particle scattered by `90^(@)`, when it approach a gold nucleus (Z=79). |
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Answer» Correct Answer - `2.27xx10^(-14)m` Here, `b=? KE=5MeV=5xx1.6xx10^(-13)J` `theta=90^(@), Z=79` `b=(Ze^(2)cot theta//2)/(4pi in_(0)(KE))` `=(9xx10^(9)xx79(1.6xx10^(-19))^(2)cot 45^(@))/(5xx1.6xx10^(-13))` `b=2.27xx10^(-14)m` |
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| 65. |
The nucear forcesA. are stronger, being roughly hundred times that of electromagnetic forcesB. have a short range dominant over a distance of about a few fermiC. are central forces, independent of the spin of the nucleonsD. are independent of the nuclear charge. |
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Answer» Correct Answer - A::B::D Nuclear forces are strongest forces, are short range forces, and are independent of the nuclear charge. |
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| 66. |
The correct relation for impact parameter, where symbols have their usual meaning isA. `(Ze^(2)cot theta//2)/((4pi in_(0))E)`B. `(2Ze^(2)cot theta)/(4pi in_(0).E)`C. `(Z^(2)ecot theta//2)/(4pi in_(0)E)`D. `((4pi in_(0))E)/(Ze^(2)cot theta//2)` |
| Answer» Correct Answer - A | |
| 67. |
Which of the following products in a hydrogen atom are independent of the principal quantum number n? The symbols have their usual meaningsA. `vxxr`B. `En`C. `Er`D. `vxxn` |
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Answer» Correct Answer - C::D `v=(2piKe^(2))/(nh) :. vn=(2piKe^(2))/h` `v_(n)` is independent of n. Also `E prop 1/(n^(2)) and r prop n^(2)` `:.` Er is independent of n. |
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| 68. |
Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approx. `10^(-15)m`. |
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Answer» Yes, the nucleus (neutrons and protons) are fundamental particles. To resolve two objects, say nucleus separated by distance d, the wavelength `lambda` of probing signal must be less than or equal to d. As `d=10^(-15)m`, therefore to detect separates parts, if any, insides a nucleon, the electron must have a wavelength `lambdale10^(-15)m`. Now, `lambda=h/p or p=h/lambda` and Kinetic energy, `K=pc=(hc)/lambda=(6.63xx10^(-34)xx3xx10^(8))/(10^(-15))"joule"` `K=(19.89xx10^(-11))/(1.6xx10^(-19))eV ~=10^(9)eV=1GeV` |
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| 69. |
Calculate the energy generated in kWh, when 100g of `._(3)Li(7)`. are converted into `._(2)He(4)` by proton bombardment. Given mass of `._(3)Li(7)=7.0183a.m.u ,` mass of `._(2)He(4)=4.0040a.m.u`, mass of `._(1)H(1)atom=1.0081a.m.u.` |
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Answer» Correct Answer - `6.5475xx10^(6)kWh` Here, E=? The nuclear reaction is `._(3)Li^(7)+._(1)H^(1)to 2 ._(2)He^(4)+Q` Mass defect, `Deltam=7.0183+1.0081-2xx4.0040` `=0.0184 a.m.u.` `Q=0.00184xx931MeV=17.13MeV` Number of nuclei in 100g of `._(3)Li^(7)` `=(6.023xx10^(23))/7xx100` `=8.6xx10^(24)` Total energy released when 100 g of `._(3)Li^(7)` is converted into Helium `=17.13xx8.6xx10^(24)MeV` `=17.13xx8.6xx1.6xx10^(-13)xx10^(24)J` As `1kWh=3.6xx10^(6)J` `:. E=(17.13xx8.6xx1.6xx10^(11))/(3.6xx10^(6))kWh` `E=6.5475xx10^(6)kWh` |
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| 70. |
Which state of the triply ionized `Be^(+++)` has the same orbital radius as that of the ground state of hydrogen? Compare the energies of two states. |
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Answer» Correct Answer - `2;4` Suppose orbital radius of groung state of hydrogen (n=1) is `r_(1)` Radius of nth orbit of hydrogen like atom is `r_(n)=(n^(2))/Z r_(1)` for `Be^(+++), Z=4` As `r_(n)=r_(1)`, therefore, `(n^(2))/4=1 or n=2` Let energy of electrons in ground state of hydrogen `=E_(1)`. Energy of electron in nth state of hydrogen like atom `E=(Z^(2))/(n^(2)) E_(1)=(4^(2))/(2^(2)) E_(1)=4E_(1)` `:. (E_(2))/(E_(1))=4` |
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| 71. |
Calculate radius of first orbit of singly ionized He atom, when radius of first orbit of hydrogen atom is 0.53Å. |
| Answer» for helium atom, `Z=2:. r_(He)=(r_H)/Z=(0.53Å)/2=0.265Å` | |
| 72. |
Calculate the half life periode of a radioactive substance if its activity drops to `1/16` th of its initial value of 30 yr. |
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Answer» Correct Answer - `7.5 yr` `T=?, N/(N_(0))=1/16, t=30years` As `N/(N_(0))=(1/2)^(n)=1/16=(1/2)^(4)` `:. n=4=t/T,T=t/4=30/4=7.5 years` |
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| 73. |
A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5MeV per nucleon. Calculate the energy Q released per fission in MeV. |
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Answer» The nuclear reaction is `X^(240)toY^(110)+Z^(130)+Q` As per the given data, `Q=110xx8.5+130xx8.5-240xx7.6` `=240(0.9)MeV=216MeV` Energy released /fission `=216MeV` |
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| 74. |
Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half life of 1 yr. After 1 yr.A. all the containers will have 5000 atoms of the materialB. all the containers will contain the same number of atoms of material but that number will only be approx. 5000C. the containers will in general have different numbers of the atoms of the material but their average will be close to 5000D. none of the containers can have more than 5000 atoms |
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Answer» Correct Answer - C Radioactivity is spontaneous self disruptive activity of the radioactive material. In t=1yr =half life of material , on the average, half the number of atoms will decay. Therefore, the containers will in general have different number of the matarials, but their average will be close to 5000. Choice (c) is correct. |
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| 75. |
(a) What are thermal neutrons in the context of nuclear fission ? (b) What role does a moderator play in a nuclear reactor ? |
| Answer» Thermal neutrons are low energy neutrons `~~1/40eV`. Neutrons are considered as ideal particles for nuclear fission, because they are neutral particles on which neither any force of attraction nor any force of repulsion acts. | |
| 76. |
Why is nuclear density same for all nuclei? |
| Answer» This is because density `= ("mass")/("volume")`, and volume of nucleus varies directly as its mass number. | |
| 77. |
Statement-1: `1am u.=933MeV` Statement-2: It follows form `E=mc^(2)`A. Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, but Statement-2 is falseD. Statement-1 is false, but Statement-2 is true. |
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Answer» Correct Answer - B Both the statement are true, but the explanation is inadequate. Infact, `1a.m.u.=1.66xx10^(-27)kg` form `E=mc^(2)=1.66xx10^(-27)(3xx10^(8))^(2) "joule"=(1.66xx9xx10^(-11))/(1.6xx10^(-3)) MeV~~933MeV` |
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| 78. |
The above is a plate of binding energy per nucleon `E_(0)` against the nuclear mass `M, A, B,C,D,E,F` correspond to different nuclei Consider four reactions:A. (i) and (iii)B. (ii) and (iv)C. (ii) and (iii)D. (i) and (iv) |
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Answer» Correct Answer - D Energy `(in)` is released only when ligther nuclei splits into ligther nuclei such as in (i) `A+BtoC+in`. Again, energy is released when a heavy nucleus splits into ligther nuclei such as in (iv) `FtoD+E+in` `:.` choice (d) is correct. |
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| 79. |
Statement-1: Density of nuclear matter is same for all nuclei Statement-2: Density has nothing to do with mass and size of nucleus.A. Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, but Statement-2 is falseD. Statement-1 is false, but Statement-2 is true. |
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Answer» Correct Answer - C Statement-1 is true, but the statement-2 is false. Infacts as `R propA^(1//3)`, volume `prop A`. That is why dependences of density (mass/volume) on A vanishes. |
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| 80. |
In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose becauseA. they will break upB. elastic collision of neutrons with heavy nuclei will not slow them downC. the net weight of the reactor would be unbearbly highD. substances with heavy nuclei do not occure in liquid or gaseous state at room temperature |
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Answer» Correct Answer - B The moderators used have light nuclei (like proton). When protons undergo perfectely elestic collision with the neutrons emitted, their velocities are exchanged, i.e., neutrons come to rest and protons move with the velocity of neutrons. Heavy nuclie will not serve the purpose because elestic collision of neutrons with heavy nuclie will not slow them down. Choice (b) is correct. |
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| 81. |
What do you understand by the fact that the binding energy of helium nucleus is 28.17MeV? |
| Answer» It means 28.17 MeV energy is required to separate 2 neutrons and 2 protons of the helium nucleus to finite distance apart. | |
| 82. |
A mixture consists of two radioactive materials `A_1` and `A_2` with half-lives of `20 s` and `10 s` respectively. Initially the mixture has `40 g` of `A_1` and `160 g` of `a_2`. The amount the two in the mixture will become equal afterA. `60s`B. `80s`C. `20s`D. `40s` |
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Answer» Correct Answer - D Here, `T_(1)=20s, T_(2)=10s, N_(01)=40g`, `N_(02)=160g`. Let the amount of the two radioactive materials become equal after t sec. form `N/(N_(0))=(1/2)^(n)=(1/2)^(t//T)`, we get `N_(1)=N_(01)(1/2)^(t//20)=40(1/2)^(t//20)` `N_(2)=N_(02)(1/2)^(t//10)=160(1/2)^(t//10)` As `N_(1)=N_(2) :. 40(1/2)^(t//20)=160(1/2)^(t//10)` or `(1/2)^(t//20)=4(1/2)^(t//20)=(1/2)^(t/10-2)` `:. t/20=t/10-2 or t=40s` |
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| 83. |
Two nuclei have mass number in the ratio 1:8. What is the ratio of their nuclear radii? |
| Answer» `(R_1)/(R_2)=((A_1)/(A_2))^(1//3)=(1/8)^(1//3)=1/2` | |
| 84. |
Biologically useful technetium nuclei (with atomic weight 99) have a half life of 6 hrs. A solution containing `10^(-12)` g of this is injected into the bladder of a patient. Find its activity in the beginning and after one hour. |
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Answer» Correct Answer - `7.026xx10^(8)hr^(-1); 6.26xx10^(8)hr^(-1)` Here, `T=6hrs., lambda=0.693/T=0.693/6 hr^(-1)` No. of atoms in `10^(-12)g` at t=0 `N_(0)=(6.023xx10^(23))/99xx10^(-12)=6.08xx10^(9)` Activity of the sample at t=0 `((dN)/(dt))_(0)=lambda N_(0)=0.693/6xx6.08xx10^(9)hr^(-1)` `=7.026xx10^(8)hr^(-1)` form `N/(N_(0))=(1/2)^(t//T)=(1/2)^(1//6)` calculate `N=5.423xx10^(9)` `:. ((dN)/(dt))_(t)=lambdaN=0.693/6xx5.423xx10^(9)` `=6.26xx10^(8)hr^(-1)` |
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| 85. |
Assuming the nuclei to be spherical in shape, how does the surface area of a nucleus of mass number `A_1` compare with that of a nucleus of mass number `A_2`? |
| Answer» `((A_1)/(A_2))=((R_1)/(R_2))^2=[((A_1)/(A_2))^(1//3)]^2=((A_1)/(A_2))^(2//3)` | |
| 86. |
The radioactive decay rate of a radioactive element is found to be `10^(3) ` disintegration per second at a cartain time . If the half life of the element is one second , the dacay rate after one second ….. And after three second is …… |
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Answer» It is known that radioactive decay rate is directly proportional to the number of nuclei left. `N_(0)` corresponds to `10^(3)` (disintegrations/sec). As half life T=1sec. , therefore (i) no. of half lives in 1 sec., `n=1` As `N=N_(0)(1/2)^(n) :. N=1000(1/2)^(1)=500 :.` No. of disintegrations/sec. after one sec. `=500` (ii) no,. of half lives in `3 sec. n=3/1=3` As `N=N_(0)(1/2)^(n) :. N=1000(1/2)^(3)=1000/8=125` `:.` No. of disintegration/sec. after three sec.`=125` |
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| 87. |
Two different radioactive elements with half lives `T_1 and T_2` have `N_1 and N_2` (undecayed) atoms respectively present at a given instant. Determine the ratio of their activities at this instant. |
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Answer» We know `A=(dN)/(dt)=lambdaN=(0.693N)/T` `:. A_1=(0.693N_1)/(T_1) and A_2=(0.693N_2)/(T_2) ` `(A_1)/(A_2)=(N_1)/(T_1)xx(T_2)/(N_2)=((N_1)/(N_2))((T_2)/(T_1))` |
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| 88. |
Obtain approximately the ratio of nuclear radii of `._26U^(56)` and `._92U^(238)`. What is the approximately ratio of their nuclear densities? |
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Answer» Here, `A_1=56, A_2=238` `(R_1)/(R_2)=((A_1)/(A_2))^(1//3)=(56/238)^(1//3)=(0.235)^(1//3)=0.617` As nuclear density is same for all nuclei, therefore, `(rho_1)/(rho_2)=1` |
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| 89. |
The nuclear radius of `._(8)O^(16)` is `3 xx10^(-15) m`. If an atomic mass unit is `1.67 xx 10^(-27) kg`, then the nuclear density is approximately. |
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Answer» Correct Answer - `2.359xx10^(17)kg//m^(3)` Here, `R=3xx10^(-15)m, rho=?` Mass, `m=16a.m.u =16xx1.66xx10^(-27)kg` `rho=m/V=m/(4/3piR^(3))=(3m)/(4piR^(3))` `=(3xx16xx1.66xx10^(-27))/(4xx3.14(3xx10^(-15))^(3))=2.359xx10^(17)kg//m^(3)` |
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| 90. |
Two nuclei P and Q have equal number of atoms at t=0. Their half lives are 3 hrs and 9 hrs respectively. Compare their rates of disintegration after 18 hours form the starts. |
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Answer» Correct Answer - `3:16` Number of half lives of P in 18 hours `=18/3=6` Number of half lives of Q in 18 hours `=18/9=2` Therefore, no. of nuclei left undecayed: `N_(1)=N_(0)(1/2)^(6)=N_(0)//64` `N_(2)=N_(0)(1/2)^(2)=N_(0)//4` The ratio of their rates of disintegration is `(R_(1))/(R_(2))=(lambda_(1)N_(1))/(lambda_(2)N_(2))=(T_(2))/(T_(1))((N_(1))/(N_(2)))=9/3xx(N_(0)//64)/(N_(0)//4)=3/16` `3:16` |
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| 91. |
There are `4sqrt2xx10^6` radioactive nuclei in a given radioactive sample. If half life of sample is 20s, how many nuclei will decay in 10seconds? |
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Answer» Here, `N_0=4sqrt2xx10^6`, `T=20s, t=10s`, form `N/(N_0)=(1/2)^(t//T)=(1/2)^(10//20)=(1/2)^(1//2)=1/(sqrt2)` `N=(N_0)/(sqrt2)=(4sqrt2)/(sqrt2)xx10^6=4xx10^6` Number of nuclie decayed=`N_0-N` `=4sqrt2xx10^6-4xx10^6` `=4xx10^6(sqrt2-1)` `=4xx0.414xx10^6=1.656xx10^6` |
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| 92. |
Show that the decay rate R of a sample of radionuclide is related to the number of radioactive nuclei N at the same instant by the expression `R=lambdaN`. |
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Answer» According to radioactive decay law. `N=N_0d^(-lambdat)`. Rate of decay, `R=-(dN)/(dt)=(-d)/(dt)(N_0e^(-lambdat))=lambdaN_0e^(-lambdat)=lambdaN`. |
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| 93. |
The natural boron is composed of two isotopes `._5B^10` and `._5B^11` having masses `10.003 u` and `11.009 u` resp. Find the relative abundance of each isotope in the natural boron if atomic mass of natural boron is 10.81u. |
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Answer» Let `._(5)B^(10)` be `x%`. Therefore `._(5)B^(11)` will be `(100-x)%`. As average atomic mass = weighted average of the masses of isotopes `:. 10.81 = (10.003x+11.009(100-x))/(100)` Calculate `x = 19.78%` `:. (100 -x) = 100 - 19.78 = 80.22%` |
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| 94. |
The binding energies of deutron `(._1H^2)` and `alpha`-particle `(._2He^4)` are 1.25 and 7.2 MeV/nucleon respectively. Which nucleus is more stable? |
| Answer» `alpha`-particle `(._2He^4)` is more stable, because a nucleus is more stable when value of binding energy per nucleon is larger. | |
| 95. |
The nucleus of an atom of `._(92)Y^(235)` initially at rest decays by emitting an `alpha` particle. The binding energy per nucleon of parent and dougther nuclei are 7.8MeV and 7.835MeV respectively and that of `alpha` particles is `7.07MeV//"nucleon"`. Assuming the dougther nucleus to be formed in the unexcited state and neglecting its share of energy in the reaction, calculate speed of emitted alpha particle. Take mass of `alpha` particle to be `6.68xx10^(-27)kg`. |
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Answer» Correct Answer - `1.573xx10^(7)m//s` The given reaction is `._(92)Y^(235) to ._(90)X^(231)+._(2)He^(4) +"energy"` Energy released during the decay process is `E=7.835xx231+4xx7.07-7.8xx235` `E=1809.885+28.28-1833.3=5.165MeV` `5.165xx1.6xx10^(-13)J` If v is speed of alpha particle emitted, then `1/2mv^(2)=E` `v=sqrt((2E)/m)` `=sqrt((2xx5.165xx1.6xx10^(-13))/(6.68xx10^(-27)))` `=sqrt(16.528/6.68)xx10^(7)m//s` `v=1.573xx10^(7)m//s` |
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| 96. |
The nucleus of an atom of `._(92)Y^(235)` initially at rest decays by emitting an `alpha` particle. The binding energy per nucleon of parent and daughter nuclei are `7.8MeV` and `7.835MeV` respectively and that of `alpha` particles is `7.07MeV//"nucleon"`. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share of energy in the reaction, calculate speed of emitted alpha particle. Take mass of `alpha` particle to be `6.68xx10^(-27)kg`. |
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Answer» The nuclear reaction is `._(92)Y^(235)to._(90)X^(231)+._2He^(4)+Q` Total B.E. of `._(92)Y^(235)=7.89xx235Mev=1833MeV` Total B.E. of `._(90)X^(231)=7.835xx231Mev=1809.9MeV` B.E. of `._(2)He^(4)=7.07xx4=28.3MeV` `:.` Q=Total BE of products-BE of parent atom=`1809.9+28.3-1833=5.2MeV` `=5.2xx1.6xx10^(-13)J=8.32xx10^(-13)J` This is the energy carried by `alpha` particle. form `1/2mv^(2)=Q=8.32xx10^(-13)` `v=sqrt((8.32xx2xx10^(-13))/m)=sqrt((16.64xx10^(-13))/(6.68xx10^(-27)))=1.58xx10^(7)m//s` |
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| 97. |
Hydrogen `(_(1)H^(1))` Deuterium `(_(1)H^(2))` singly omised helium `(_(1)He^(1))` and doubly ionised lithium `(_(1)Li^(6))^(++)` all have one electron around the nucleus Consider an electron transition from `n = 2 to n = 1` if the wavelength of emitted radiartion are `lambda_(1),lambda_(2), lambda_(3),and lambda_(4)`, repectivelly then approximetely which one of the following is correct ?A. `lambda_(1)= lambda_(2)=4lambda_(3)=9lambda_(4)`B. `lambda_(1)=2 lambda_(2)=3lambda_(3)=4lambda_(4)`C. `4lambda_(1)=2 lambda_(2)=2lambda_(3)=lambda_(4)`D. `lambda_(1)=2 lambda_(2)=2lambda_(3)=lambda_(4)` |
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Answer» Correct Answer - A form `1/lambda=RZ^(2)(1/(1^(2))-1/(2^(2)))=3/4 RZ^(2) :. lambda=4/(3RZ^(2))` `lambda_(1)=4/(2R)` , `lambda_(2)=4/(3R)` , `lambda_(3)=4/(3R(2^(2)))=4/(12R)` , `lambda_(4)=4/(3R(3^(2)))=4/(27R)` , `lambda_(1)=lambda_(2)= 4lambda_(3) =9lambda_(4)`. |
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| 98. |
Hydrogen atom in ground state is excited by a monochromatic radiation of `lambda = 975 Å`. Number of spectral lines in the resulting spectrum emitted will beA. `3`B. `2`C. `6`D. `10` |
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Answer» Correct Answer - C Here, `lambda=975Å=975xx10^(-10)m` `E=(hc)/lambda=(6.63xx10^(-34)xx3xx10^(8))/(975xx10^(-10)) "joule"` `=(6.63xx3xx10^(-16))/(975xx1.6xx10^(-19))eV=12.75 eV` `:. E_(n)=-13.6+12.75=-8.5 eV=-13.6/(n^(2))` Which given n=4, i.e, hydrogen atom will be excited to n=4. Number of spectral lines `=(n(n+1))/2=(4(4-1))/2=6` |
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| 99. |
`._(2)He^(3) and He_(1)^(3)` nuclei have the same mass number. Do they have the same binding energy? |
| Answer» No. `._(2)He^(3)` and `._(1)He^(3)` nuclei have the same mass number, but the binding energy of `._(1)He^(3)` has 1 proton and 2n. The repulsive force between protons is missing in `._(1)He^(3)` . | |
| 100. |
The de - Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature `T ("kelvin")` and mass `m`, isA. `lambda=h/(sqrt(3mKT))`B. `lambda=h/(sqrt(6mKT))`C. `lambda=h/(sqrt(5mKT))`D. `lambda=h/(sqrt(2mKT))` |
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Answer» Correct Answer - A As `1/2mv^(2)=3/2kT :. mv^(2)=3kT` or `m^(2)v^(2)=3mkT` or `mv=sqrt(3mkT)` `lambda=h/(mv)=h/(sqrt(3mkT))` |
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