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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.
| 51. |
Making use of the tables of atomic masses, determine the energies of the following reaction:(a) `Li^(7)(p,n)Be^(7)` , (b)`Be^(9)(n,gamma)Be^(10)`, (c )`Li^(7)(alpha,n)B^(10)`, (d) `O^(16)(d,alpha)N^(14)`. |
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Answer» (a) The reaction is `Li^(7)(p,n)Be^(7)` and the energy of reaction is `Q=(M_(Be)^(7)+M_(Li)^(7))c^(2)+(M_(p)-M_(n))c^(2)` `=(Delta_(Li_(7))-Delta_(Be)^(7))c^(2)+Delta_(p)-Delta_(n)` `=[0.01601+0.00738-0.01693-0.00867]am uxxc^(2)` `= -1.64MeV` (b) The reaction is `Be^(9)(n, gamma)Be^(10)` Mass of `gamma` is taken zero.Then `Q=(M_(Be)^(9)+M_(n)-M_(Be)^(10))c^(2)` `=(Delta_(Be)^(9)+Delta_(n)-Delta_(Be)^(10))c^(2)` `=(0.01219+0.00867-0.01354)overset//, am uoverset//xxc^(2)` `=6.81MeV` (c )The reaction is `Li^(7)(alpha,n)B^(10)`. The energy is `Q=(Delta_(Li)^(7)+Delta_(alpha)-Delta_(n)-Delta_(B)^(10))c^(2)` `=(0.01601+0.00260-0.00867-.01294)am uxxc^(2)` `= -2.79MeV` (d) The reaction is `O^(16)(d, alpha)N^(14)`. The energy of reaction is `Q=(Delta_(O)^(16)+Delta_(d)-Delta_(alpha)-Delta_(N)^(14))c^(2)` `=(-0.00509+0.01410-0.00260-0.00307)am uxxc^(2)` `=2.79MeV` |
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| 52. |
Taking the value of atomic masses from the tables, calculate the kinetic energies of a positron and a neutrino emitted by `c^(11)` nucleus for the case when the daughter nucleus does not recoil. |
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Answer» We assume that the parent nucleus is at rest. Then since the daughter nucleus does not recoil, we have `vec(P)= -vec(P)_(v)` i.e., positron & `v` momentum are equal and opposite. On the other hand `sqrt(c^(2)p^(2)+m_(e )^(2)c^(4))+cp=Q=` total energy released. (Here we have used the fact that energy of the neutrino is `|vec(P)_(v)|=cp`) Now `Q=[ ("Mass of "C^(||)"nucleus")-("Mass of "B^(||)"nucleus")]c^(2)` `=["Mass of " C^(||)"atom-Mass of " B^(||)"atom"-m_(e )]c^(2)` `=0.00213 am uxxc^(2)-m_(e )c^(2)` `=(0.00213xx931-0.511)MeV= 1.47MeV` Then `c^(2)p^(2)+(0.511)^(2)=(1.470cp)^(2)=(1.47)^(2)-(1.47)^(2)-2.94cp+c^(2)p^(2)` Thus `cp= 0.646 MeV=` energy of neutrino Also `K.E`. of electron `=1.47-0.646-0.511= 0.313MeV` |
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| 53. |
Taking the values of atomic masses from the tables, find the maximum kinetic energy of beta-particles emitted by `Be^(10)` nuclei formed directly in the ground state. |
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Answer» The reaction is `Be^(10) rarr B^(10)+e^(-)+overset(-)v_(e )` For maximum `KE` of electrons we can put the energy of `vec(v)_(e )` to be zero. The atomic massed are `Be^(10)=10.016711 amu` `B^(10)= 10.016114 amu` So the `K.E` of electrons is (see previous problem) `597xx10^(-6) amuxxc^(2)= 0.56MeV` The momentum of electrons with this `K.E` is `0.941(MeV)/(c )` and the recoil energy of the daughter is `((0.941)^(2))/(2xxm_(d)c^(2))=((0.941)^(2))/(2xx10xx938)MeV= 47.2eV` |
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| 54. |
Taking the value of atomic masses from the tables, calculate the energy per nucleon which is liberated in the nuclear reaction `Li^(6)+H^(2)rarr2He^(4)`. Compare the obtained magnitude with the energy per nucleon liberated in the fission of `U^(235)` nucleus. |
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Answer» The energy realesed in the reaction `Li^(6)+H^(2)rarr2He^(4)` is `Delta_(Li^(6))+Delta_(H^(2))-2Delta_(He^(4))` `=0.01513+0.01410-2xx0.00260am u` `=0.02403 am u= 22.37MeV` (This resut for change in `B.E`. Is correct because the contribution of `Delta_(n) & Delta_(H)` cancels out by conservation law for protons & neutrons). Energy per nucleon is then `(22.37)/(8)= 2.796MeV//n ucl eon`. This should be compared with the value `(200)/(235)=0.85MeV//"nucleon"` |
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| 55. |
Find the energy of the reaction `Li^(7)+prarr2He^(4)` if the binding energies per nucleon in `Li^(7)` and `He^(4)` nuclei are known to be equal to `5.60` and `7.06 MeV` respectively. |
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Answer» The energy of reaction `Li^(7)+prarr2He^(4)` is , `2xxB.E of He^(4)-B.E of Li^(7)` `8epsilon_(alpha)-7epsilon_(Li)= 8xx7.06-7xx5.60= 17.3MeV` |
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| 56. |
Derive the formula expressing molar heat capacity of a unidimensional crystal, a chain of identical atoms, as a function of temperature `T` if the Debye temperature of the chain is equal ot `Theta`. Simplyfie the octained expression for the case `Tgt Theta` |
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Answer» In the Debye approximation the number of modes per unit frequency interval is given by `dN= (1)/(pi v)d omega 0 le omega le(k Theta)/(ħ)` But `(k Theta)/(ħ)= pin_(0)v` Thus `dN=(l)/(pi v)d omega, 0 le omega le pin_(0)v` The enrgy per mode is `lt E ge(1)/(2)ħ omega+(ħ omega)/(e^(ħ omega//kT)-1)` The total interval energy of the chain is `U=(l)/(pi v) int_(0)^(pin_(0)v) (1)/(2)ħ omega d omega` `+(l)/(piv)int_(0)^(pi n_(0)v)( ħ omega)/(e^( ħ omega//kT)-1) domega=(l ħ)/(4pi v)(pi n_(0)v)^(2)+(l ħ)/(4 pi v)(pi n_(0)v)^(2)+(l)/(pi v ħ)(kT)^(2) int_(0)^(Theta//T)(xdx)/(e^(x)-1)= ln_(0)k.( ħ)/(k)(pi n_(0)v).(1)/(4)` `+ln_(0)k(T^(2))/((pi n_(0)vħ//k)) int_(0)^(Theta//T)(x dx)/(e^(x)-1)` We put `ln_(0)k=R` for `1` mole of the chain Then `U=R Theta{(1)/(4)+((I)/(Theta))^(2) int_(0)^(Theta//T)(x dx)/(e^(x)-1)}` Hence the molar heat capacity is by defferentiation `C_(v)=((delU)/(delT))=R[2((T)/(Theta))int_(0)^(Theta//T)(xdx)/(e^(x)-1)-(Theta//T)/(e^(e//T)-1)]` when `T gt gt Theta, C_(v)~~R` |
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| 57. |
Derive Eq. (6.4c), making use of the Boltzmann distribution. From Eq.(6.4c) obtain the expression for molar vibration heat capacity `C_("V vib")` of diatomic gas. Calculate `C_("V vib")` for `cl_(2)` gas at the temperature `300K`. The natural vibration frequency of these molecules is equal to `1,064.10^(14)s^(-1)`. |
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Answer» by defination `lt E gt =(Sigma ^(E_(V)e^(-E_(v)//kT)))/(Sigma exp(-E_(v)//kT)) = ((del)/(delbeta)Sigma_(v=0)^(oo)e^(betaE_(v)))/(Sigma_(v=0)^(oo)e^(-betaE_(v)))` `=-(del)/(del beta) In Sigma_(v=0)^(oo)e^(-beta(v+1//2)ħomega), beta=(1)/(KT)` `=-(del)/(delbeta) In e^(-1//2beta ħ omega)(1)/(1-e^(-beta ħomega))` `= (del)/(del beta)[-(1)/(2) ħomegabeta-In(1-e^(-beta ħ onrga))]` `(1)/(2) ħ omega+( ħ omega)/(e^( ħ oega//kT)-1)` Thus for one gm mole of diatomic gas `C_("v vib")=N(dellt E gt)/(delT)=(R((ħ omega)/(KT))^(2)e^(ħ omega//kT))/((e^(ħ omega//kT)-1)^(2))` where `R=Nk` is the gas constant. In the present case `(ħ omega)/(KT)= 2.7088` and `C_("V 1 vib")= 0.56R` |
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| 58. |
A free sationary `Ir^(191)` nucleus with excitation energy `E=129 keV` passes to the ground state, emitting a gamma quantum. Calculate the fractional change of gamma quanta energy due to recoil of the nucleus. |
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Answer» With recoil neglected, the `gamma` quantum will have `129keV` enrgy. To a first approximation, its momenturm will be `129keV//c` and the energy of recoil will be `((0.129)^(2))/(2xx191xx931)MeV= 4.18xx10^(-8)MeV` In the next approximation we therefore write `E_(gamma)overset~-.129-8.2xx10^(-8)MeV` Therefore `(deltaE)/(E_(gamma))=3.63xx10^(-7)` |
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| 59. |
According to the Bohr-Sommerfeld postulate the periodic motion of a particle in a potential field must satisfy the following qyantization rule: `ointp dq= 2pi ħn`, where `q` and `p` are generalized coordinate and momenum of the particle, `n` are integers. Making use of this rule, find the permitted values of energy for a particle of mass `m` moving (a) In a uniimensional rectangular potential well of width `l` (b) along a circule of radius `r`, (c ) in a unidimentional potential field `U=alphax^(2)//2`, where `alpha` is a positive constant: (d) along a round orbit in a central field, where the potential enargy of the particle is equal to `U= -alpha//r`(`alpha` is a positve constant) |
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Answer» (a) if we measure energy from the bottom of the well, then `V(x) = 0` inside the walls. Then the quantization condition reads `oint p d x = 2 l p = 2 pi nħ` or `p = pi ħ//l` Hence `E_(n) = (p^(2))/(2 m) = (pi^(2) n^(2) ħ)/(2 m l)`. `oint p d x = 2 l p` because we have to consider the integral form `- (1)/(2)` to `(1)/(2)` and then back to `-(1)/(2)`. (b) Here, `oint p d x = 2 pi r p = 2 pi n ħ` or `p = (n ħ)/(r )` Hence `E_(n) = (n^(2) ħ^(2))/(2 m r^(2))` (c ) By energy conservation `(p^(2))/(2 m) + (1)/(2) alpha x^(2) = E` so `p = sqrt(2m E - m alpha x^(2))` Then `oint p d x = oint sqrt(2m E - m alpha x^(2) dx)` `= 2sqrt(m alpha) int_(-(sqrt( 2E))/(alpha))^(sqrt(2E)/(alpha)) sqrt((2 E)/(alpha) - x^(2)) dx` The integral is `int_(-a)^(a)sqrt(a^(2) - x^(2)) dx = a^(2) int_(-x//2)^(x//2) cos^(2) theta d theta` `= (a^(2))/(2) int_(-x//2)^(x//2)(1 + cos 2 theta) d theta = a^(2)(pi)/(2)`. Thus `oint p d x = pi sqrt(m a). (2 E)/(alpha) = E.2 pi sqrt((m)/(alpha)) = 2 pi n ħ` Hence `E_(n) = n ħsqrt((alpha)/(m))`. ( b) It is required to find the energy levels of the circular orbit for the rotential `U( r) = -(alpha)/(r )` In a circular orbit, the particle only has tangible velocity and the qunatization condition reads `oint p d x = m v. 2 pi r = 2 pi n ħ` so `m v r = M = n ħ` The energy of the particle is `E = (n^(2) ħ^(2))/(2 m r^(2)) - (alpha)/(r )` Equilibrium requires that the energy as a function of `r` be minimum. Thus `(n^(2) ħ^(2))/(mr^(3))=(alpha)/(r^(2)) or r=(n^(2) ħ^(2))/(m alpha)` Hence `E_(n)= -(malpha^(2))/(2n^(2) ħ^(2))` |
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| 60. |
A stationary helium ion emits a photon corresponding to the first line of Lyman series. That photon liberates a photoelectron form a stationary hydrogen atom in ground state. Find the velocity of photoelectron. Take mass of electron `=9.11xx10^(-31)kg` and ionisation energy of hydrogen atom=13.6ev. |
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Answer» We neglect recoil effects. The energy of the first Lyman line photon emitted by `He^(+)` is `4 ħR(1-(1)/(4))=3 ħR` The velocity `v` of the photoelectron that this photon liberates is given by `3 ħR=(1)/(2)mv^(2)+ ħR` where ` ħ R` on the right is the binding energy of the `n=1` electron in `H` atom. Thus `v=sqrt((4 ħR)/(m))=2sqrt(( ħR)/(m))=3.1xx10^(6)m//s` Here `m` is the mass of the direction. |
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| 61. |
A parallel stream of electrons accelerated by a potential difference `V= 25 V` falls normally on a diaphragm with two narrow slits separated by a distance `d=50 mu m`. Calculate the distance betweeb neighbouring maxima of the diffraction patterns on a screen located at a distance `=100` cm from the slits. |
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Answer» From the Young slit formula `Deltax=(l lambda)/(d)=(l)/(d).(2piħ)/(sqrt(2meV))` Substitution gives `Deltax= 4.90 mu m`. |
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| 62. |
Determine the spectral symbol of an atomic singlet term if the total splitting of that term in a weak magnetic field of induction `B= 3.0KG` amounts to `DeltaE= 104 mu eV`. |
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Answer» For single term `S=0, L=J,g=1` Then the total splitting is `deltaE= 2Jmu_(B)B` Substitution gives `J=3(=deltaE//2mu_(B)B)` The term is `.^(1)F_(3)`. |
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| 63. |
An atom is located in a magnetic firld of induction `B = 2.50kG`. Find the value of the total splitting of the following terms (expressed in `eV` units): (a) `.^(1)D`, (b)`.^(3)F_(4)`. |
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Answer» (a) For the `.^(1)D_(2)` term `g=1+(2xx3+0-2xx3)/(2xx2xx3)=1` and `Delta E= -mu_(B)M_(J)B` `M_(J)= -2,-1,0+1,+2`. Thus the spliting is `deltaE=4mu_(B)` Substitution gives `deltaE= 57.9mu eV` (b) For the `.^(3)F_(4)` term `g=1+(4xx5+1xx2-3xx4)/(2xx4xx5)=1+(10)/(40)=(5)/(4)` and `Delta=-(5)/(4)mu_(B) BM_(J)` where `M_(J)= -4 t o +4`. Thus `deltaE=(5)/(4)mu_(B) Bxx8= 10mu_(B)(= 2g Jmu_(B))` Substitution gives `deltaE=144.7 mueV` |
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| 64. |
The spectral series of the hydrogen spectrum that lies in the ultraviolet region is theA. Balmer seriesB. Pfund seriesC. Paschen seriesD. Lyman series |
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Answer» Correct Answer - D |
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| 65. |
For atoms of light and heavy hydrogen `(H and D)` fine the difference, (a) between the binding energies of their electrons in the ground state. (b) between the wavelength of first lines of the Lyman series. |
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Answer» The difference between the binding energies is `DeltaE_(b)=E_(b)(D)-E_(b)(H)` `=Delta(m)/(1+(m)/(M))(e^(4))/(2 ħ^(2))+(m)/(1+(m)/(2M))(e^(4))/(2 ħ^(2))` `=(me^(4))/(2 ħ^(2))((m)/(2M))` Substitution gives `Delta E_(b)= 3.7 meV` For the first line the Lyman series `(2pi ħC)/(lambda)= ħR((1)/(4)-(1)/(4))=(3)/(4) ħR` or `lambda=(8pi c)/(3R)=(8pi c)/(3R)=(8 pi ħc)/(3E_(b))` Hence `DeltaE_(b)=E_(b)(D)-E_(b)(H)` `=Delta(m)/(1+(m)/(M))(e^(4))/(2 ħ^(2))+(m)/(1+(m)/(2M))(e^(4))/(2 ħ^(2))` `=(me^(4))/(2 ħ^(2))((m)/(2M))` Substitution gives `Delta E_(b)= 3.7 meV` For the first line the Lyman series `(2pi ħC)/(lambda)= ħR((1)/(4)-(1)/(4))=(3)/(4) ħR` or `lambda=(8pi c)/(3R)=(8pi c)/(3R)=(8 pi ħc)/(3E_(b))` Hence `lambda_(H)-lambda_(D)=(8 pi ħc)/(3)((1)/(E_(b)(H))-(1)/(E_(b)(D)))` `=(8pi ħc)/(3((me^(4))/(2 ħ^(2)))).(m)/(2M)` `=(m)/(2M)xxlambda_(1)` (where `lambda_(1)` is the wavelength of the first line of Lyman series without considering nuclear motion). Substitution gives (see.21 for `lambda_(1)`) using `lambda_(1)= 121nm` `Delta lambda=33p m` |
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| 66. |
Calculate the separation between the particles of a system in the ground state, the corresponding binding energy, and the wavelength of the first line of Lyman series,if such a system is (a) a mesonic hydrogen atom whose nucleus is a proton (in a mesonic atom an electron is replaced by a meson whose charge is the same and mass is `207` si that of an electron), (b) a positronium consisting of an electron and positron revolving around their common centre of masses. |
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Answer» In a mesonic system, the reduced mass of the system is related to the mass of the meson `(m_(mu))` and proton `(m_(p))` by `mu=(m_(mu)m_(p))/(m_(mu)+M_(p))= 186.04 m_(e )` Then, separation between the particles in the ground state `=( ħ^(2))/(mue^(2))` `=(1)/(186) ( ħ^(2))/(m e^(2))` `E_(b)=(meason) =(mu e^(4))/(2 ħ^(2))= 186xx13.65eV` `=2.54 keV` `lambda_(1)=(8 pi ħc)/(3E_(b)(meason))=(lambda_(1)(Hydr og e n))/(186)= 0.65nm` (on using `lambda_(1)(Hydrogr en)=121nm)` (b) In the postitronium `mu=(m_(e )^(2))/(2m_(e ))=(m_(e ))/(2)` The sepration between the particles is the ground state `=2(ħ^(2))/(m_(e )e^(2))= 105.8p m` `E_(b)(positronium)=(m_(e))/(2).(e^(4))/(2ħ^(2))=(1)/(2)E_(b)(H)=6.8eV` `lambda_(1)(postironium)=(2lambda_(1)(Hydrog en)=0.243nm` |
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| 67. |
Evaluate the maximum value of energy and momentum of a phonon (acousite quantum) in copper whose Debyed temperature is equal to `300K`. |
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Answer» The maximum energy of the phonon is `ħ omega_(m)= kTheta= 28.4meV` On substituting `Theta= 330K`. To get corresponding value of the maximum moment we must have the dispersion relation `omega= omega (vec(K))`. For small `(vec(k))` we know `omega=|vec(k)|` where `v` is the velocity of sound in the crystal. For an order of magnitude estimate we continue to use this result for high`|vec(k)|`. Then we estimate `v` from the values of the models of elasticity and density `v~sqrt((E )/(rho))` We write `E~ 100GPa, rho= 8.9xx109^(3)kg//m^(3)` Then `v~3xx10^(3)m//s` Hence `ħ|vec(k)|_(max)~( ħ omega_(m))/(v)~ 1.5xx10^(-19) gm cms^(-1)` |
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| 68. |
Atomic hydrogen is in thermodynamics equilibrium with its radiation. Find: (a) the ratio of probabilites of induced and spontaneous radiation of the atoms from the level `2P` at a temperature `T=3000K`, (b) the temperature at which these probabilites become equal. |
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Answer» Atomic hydrogen is in thermodynamics equilibrium with its radiation, Find: (a) the ratio to probabilites of induced and spontaneous radiations of the atom from the level `2P` at a temperature `T=3000K`, (b) the temperature at which these probabilites become equal. |
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| 69. |
A particle of mass `m` strikes a stationary nucleus of mass `M` and activated an endoergic reaction. Demonstrate that the threshold (minimal) kinetic energy required to initiate this reaction is defined by Eq.(6.6d). |
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Answer» Energy required is minimum when the reaction products all move in the direction of the incident particle with the same velocity (so that the combination is at rest in the centre of mass frame.) We then have `sqrt(2mT_(th))= (m+M)v` (Total mass is constant in the nonrelativistic limit). `T_(th)-|Q|=(1)/(2)(m+M)v^(2)=(mT_(th))/(m+M)` or `T_(th)(M)/(m+M)=|Q|` Hence `T_(th)=(1+(m)/(M))|Q|` |
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| 70. |
Up to what temperature has one to heat classiacal electronic gas to make the mean enrgy of its electrons equal to that of free electrons in copper at `T=0`? Only one free electron is supposed to correspond to each other atom. |
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Answer» The mean `K.E` of electrons in a Fermi gas is `(3)/(5)E_(F)`. This muat equal `(3)/(2)kT`. Thus `T=(2E_(F))/(5k)` we calculate, `E_(F)` first. For `C_(u)` `n=(N_(A))/(M//rho)=(rhoN_(A))/(M)= "8.442xx10^(22)per c.c` Then `E_(F)= 7.01eV` and `T= 3.25xx10^(4)K` |
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| 71. |
The ground state energy of hydrogen atom is `-13.6 eV`. What is the potential energy of the electron in this stateA. `0 eV`B. `-27.2 eV`C. `1 eV`D. `2 eV` |
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Answer» Correct Answer - B |
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| 72. |
The radius of hydrogen atom in its ground state is `5.3 xx 10^(-11)m`. After collision with an electron it is found to have a radius of `21.2 xx 10^(-11)m`. What is the principle quantum number of `n` of the final state of the atom ?A. `n = 4`B. `n = 2`C. `n = 16`D. `n = 3` |
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Answer» Correct Answer - B |
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| 73. |
Calculate the binding energy of a `K` electron in vanidium whose `L` absorption edgerhas the wavelength `lambda_(L)=2.4nm`. |
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Answer» From the diagram above we see that the binding energy `E_(b)` of a `K` electron is the sum of the energy of a `K_(alpha)` line and the energy corresponding to the `L` edge of absorption spectrum `E_(b)=(2pi ħc)/(lambda_(L))+(3)/(4)ħR(Ƶ1)^(2)` For vanndium `Ƶ=23` and the energy of `K_(alpha)` line of vanadium has been calculated in problem 134(b). Using `(2 pi ħc)/(lambda_(L))= 0.51 keV` for `lambda_(L)=2.4nm` we get `E_(b)=5.46keV` |
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| 74. |
Find the velocity of photoelectrons liberated by electromagnetic radiation of wavelength `lambda=18.0nm` from stationary `He^(+)` ions in the ground state. |
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Answer» By conservation of enery `(1)/(2)mv^(2)=(2pi ħ c)/(lambda)-E_(b)` where `E_(b)=4 ħ R` is the binding energy of the electron in the ground state of `He^(+)`. (Recoil of `He^(++)` nucleus is neglected). Then `v=sqrt((2)/(m)((2pi ħc)/(lambda)-E_(b)))` Substitution gives `v= 2.25xx10^(6)m//s` |
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| 75. |
Find the kinetic energy and the velocity of the photoelectrons liberated by `K_(alpha)` radiation of zinc from the `K` shell of iron whose `K` band absorption edge wavelength is `lambda_(k)=174p m`. |
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Answer» The energy of the `K_(alpha)` radiation of Ƶn is ` ħ omega=(3)/(4) ħR(Ƶ-1)^(2)` where `Ƶ=` atomic number of Zinc`= 30`. The binding energy of the `K` electrons in iron is obtained from the wavelength of `K` absorption edge as `E_(k)= 2pi ħc//lambda_(k)` Hence by Einstein equation `T=(3)/(4) ħR(Ƶ-1)^(2)-(2pi ħc)/(lambda_(k))` substitution gives `T=1.463keV` This corresponds to a velocity of the photo electrons of `v= 2.27xx10^(6)m//s` |
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| 76. |
How many neutrons are there in the hundredth generation if the fission process starts with `N_(0)= 1000` neutrons and takes place in a medium with multiplication constant `k=1.05`? |
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Answer» No. of nuclei in the first generation=No. of nuclei initially`=N_(0)` `N_(0)` in the second generation`=N_(0)xx` multiplication factor `=N_(0).k` `N_(0)` in the `3rd` generation `=N_(0).k.k=N_(0)k^(2)` `N_(0)` in the nth generation `=N_(0)k^(n-1)` Substitution gives `1.25xx10^(5)` neutrons |
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| 77. |
In a thermal reactor the mean lifetime of one generation of thermal neutrons is `tau=0.10 s`. Assuming the multiplication constant to be equal to `k=1.010`, find : (a) how many times the number of neutrons in the reactor, and consequently its power, will increase over `t=1.0 mi n` , (b) the period `T` of the reactor, i.e., the time period over which its power increases `e-fold. |
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Answer» (a) This number is `K^(n-1)` where `n=` no. of generations in time `t=t//T` Substitution gives `388`. (b) We write `k^(n-1)=e^(((T)/(tau)-1)In k)` or `(T)/(tau)-1=(1)/(In k)` and `T=tau(1+(1)/(In k))=10.15 sec` |
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| 78. |
What amount of energy (in eV) should be added to an electron to reduce its de-Broglie its de-Broglie wavelength from 100 to 50 pm? Given `h=6.62xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg`. |
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Answer» From `lambda=(2pi ħ)/(P)=(2pi ħ)/(sqrt(2mT))` we find `T=(4pi^(2) ħ^(2))/(2 m lambda^(2))=(2pi^(2) ħ^(2))/(m lambda^(2))` Thus `T_(2)-T_(1)=(2pi^(2) ħ^(2))/(m)((1)/(lambda_(2)^(2))-(1)/(lambda_(1)^(2)))` Substitution gives `Delta T= 451 eV= 0.451ke V`. |
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| 79. |
The de-Broglie wavelength of an electron in the first Bohr orbit isA. Equal to one fourth the circumference of the first orbitB. Equal to half the circumference of the first orbitC. Equal to twice the circumference of the first orbitD. Equal to the circumference of the first orbit |
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Answer» Correct Answer - D |
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| 80. |
Radius of first Bohr orbit is r . What is the radius of `2^(nd)` Bohr orbit?A. `8r`B. `2r`C. `4r`D. `2 sqrt(2r)` |
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Answer» Correct Answer - C |
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| 81. |
Assuming that the splitting of a `U^(235)` nucleus liberates of `U^(235)` isotope, and the mass of cal with calorific value of `30 kJ//g` which is equivalent to that the to that for one `kg` of `U^(235)`, (b) the mass of `U^(235)` isotope split during the explosion of the atomic bomb with `30 kt` trotyl equivalent if the calorific value of troty is `4.1 k J//g`. |
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Answer» (a) the energy liberated in the fission of `1kg of U^(235)` is `(1000)/(235)xx6.023xx10^(23)xx200MeV=8.21xx10^(10)kJ` The mass of coal with equivalent calorific value is `(8.21xx10^(10))/(30000) kg= 2.74xx10^(6)kg` (b) The required mass, is `(30xx10^(9)xx4.1xx10^(3))/(200xx1.602xx10^(-13)xx6.023xx10^(23))xx(235)/(1000)kg =1.49kg` |
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| 82. |
Which of the following particles are constituents of the nucleusA. Protons and electronsB. Protons and neutronsC. Neutrons and electronsD. Neutrons and positrons |
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Answer» Correct Answer - B |
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| 83. |
Size of nucleus is of the order ofA. `10^(-10)m`B. `10^(-15)m`C. `10^(-12)m`D. `10^(-19) m` |
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Answer» Correct Answer - B |
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| 84. |
What amount of heat is liberated during the formation of one gram of `He^(4)` from deuterium `H^(2)`? What mess of coal with calorific value of `30kJ//g` is thermally equivalent to the magnitude obrained? |
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Answer» The reaction is (in effect). `H^(2)+H^(2) rarrHe^(4)+Q` Then `Q=2 Delta_(H)^(2)-Delta_(He^(4))+Q` `=0.02820-0.00260` `=0.02560amu=23MeV` Hence the energy realeased in `1 gm`of `He^(4)` is `(6.023xx10^(23))/(4)xx23.8xx16.03xx10^(-13)Jou l e=5.75xx10^(8)kJ` This energy can be derived from `(5.75xx10^(8))/(30000)kg= 1.9xx10^(4)`kg of Coal. |
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| 85. |
The particles which can be added to the nucleus of an atom without changing its chemical properties areA. ElectronsB. ProtonsC. NeutronsD. None of the above |
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Answer» Correct Answer - C |
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| 86. |
Which of the following has the mass closest in value to that of the positron (1 a.m.u. `= 931 MeV` )A. ProtonB. ElectronC. PhotonD. Neutrino |
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Answer» Correct Answer - B |
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| 87. |
The mass of a nucleus can be either less than or more than the sum of the masses of nucleons present in it. The whole mass of the atom is considered in the nucleus.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If assertion is false but reason is true |
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Answer» Correct Answer - D |
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| 88. |
If the binding energy of the deuterium is `2.23 MeV`. The mass defect given in a.m.u. is.A. `-0.0024`B. `-0.0012`C. 0.0012D. 0.0024 |
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Answer» Correct Answer - D |
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| 89. |
The neutron was discovered byA. Marie CurieB. Pierre CurieC. James ChadwickD. Rutherford |
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Answer» Correct Answer - C |
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| 90. |
The mass number of a nucleus is.A. Always less than its atomic numberB. Always more than its atomic numberC. Always equal to its atomic numberD. Sometimes more than and sometimes equal to its atomic number |
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Answer» Correct Answer - D |
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| 91. |
Nuclear binding energy is equivalent toA. Mass of protonB. Mass of neutronC. Mass of nucleusD. Mass defect of nucleus |
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Answer» Correct Answer - D |
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| 92. |
The energy equivalent of 1 kilogram of matter is aboutA. `10^(-15) J`B. `1 J`C. `10^(-12) J`D. `10^(17) J` |
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Answer» Correct Answer - D |
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| 93. |
Find the energy `Q` liberated in `beta^(-_(-))` and `beta^(+)-` decays and in `K`- caputure if the masses of the parent atom `m_(p)`, the daughter atom `M_(d)` and an electron `m` are known. |
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Answer» In `beta^(-)` decay `Z^(X^(A))rarr_(Z+1)Y^(A)+e^(_)+Q` `Q=(M_(x)-M_(y)-m_(e ))c^(2)` `=[(M_(x)+ZM_(e )c^(2))-(M_(y)+Zm_(e )+m_(e ))]c^(2)` `=(M_(p)-M_(d))c^(2)` since `M_(p),M_(d)` are the masses of the atoms. The binding energy of the electrons in ignored. In `K` capture `e_(k)^(-)+._(Z)X^(A) rarr_(z-1)Y^(A)+Q` `Q=(M_(X)-M_(Y))c^(2)+m_(e )c^(2)` `=(M_(x)^(c^(2))+Zm_(e )c^(2))-(M_(Y)c^(2)+(Z-1)m_(e )c^(2))` `=c^(2)(M_(p)-M_(d))` In `beta^(+) decay _(Z)X^(A) rarr_(Z-1)Y^(A)+e^(+)+Q` Then `Q= (M_(x)-M_(y)-m_(e ))c^(2)` `[M_(x)+Zm_(e )]c^(2)-[M_(y)+(Z-1)m_(e )]c^(2)-2m_(e )c^(2)` `=(M_(p)-M_(d)-2m_(e ))c^(2)` |
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| 94. |
From the table of atomic masses determine the velocity of a nucleus appearing as a result of `K`-capture in a `Be^(7)` atom provided the daughter nucleus turns out to be in the ground state. |
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Answer» The process is `e_(k)^(-)Be^(7) rarrLi^(7)+v` The energy available in the process is `Q=c^(2) ("Mass of " Be^(7) "atom-Mass of " li^(7) "atom")` `= 0.00092xx931MeV= 0.86MeV` The momentum of a `K` electron is negligible. So in the rest frame of the `Be^(7)` atom, most of the energy is taken by neutrino whose momentum is very nearly `0.86MeV//c` The momentum of the recoiling nucleus is equal and opposite. The velocity of recoil is `(0.86MeV//c)/(M_(Li))=cxx(0.86)/(7xx931)= 3.96xx10^(6)cm//s` |
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| 95. |
The graph which represents the correct variation of logarithm of activity (log A) versus time, in figure is. .A. AB. BC. CD. D |
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Answer» Correct Answer - D |
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| 96. |
The graph between log `R` and log `A` wher `R` is the nuclear radius and `A` is the mass of is.A. B. C. D. |
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Answer» Correct Answer - A |
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| 97. |
The charge density in a nucleus varies with distance from the centre of the nucleus according to the curve in Fig.A. B. C. D. |
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Answer» Correct Answer - C |
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| 98. |
Orbital acceleration of electron isA. `(n^(2)h^(2))/(4pi^(2) m^(2) r^(3))`B. `(n^(2)h^(2))/(2n^(2) r^(3))`C. `(4n^(2) h^(2))/(pi^(2) m^(2) r^(3))`D. `(4n^(2)h^(2))/(4pi^(2) m^(2) r^(3))` |
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Answer» Correct Answer - A |
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| 99. |
Assertion: The positively changed nucleus of an atom has a radius of almost `10^(-15)m`. Reason: In `alpha`-particle scattering experiment the distance of closest apporach for `alpha`-particles is `~~10^(-15)m`.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - A |
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| 100. |
Assertion: The electron in the hydrogen atom passes from energy level `n = 4` to the `n = 1` level. The maximum and minimum number of photon that can be emitted are six and one respectively. Reason: The photons are emitted when electron make a transtition from the higher energy state to the lower energy state.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - B |
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