320 + Interview Questions in NUCLEI IN GENERAL AWARENESS Page 5 InterviewSolution

201.	If `m,m_n` and `m_p` are masses of `._Z X^A` nucleus, neutron and proton respectively.A. `m=(A-Z) m_p+Zm_p`B. `m=(A-Z)m_p+Zm_n`C. `m lt (A-Z) m_n+Zm_p`D. `m lt (A-Z) m_n+Zm_p`
Answer» Correct Answer - C

Discussion

202.	Why is it found experimentally difficult to detect neutrinos in nuclear β-decay ?
Answer» Neutrinos are neutral (chargeless), (almost) I massless particles., that hardly interact with matter Alternatively, The neutrinos can penetrate large quantity of matter without any interaction.

202.

Why is it found experimentally difficult to detect neutrinos in nuclear β-decay ?

Answer»

Neutrinos are neutral (chargeless), (almost) I massless particles., that hardly interact with matter

Alternatively,

The neutrinos can penetrate large quantity of matter without any interaction.

Discussion

203.	What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE / A) in the range of mass number 'A' lying 30 <A<170 ?
Answer» Saturation / short range nature of nuclear forces.

Discussion

204.	In a typical nuclear reaction e.g.21H + 21H → 32He + n + 3.27 Me V,although number of nucleons is conserved, yet energy is released. How ? Explain
Answer» Since the total initial mass of nuclei on the left side of reaction is.greater than the total,final mass nucleus on the right hand side, this difference of mass appears as the energy released.

Discussion

205.	Find the binding energy of `._(26)^(56)Fe`. Atomic mass of Fe is 55.9349u and that of Hydrogen is 1.00783u and mass of neutron is 1.00876u.
Answer» Mass of hydrogen atom `m_(p)=1.00876u , m_(n)=1.00867u` `Z=26, A = 56` Mass of Iron atom M = 55.9349u (i) Mass defect `Delta M` `= [Zm_(P)+(A-Z)m_(n)-M]` `=[26xx1.00876+(56-26)(1.00867)-55.9349]u` `therefore Delta m = 0.55296u` (ii) BE of nucleus `= Delta MC^(2)` `= Delta M xx 931.5 MeV` `= 0.55296(931.5) MeV` `= 515.08 MeV`

Discussion

206.	Find the energy required to split `._(8)^(16)O` nucleus into four `alpha` - particles. The mass of an `alpha`- particle is 4.002603u and that of oxygen is 15.994915u.
Answer» The energy required to split O = [Total mass of the products - Total mass of the reactants] `c^(2)` Mass of four `._(2)^(4)He` - Mass of `._(8)^(16)O xx c^(2)` `=[(4xx4.002603)-15.994915]u xx c^(2)` `= [1.6010412 - 15.994915] u xx c^(2)` `= (0.015497) 931.5 MeV = 14.43 MeV`

Discussion

207.	How much energy is required to separate the typical middle mass nucleus `._(50)^(120)Sn` into its constituent nucleons ? (Mass of `._(50)^(120)Sn=119.902199u`, mass of proton = 1.007825u and mass of neutron = 1.008665u)
Answer» `m_(P)=1.007825u` `m_(n)=1.008665u` For Sn, Z = 50, A = 120, M = 119.902199u (i) Mass defect `Delta m` `= [Zm_(P)+(A-Z)m_(n)-M]u` `= 50(1.007825)+(120-50)[(1.008665) - 119.902199]` `= [50xx1.007825+70xx1.008665 - 119.902199]u` `= [50.39125+70.60655-119.902199]u` `Delta M = [120.9978-119.902199]` `= 1.095601 u` (ii) Energy required to separate the nucleons = B.E of the nucleus `BE=Delta Mc^(2)=Delta M xx931.5 MeV` `= 1.095601xx931.5 MeV` `= 1020.5 MeV`

Discussion

208.	Calculate the binding energy of an `alpha`-particle. Given that mass of proton = 1.0073 u, mass of neutron = 1.0087 u, and mass of `alpha`- particle = 4.0015 u.
Answer» For `._(2)He^(4), A = 4, Z = 2, m_(P)=1.0073u` `m_(n)=1.0087u, m_(n)=4.0015 u` (i) `Delta M` `= [Zm_(P)+(A-Z)m_(n)-M]` `= [2(1.0073)+(402)(1.0087)-4.00260]` `= [2xx1.0073+2xx1.0087-4.00260]` `= (2.0146+2.0174)-4.0015` `Delta M = [4.032-4.0015] = 0.0305 u` (ii) `BE=Delta xx c^(2)=Delta M xx 931.5 MeV` `= 0.0305xx931.5 MeV` `= 0.0305xx931.5` `therefore BE=28.41 MeV`

Discussion

209.	The half-life of radon is 3.8 days. After how many days `19/20` of the sample will decay ?
Answer» Correct Answer - 16.4 days `N=(1-19/20)N_0rArr N/N_0=1/20rArr 1/20=(1/2)^n rArr n="log20"/"log2", t=nxxT_(1//2)`

Discussion

210.	Radon has 3.8 days as its half-life . How much radon will be left out of 15 mg mass after 38 days ?A. 1.05 mgB. 0.015 mgC. 0.231 mgD. 0.50 mg
Answer» Correct Answer - b Here, `n=1/T_(1//2)=38/3.8=10` `N=N_0(1/2)^n=m=m_0(1/2)^n =15 mg xx (1/2)^10` `=15/1024`=0.015 mg

Discussion

211.	Write nuclear reaction equations for `alpha`- decay of `._(88)^(226)Ra`
Answer» `._(88)Ra^(226)rarr ._(86)Rn^(222)+._(2)He^(4)`

Discussion

212.	1 mg redium has `2.68xx10^18` atoms. Its half life is 1620 years. How many radium atoms will disintegrate from 1 mg of pure radium in 3240 years ?A. `2.01xx10^9`B. `2.01xx10^18`C. `1.01xx10^9`D. `1.01xx10^18`
Answer» Correct Answer - b `n=t/T_(1//2)=3240/1620`=2 As `N/N_0=m/m_0=(1/2)^n` Mass of radium left after 2 half lives is `m=m_0(1/2)^n = 1xx(1/2)^2 =1/4` = 0.25 mg Mass of radium disintegrated =1-0.25 =0.75 mg Number of radium atoms disintegrated = 0.75 x 2.68 x `10^18=2.01xx10^18`

Discussion

213.	Find the Q-value and the kinetic energy of the emitted `alpha` - particle in the `alpha` - decay of `._(88)^(226)Ra` Given m `(._(88)^(226)Ra)=226.02540 u`, `m (._(86)^(222)Rn)=222.01750 u`, `m (._(86)^(222)Rn)=220.01137 u, m (._(84)^(216)Po)` `= 216.00189 u`.
Answer» `._(88)Ra^(226)rarr ._(85)Rn^(222)+._(2)He^(4)` Q value `[m(._(88)Ra^(226))-m(._(86)Rn^(222))-m_(alpha)]xx931.5 MeV` `= [226.02540-222.01750-4.00260]xx931.5 MeV` `Q = 0.0053xx931.5 MeV = 4.94 MeV` K.E of a particle `= ((A-4)Q)/(A)=(226-4)/(226)xx4.94 = 4.85 MeV`

Discussion

214.	The kinetic energy of `alpha`-particle emitted in the `alpha`-decay of `""_(88)Ra^(266)` is [given, mass number of Ra = 222 u]A. `5.201` Me VB. `3.301` Me VC. `6.023` Me VD. `4.871` Me V
Answer» Correct Answer - D (d) the nuclear should be `""_(88)Ra^(226) to ""_(86)Rn ^(222)+ alpha ` Energy corresponding mass defect `Q =[{:("(mass number of Ra )"),("-(Mass number of Ru )"),("-m (4He )"):}]Uxx931.5Me V = 4.871MeV `

Discussion

215.	The half - life period of a radioactive substance is 20 days. What is the time taken for `7//8^(th)` of its original mass to disintegrate ?
Answer» Half life period = 20 days In this problem, `(1)/(2^(n))=("Quantity remaining")/("Initial quantity")` `=(1)/(8)=(1)/(2^(3))` `therefore n =3` `therefore` Time taken to disintegrate `= n xx` Half life time `= 3xx20 = 60` days

Discussion

216.	A nucleus `X^232` initially at rest undergoes `alpha` decay, the `alpha`-particle produced in above process is found to move in a circular track of radius 0.22 m in a uniform magnetic field of 1.5 T. Find Q value of reaction.
Answer» Correct Answer - 5.5 MeV

Discussion

217.	A nitrogen nucleus `7^(N^(14))` absorbs a neutron and can transfrom into lithium nucleus `3^(Li^(7))` under suitable conditions, after emitting :A. 4 protons and 3 neutronsB. `1alpha` particle, 4 protons and 2 negative `beta` particleC. 4 protons and 4 neutronsD. 5 protons and 1 negative `beta` particle
Answer» Correct Answer - B::C

Discussion

218.	A certain substance decays to 1/232 of its initial activity in 25 days. Calculate its half-life.
Answer» Fraction of substance decays `= ("Quantity remains")/("Initial quantity")` `=(1)/(2^(n))=(1)/(32)=(1)/(2^(5))` `therefore n = 5` Duration of time = 2 days We know `(n)=("Duration of time")/("Half life time")` `therefore` Half life time `=("Duration of time")/(n)` `(25)/(5)=5` days

Discussion

219.	Bombardment of lithium with protons gives rise to the following reaction : `._(3)^(7)Li + ._(1)^(1)H rarr 2[._(2)^(4)He]+Q`. Find the Q - value of the reaction. The atomi mases of lithium, proton and helium are 7.016u, 1.0084 and 4.004 u respectively.
Answer» Mass of Lithrium = 7.016 u `m_(P)=1.008 u` Mass of Helium = 4.004 u, `u=931.5 MeV` Q = [Total mass of the reactants - Total mass of the products `c^(2)` = [Mass of Lithium `+ m_(P) -` (2 `xx` mass of Helium)] `xx 931.5 MeV` `= [7.016+1.008-2(4.004)]xx931.5 MeV` `=[8.024-8.008]xx931.5 MeV` `therefore` Energy `Q=0.016xx931.5` `= 14.904 MeV`

Discussion

220.	Plutonium decays with a half - life of 24,000 years. If plutonium is stored for 72,000 years, what fraction of it remains ?
Answer» Half life of plutonium, T = 24000 years, Stored time of plutonium, t = 72000 years no. of half lives, `n = (t)/(T)=(72000)/(24000)=3`, Fraction of plutonium remains `=(N)/(N_(0))=(1)/(2^(n))=(1)/(2^(3))=(1)/(8)`

Discussion

221.	The variation of decay rate of two radioactive samples A and B with time is shown in fig. Which of the following statements are true?A. Decay constant of A is greater than that of B, hence A always decays faster than B.B. Decay constant of A is greater than that of B, but it does not always decays faster than B.C. Decay constant of B is smaller than that of A but still its decay rate becomes equal to that of A at a later instant.D. Both (b) and (c )
Answer» Correct Answer - d

Discussion

222.	Give the three types of radioactive decay occur in nature.
Answer» Three types of radioactive decay occur in nature : (i) α-decay in which a helium nucleus ⁴₂He is emitted; (ii) β-decay in which electrons or positrons (particles with the same mass as electrons, but with a charge exactly opposite to that of electron) are emitted; (iii) γ-decay in which high energy (hundreds of keV or more) photons are emitted. Each of these decay will be considered in subsequent sub-sections.

222.

Give the three types of radioactive decay occur in nature.

Answer»

Three types of radioactive decay occur in nature :

(i) α-decay in which a helium nucleus ⁴₂He is emitted;

(ii) β-decay in which electrons or positrons (particles with the same mass as electrons, but with a charge exactly opposite to that of electron) are emitted;

(iii) γ-decay in which high energy (hundreds of keV or more) photons are emitted. Each of these decay will be considered in subsequent sub-sections.

Discussion

223.	In r-ray emission from a nucleus:(A) Both neutron number and proton number change(B) There is no change in neutron and proton number(C) Only the neutron number changes(D) only the proton number changes
Answer» The answer is (B) There is no change in neutron and proton number

Discussion

224.	Nuclear fission can be explained by:(A) proton-proton cycle(B) shell model of nucleus(C) liquid drop model of nucleus(D) independent of nuclear model of nucleus.
Answer» The answer is (C) liquid drop model of nucleus

Discussion

225.	A `.^7Li` target is bombarded with a proton beam current of `10^-4` A for `1` hour to produce `.^7 Be` of activity `1.8xx10^8` disintegrations per second. Assuming that `.^7Be` radioactive nucleus is produced by bombarding `1000` protons, determine its half-life.
Answer» charge , q=It = `10^(-4)xx1xx3600 = 0.36C` number of protons ` =(q)/( E) =(0.36)/(1.6xx10^(-19))=2.25xx10^(18)` number of `""^(7)Be` nuclei `=(2.25x10^(18))/(1000)=2.25xx10^(15)` Activity ,R `=("in 2)/(T_(1//2))N,` where ` R= 2.8xx10^(8)` dps (given ) half - life of substance , `T_(1//2)=("N in"2)/(R) =(2.25xx10^(15)xx0.693)/(2.8xx10^(8))` `=0.556xx10^(7)s`

Discussion

226.	Consider an n-p-n transistor with its base - emitter junction forward biased and collector base junction reverse biased . Which of the following statements are true?A. Electrons crossover from emitter to collectorB. Holes move from base to collectorC. Electrons move form emitter to baseD. Electrons from emitter mive out to base without going to the collector.
Answer» Correct Answer - A::C::D Here emitter-base junction is forward biased i.e., the positive pole of emitter base battery connected to base and its negative pole of emitter, Also, the collector base junction reverse biased, i.e., the positive pole of the collector base battery is connected to collection and negative pole of base. Thus, electron move from emmiter to bae and crossover from emitter to collector.

Discussion

227.	A nucleus has atomic number 11 and mass number 24. State the number of electrons, protons and neutrons in the nucleus
Answer» Correct Answer - 13 Electrons do not exist in the nucleus . Number of protons in nucleus (Z) =11 Number of neutrons in nucleus (N) =A-Z

Discussion

228.	Two stable isotopes of lithium `._3^6Li` and `._3^7Li` have respective abundances of 7.5% and 92.5% . These isotopes have masses 6.0152 u and 7.016004 u respectively. Find the atomic weight of lithiumA. 6.941 uB. 3.321 uC. 2.561 uD. 0.621 u
Answer» Correct Answer - a Atomic weight = average weight of the isotopes `=(6.01512xx7.5xx7.01600xx92.5)/((7.5+92.5))=(45.1134+648.98)/100` =6.941 u

Discussion

229.	Two stable isotopes of lithium `._(3)^(6)Li` and `._(3)^(7)Li` have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and7.01600 u, respectively. Find the atomic mass of lithium.
Answer» Atomic weight = Weight average of the isotopes. `= (6.01512xx7.5+7.01600xx92.5)/((7.5+92.5))` `= (45.1134+648.98)/(100)` `= 6.9414`

Discussion

230.	Two stable isotopes of lithium `._3^6Li` and `._3^7Li` have respective abundances of 7.5% and 92.5% . These isotopes have masses 6.0152 u and 7.016004 u respectively. Find the atomic weight of lithium
Answer» Correct Answer - 6.941 u Average atomic mass =`(p_1m_1+p_2m_2)/(p_1+p_2)`

Discussion

231.	Sample of two radioactive nuclides `A` and `B` are taken. `lambda_(A)` and `lambda_(B)` are the disintergration constants of `A` and `B` respectively. In which of the following cases, the two sample can simultaneously have the same decay rate at any time ?A. Initial rate of decay of A is twice the initial rate of decay of B and `lambda_A=lambda_B`.B. Initial rate of decay of A is twice the initial rate of decay of B and `lambda_Agtlambda_B`.C. Initial rate of decay of B is twice the initial rate of decay of A and `lambda_Agtlambda_B`.D. Initial rate of decay of B is twice the initial rate of decay of A at t=2h and `lambda_B=lambda_A`
Answer» Correct Answer - b

Discussion

232.	Binding energy per nucleons vs mass curve for nucleus is shown in the figure `W,X,Y`and `Z` are four nuclei indicated on the curve . The process that would release energy is A. `Yto2Z`B. `WtoX+Y`C. `Wto2Y`D. `XtoY+Z`
Answer» Correct Answer - C Energy is released in a process when total binding energy (BE) of the nucleus is increased or we can say when total BE of products is more than the reactants. By calculation we can see that only in case of option (c ), this happens. Given, `W to 2Y` BE of reactants `=120xx7.5=900 MeV` and BE of products `=90xx8+60xx8.5=1230MeV` i.e., BE of products `gt` BE of reactants.

Discussion

233.	In a radioactive sample, the fraction of initial number of redioactive nuclie, which remains undecayed after n mean lives isA. `(1)/(e^(n))`B. `e^(n)`C. `1-(1)/(e^(n))`D. `((1)/(e-1))^(n)`
Answer» Correct Answer - A `(N)/(N_(0))=e^(-lamda((n)/(lamda)))=(1)/(e^(n))" "(therefore"mean life", t=(1)/(lamda))`

Discussion

234.	Tritium is an isotope of hydrogen whose nucleus triton contains 2 neutrons and 1 proton . Free neutrons decay into `p+bar(e) +bar(n).` If one of the neutrons in Triton decays , it would transform into `He^(3)` nucleus. This does not happen. This is becauseA. triton energy is less than that of a `He^(3)` nucleusB. the electron created in the beta decay proess connot remain in the nucleusC. Both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a `He^(3)` nucleusD. Ferr neutrons decay due to external perturbations which is absent in triton nucleus
Answer» Correct Answer - A Triitum `(""_(1)H^(3))` contain 1 proton and 2 neutrons. A neutron decays as, `nto p+ bare+barv,` the nucleus may have 2 protons and one neutron i.e., tritium will transfrom into a `He^(3)` (2 protons and 1 neutron). Triton energy is less than that of a `He^(3)` nucleus i.e., transformation is not allowed energetically.

Discussion

235.	Tritium is an isotope of hydrogen whose nucleus Triton contains 2 neutrons and 1 proton. Free neutrons decay into p + bar e + bar ν . If one of the neutrons in Triton decays, it would transform into He3 nucleus. This does not happen. This is because (a) Triton energy is less than that of a He3 nucleus. (b) the electron created in the beta decay process cannot remain in the nucleus. (c) both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a He3 nucleus. (d) because free neutrons decay due to external perturbations which is absent in a triton nucleus.
Answer» (a) Triton energy is less than that of a He³ nucleus.

Discussion

236.	Heavy stable nucle have more neutrons than protons. This is because of the fact that (a) neutrons are heavier than protons. (b) electrostatic force between protons are repulsive. (c) neutrons decay into protons through beta decay. (d) nuclear forces between neutrons are weaker than that between protons.
Answer» (b) electrostatic force between protons are repulsive.

Discussion

237.	If the binding energy per nucleon of deuterium is 1.115 MeV, its mass defect in atomic mass unit isA. 0.0048B. 0.0024C. 0.0012D. 0.0006
Answer» Correct Answer - b Binding energy per nucleon is `E_"bn"=E_b/A` For deuterium , A=2 `therefore ` 1.115 MeV=`E_b/2` or `E_b=2xx1.115` MeV or `E_b=DeltaMc^2` Mass defect, `DeltaM=(2xx1.115 )/931.5 u " " [because 1 u = 931.5 MeV//c^2]` =0.0024 u

Discussion

238.	Statement-1:Light nuclei having equal number of protons and neutrons are more stable Statement-2:In heavy nuclei, there is an excess of neutrons due to comlomb repulsion between protons .A. Statement-1 is True , Statement-2 is True ,Statement-2 is a correct explanation for Statement-1B. Statement-1 is True , Statement-2 is True ,Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False, Statement-2 is True
Answer» Correct Answer - B

Discussion

239.	A certain element has half life period of 30 days. Find its average life
Answer» `T_(1//2)`=30 days `T_"av"=1/lambda=1.44 T_(1//2)` = 1.44 x 30 days =43.2 days

Discussion

240.	Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of `._(8)^(16)O` in `MeV//c^(2)`.
Answer» `1u = 1.6605xx10^(-27)kg` To convert it into energy units, we multiply it by `c^(2)` and find that energy equivalent `= 1.6605xx10^(-27)xx(2.9979xx10^(8))^(2)"kg m"^(2)//s^(2)` `= 1.4924xx10^(-10)J` `=(1.4924xx10^(-10))/(1.602xx10^(-19))eV` `= 0.9315xx10^(9) eV=931.5 MeV` or, l u `= 931.5 MeV//c^(2)` For `._(8)^(16)O, Delta M = 0.13691 u = 0.13691xx931.5 MeV//c^(2)=127.5 MeV//c^(2)` The energy needed to separate `._(8)^(16)O` into its constituents is thus `127.5 MeV//c^(2)`.

Discussion

241.	Fission of nuclei is possible because the binding energy per nuclei in themA. increases with mass number at low mass numbersB. decreases with mass number at low mass numbersC. increases with mass number at high mass numbersD. decreases with mass number at high mass numbers
Answer» Correct Answer - d For nuclei having A gt 56 , binding energy per nucleon gradually decreases.

Discussion

242.	Assertion: In the process of nuclear fission, the fragments emit two or three neutrons as soon as they are formed and subsequently emit particles. Reason : As the fragments contain an excess of neutrons over protons, emission of neutrons and particles bring their neutron/proton ratio the to stable valuesA. Statement-1 is True , Statement-2 is True ,Statement-2 is a correct explanation for Statement-1B. Statement-1 is True , Statement-2 is True ,Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False, Statement-2 is True
Answer» Correct Answer - A

Discussion

243.	Obtain the binding energy of the nuclei `._(26)^(56)Fe` and `._(83)^(209)Bi`in units of MeV from the following data : `m (._(26)^(56)Fe)= 55.934939 u, m (._(83)^(209)Bi)=208.980388 u`
Answer» (i) `._(26)F^(56)` nucleus contains 26 protons and (56 - 26) = 30 neutrons Mass of 26 proton `= 26xx1.007825` `= 26.26345 a.m.u` Mass of 30 neutrons `= 30xx1.008665` `= 30.25995 amu` Total mass of 56 nucleons `= 56.46340 a.m.u` Mass of `._(26)F^(56)` Nucleus `= 55.934939 a.m.u` Mass defect `Delta m = 56.46340-55.934939` `= 0.528461 a.m.u` Total binding energy `= 0.528461xx931.5 MeV` `= 492.26 MeV` Average B.E per nucleon `= (492.26)/(56)` `= 8.790 MeV`. (ii) `._(83)Bi^(209)` nucleus contains 83 protons and (209-83) = 126 neutrons Mass of 83 protons `= 83xx1.007825` `= 83.649475 a.m.u` Mass of 126 Neutrons `= 126xx1.008665` `= 127.09170 a.m.u` Total mass of nucleons `= 210.741260 a.m.u` Mass of `._(83)Bi^(209)` nucleus = 208.986388 a.m.u Mass defect `Delta m = 210.741260-208.980388` = 1.760872 Total B.E `= 1.760872xx931.5 MeV` = 1640.26 MeV Average B.E per nucleon `= (1640.26)/(209)` = 7.848 MeV Hence `._(26)Fe^(56)` has greater B.E per nucleon than `._(83)Bi^(209)`

Discussion

244.	Obtain the binding energy of the nuclei `._26Fe^(56)` and `._83Bi^(209)` in units of MeV from the following data: `m(._26Fe^(56))=55.934939a.m.u.` , `m=(._83Bi^(209))=208.980388 a m u`. Which nucleus has greater binding energy per nucleon? Take `1a.m.u =931.5MeV`
Answer» Atomic mass of `""_(26)^(56)Fe, m_(1)=55.934939" "u` `""_(26)^(56)Fe` nucleus has 26 protons and (56 − 26) = 30 neutrons Hence, the mass defect of the nucleus, `trianglem=26xxm_(H)+30xxm_(n)-m_(1)` where, Mass of a proton, `m_(H)=1.007825" "u` Mass of a neutron, `m_(n)=1.008665" "u` `therefore triangle m=26xx1.007825+30xx1.008665-55.934939` `=26.20345+30.25995-55.934939` `=0.528461" "u` `But" "1u=931.5 MeV//C^(2)` The binding energy of this nucleus is given as: `E_(b1)=triangle mc^(2)` Where, c = Speed of light `therefore E_(b1)=0.528461xx931.5((MeV)/(c^(2)))xxc^(2)` `=492.26 MeV` Average binding energy per nucleon `=492.26/56=8.79 Mev` Atomic mass of `_(83)^(209)Bi,m_(2)=208.980388" "u` `_(83)^(209)Bi` nucleus has 83 protons and `(209 − 83)` 126 neutrons. Hence, the mass defect of this nucleus is given as: `triangle m=83xxm_(H)+126xxm_(n)-m_(2)` Where, Mass of a proton, `m_(H)=1.007825" "u` Mass of a neutron, `m_(n)=1.008665" "u` `trianglem=83xx1.007825+126xx1.008665-208.980388` `=83.649475+127.091790-208.980388` `=1.760877" "u` `But 1u=931.5MeV//c^(2)` `therefore triangle m=1.760877xx931.5 MeV//c^(2)` Hence, the binding energy of this nucleus is given as: `E_(b2)=triangle mc^(2)` `=1.760877xx931.5((MeV)/(c^(2)))xxc^(2)` `=1640.26Mev` Average bindingenergy per nucleon `=(1640.26)/(209)=7.848Mev`

Discussion

245.	The equivalent energy of 1 g of substance isA. `9xx10^13` JB. `6xx10^12` JC. `3xx10^13` JD. `6xx10^13` J
Answer» Correct Answer - a Using, `E=mc^2` Here, m=1 g=`1 xx10^(-3)` kg , c=`3xx10^8 m s^(-1)` `therefore E=10^(-3) xx9xx10^16 = 9xx10^13` J

Discussion

246.	if in a nuclear fusion reaction, mass defect to 0.3% , then energy released in fusion of 1 kg massA. `27xx10^10` JB. `27xx10^11` JC. `27xx10^12` JD. `27xx10^13` J
Answer» Correct Answer - d Here, `Deltam`=0.3% of 1 kg `=0.3/100 kg =3xx10^(-3)` kg `therefore E=(Deltam) c^2=3xx10^(-3)xx(3xx10^8)^2=27xx10^13` J

Discussion

247.	The radius of a spherical nucleus as measured by electron scattering is 3.6 fm. What is the mass number of the nucleus most likely to be ?A. 27B. 40C. 56D. 120
Answer» Correct Answer - a Nuclear radius , `R=R_0(A)^(1//3)` where A is the mass number of a nucleus . Given, R=3.6 fm `therefore` 3.6 fm=(1.2 fm) `(A^(1//3)) " " [because R_0=1.2 fm]` or A=`(3)^3`=27

Discussion

248.	How much mass has to be converted into energy to produce electric power of 500 MW for one hour ?A. `2xx10^(-5)` kgB. `1xx10^(-5)` kgC. `3xx10^(-5)` kgD. `4xx10^(-5)` kg
Answer» Correct Answer - a Here, P=500 MW=`5xx10^8` W,t=1 h=3600 s Energy produced , E=P x t = `5xx10^8xx3600=18xx10^11` J As `E=mc^2` `therefore m=E/c^2=(18xx10^11)/((3xx10^8)^2)=(18xx10^11)/(9xx10^16)=2xx10^(-5)` kg

Discussion

249.	The ratio of the nuclear radii of the gold isotope `._79^197Au` and silver isotope `._47^107Ag` isA. 1.23B. 0.216C. 2.13D. 3.46
Answer» Correct Answer - a Here, `A_1`=197 and `A_2`=107 `therefore R_1/R_2=(A_1/A_2)^(1//3)=(197/107)^(1//3)`=1.225

Discussion

250.	If the nucleus of `._13Al^27` has a nuclear radius of about 3.6 fm, then `._52Te^125` would have its radius approximately asA. 9.6 fmB. 12 fmC. 4.8 fmD. 6 fm
Answer» Correct Answer - d Here, `A_1`=27 , `A_2`=125 , `R_1`=3.6 fm As, `R_2/R_1=(A_2/A_1)^(1//3)=(125/27)^(1//3)=5/3` `therefore R_2=5/3R_1=5/3xx3.6=6` fm

Discussion

Explore topic-wise InterviewSolutions in .

If `m,m_n` and `m_p` are masses of `._Z X^A` nucleus, neutron and proton respectively.A. `m=(A-Z) m_p+Zm_p`B. `m=(A-Z)m_p+Zm_n`C. `m lt (A-Z) m_n+Zm_p`D. `m lt (A-Z) m_n+Zm_p`

Why is it found experimentally difficult to detect neutrinos in nuclear β-decay ?

What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE / A) in the range of mass number 'A' lying 30 &lt;A&lt;170 ?

In a typical nuclear reaction e.g.21H + 21H → 32He + n + 3.27 Me V,although number of nucleons is conserved, yet energy is released. How ? Explain

Find the binding energy of `._(26)^(56)Fe`. Atomic mass of Fe is 55.9349u and that of Hydrogen is 1.00783u and mass of neutron is 1.00876u.

Find the energy required to split `._(8)^(16)O` nucleus into four `alpha` - particles. The mass of an `alpha`- particle is 4.002603u and that of oxygen is 15.994915u.

How much energy is required to separate the typical middle mass nucleus `._(50)^(120)Sn` into its constituent nucleons ? (Mass of `._(50)^(120)Sn=119.902199u`, mass of proton = 1.007825u and mass of neutron = 1.008665u)

Calculate the binding energy of an `alpha`-particle. Given that mass of proton = 1.0073 u, mass of neutron = 1.0087 u, and mass of `alpha`- particle = 4.0015 u.

The half-life of radon is 3.8 days. After how many days `19/20` of the sample will decay ?

Radon has 3.8 days as its half-life . How much radon will be left out of 15 mg mass after 38 days ?A. 1.05 mgB. 0.015 mgC. 0.231 mgD. 0.50 mg

Write nuclear reaction equations for `alpha`- decay of `._(88)^(226)Ra`

1 mg redium has `2.68xx10^18` atoms. Its half life is 1620 years. How many radium atoms will disintegrate from 1 mg of pure radium in 3240 years ?A. `2.01xx10^9`B. `2.01xx10^18`C. `1.01xx10^9`D. `1.01xx10^18`

Find the Q-value and the kinetic energy of the emitted `alpha` - particle in the `alpha` - decay of `._(88)^(226)Ra` Given m `(._(88)^(226)Ra)=226.02540 u`, `m (._(86)^(222)Rn)=222.01750 u`, `m (._(86)^(222)Rn)=220.01137 u, m (._(84)^(216)Po)` `= 216.00189 u`.

The kinetic energy of `alpha`-particle emitted in the `alpha`-decay of `""_(88)Ra^(266)` is [given, mass number of Ra = 222 u]A. `5.201` Me VB. `3.301` Me VC. `6.023` Me VD. `4.871` Me V

The half - life period of a radioactive substance is 20 days. What is the time taken for `7//8^(th)` of its original mass to disintegrate ?

A nucleus `X^232` initially at rest undergoes `alpha` decay, the `alpha`-particle produced in above process is found to move in a circular track of radius 0.22 m in a uniform magnetic field of 1.5 T. Find Q value of reaction.

A certain substance decays to 1/232 of its initial activity in 25 days. Calculate its half-life.

Bombardment of lithium with protons gives rise to the following reaction : `._(3)^(7)Li + ._(1)^(1)H rarr 2[._(2)^(4)He]+Q`. Find the Q - value of the reaction. The atomi mases of lithium, proton and helium are 7.016u, 1.0084 and 4.004 u respectively.

Plutonium decays with a half - life of 24,000 years. If plutonium is stored for 72,000 years, what fraction of it remains ?

Give the three types of radioactive decay occur in nature.

In r-ray emission from a nucleus:(A) Both neutron number and proton number change(B) There is no change in neutron and proton number(C) Only the neutron number changes(D) only the proton number changes

Nuclear fission can be explained by:(A) proton-proton cycle(B) shell model of nucleus(C) liquid drop model of nucleus(D) independent of nuclear model of nucleus.

A `.^7Li` target is bombarded with a proton beam current of `10^-4` A for `1` hour to produce `.^7 Be` of activity `1.8xx10^8` disintegrations per second. Assuming that `.^7Be` radioactive nucleus is produced by bombarding `1000` protons, determine its half-life.

A nucleus has atomic number 11 and mass number 24. State the number of electrons, protons and neutrons in the nucleus

Two stable isotopes of lithium `._3^6Li` and `._3^7Li` have respective abundances of 7.5% and 92.5% . These isotopes have masses 6.0152 u and 7.016004 u respectively. Find the atomic weight of lithiumA. 6.941 uB. 3.321 uC. 2.561 uD. 0.621 u

Two stable isotopes of lithium `._(3)^(6)Li` and `._(3)^(7)Li` have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and7.01600 u, respectively. Find the atomic mass of lithium.

Two stable isotopes of lithium `._3^6Li` and `._3^7Li` have respective abundances of 7.5% and 92.5% . These isotopes have masses 6.0152 u and 7.016004 u respectively. Find the atomic weight of lithium

Binding energy per nucleons vs mass curve for nucleus is shown in the figure `W,X,Y`and `Z` are four nuclei indicated on the curve . The process that would release energy is A. `Yto2Z`B. `WtoX+Y`C. `Wto2Y`D. `XtoY+Z`

In a radioactive sample, the fraction of initial number of redioactive nuclie, which remains undecayed after n mean lives isA. `(1)/(e^(n))`B. `e^(n)`C. `1-(1)/(e^(n))`D. `((1)/(e-1))^(n)`

If the binding energy per nucleon of deuterium is 1.115 MeV, its mass defect in atomic mass unit isA. 0.0048B. 0.0024C. 0.0012D. 0.0006

A certain element has half life period of 30 days. Find its average life

Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of `._(8)^(16)O` in `MeV//c^(2)`.

Fission of nuclei is possible because the binding energy per nuclei in themA. increases with mass number at low mass numbersB. decreases with mass number at low mass numbersC. increases with mass number at high mass numbersD. decreases with mass number at high mass numbers

Obtain the binding energy of the nuclei `._(26)^(56)Fe` and `._(83)^(209)Bi`in units of MeV from the following data : `m (._(26)^(56)Fe)= 55.934939 u, m (._(83)^(209)Bi)=208.980388 u`

Obtain the binding energy of the nuclei `._26Fe^(56)` and `._83Bi^(209)` in units of MeV from the following data: `m(._26Fe^(56))=55.934939a.m.u.` , `m=(._83Bi^(209))=208.980388 a m u`. Which nucleus has greater binding energy per nucleon? Take `1a.m.u =931.5MeV`

The equivalent energy of 1 g of substance isA. `9xx10^13` JB. `6xx10^12` JC. `3xx10^13` JD. `6xx10^13` J

if in a nuclear fusion reaction, mass defect to 0.3% , then energy released in fusion of 1 kg massA. `27xx10^10` JB. `27xx10^11` JC. `27xx10^12` JD. `27xx10^13` J

The radius of a spherical nucleus as measured by electron scattering is 3.6 fm. What is the mass number of the nucleus most likely to be ?A. 27B. 40C. 56D. 120

How much mass has to be converted into energy to produce electric power of 500 MW for one hour ?A. `2xx10^(-5)` kgB. `1xx10^(-5)` kgC. `3xx10^(-5)` kgD. `4xx10^(-5)` kg

The ratio of the nuclear radii of the gold isotope `._79^197Au` and silver isotope `._47^107Ag` isA. 1.23B. 0.216C. 2.13D. 3.46

If the nucleus of `._13Al^27` has a nuclear radius of about 3.6 fm, then `._52Te^125` would have its radius approximately asA. 9.6 fmB. 12 fmC. 4.8 fmD. 6 fm

What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE / A) in the range of mass number 'A' lying 30 <A<170 ?