InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Using Eq.(6.2e), find the probability `D` of a particle of mass `m` and energy `E` tunneling through the potential barrier shown in fig. where `U(x)=U_(0)(1-x^(2)//l^(2))`. |
|
Answer» The potential is `U(x)=U_(0)(1-(x^(2))/(l^(2)))`. The turning points are `(E )/(U_(0))=1-(x^(2))/(l^(2)) or x=+-lsqrt(1-(E )/(U_(0)))` Then `D~~exp[-(4)/( ħ)int_(0)^(lsqrt(1-(E//U_(0))))sqrt(2m{U_(0))(1-(x^(2))/(l^(2)))-E}}dx]` `=exp=exp[-(4)/(ħ)int_(0)^(lsqrt(1-(E//U_(0))))sqrt(2mU_(0))sqrt(1-(E)/(U_(0))-(x^(2))/(l^(2))dx)` `=exp[-(4l)/(ħ)sqrt(2mV_(0)) int_(0)^(x_(0))sqrt(x_(0)^(2)-x^(2))dx], x_(0)=sqrt(1-E//V_(0))` The intergal is `int_(0)^(x_(0))sqrt(x_(0)^(2)-x^(2))dx=x_(0)^(2)int_(0)^(x//2)cos^(2) thetad theta=(pi)/(4)x_(0)^(2)` Thus `D~~exp[-(pil)/(ħ)sqrt(2mU_(0))(1-(E)/(U_(0)))]` `=exp[-(pil)/(ħ)sqrt((2m)/(U_(0)))(U_(0)-E)]` |
|
| 2. |
Employing Eq.(6.2e), find the probability `D` of an electron with energy `E` tunneling through a potential barrier of width `l` and height `U_(0)` provided the barrier is shaped as shown: (a) In fig 6.4 (b) in figure 6.5. |
|
Answer» The formula is `D~~exp[-(2)/( ħ)int_(x_(1))^(x_(2))sqrt(2m(V(x)-E)dx)]` Here `V(x_(2))=V(x_(1))=E` and `V(x)gtE` in the region `x_(2)gt xgt x_(1)`. (a) For the problem, the intergral is trivial `D~~exp[-(2l)/( ħ)sqrt(2m(U_(0)-E))]` (b) We can without loss of generality take `x=0` at the point the potential begins to climb. Then `U(x)= ,{{:(0x,lt,0,,),(U_(0),(x)/(l),0 x lt l,,),(0,xgtl,,,):}` Then `D~~exp[-(2)/ (ħ)int_(l(E)/(U_(0)))^(l)sqrt(2m(U_(0)(x)/(l)-E)dx)]` `=exp[-(2)/ (ħ)sqrt((2mU_(0))/(l))^(l)int_(x_(0))^(l)sqrt(x-x_(0))dx]x_(0)=l(E )/(U_(0))` `=epx[-(2)/( ħ)sqrt((2mU_(0))/(l))(2)/(3)(x-x_(0))^(3//2):|_(x_(0))^(l)]` `=exp[-(4)/(3 ħ)sqrt((2mU_(0))/(l))(l-l(E )/(U_(0)))^(3//2)]` `= exp[-(4l)/( 3ħU_(0))(U_(0)-E)^(3//2)sqrt(2m`)] |
|
| 3. |
Employing the uncertainty principle, estimate the minimum kinetic energy of an electron confined within a region whose sizes is `l=0.20nm`. |
|
Answer» Clearly `Deltax le l" so" Delta p_(x),ge(ħ)/(l)` Now `p_(x)geDeltaP_(x)` and so `T=(p_(x)^(2))/(2m) ge (ħ^(2))/(2ml^(2))` Thus `T_(min)=(ħ^(2))/(2m l^(2))~=0.95eV` |
|
| 4. |
In Bohr’s model of hydrogen atom, which of the following pairs of quantities are quantizedA. Energy and linear momentumB. Linear and angular momentumC. Energy and angular momentumD. None of the above |
|
Answer» Correct Answer - C |
|
| 5. |
when a hydrogen atom is raised from the ground state to an excited stateA. P.E. increases and K.E. decreasesB. P.E. decreases and K.E. increasesC. Both kinetic energy and potential energy increaseD. Both K.E. and P.E. decrease |
|
Answer» Correct Answer - A |
|
| 6. |
In Bohr model of the hydrogen atom, the lowest orbit corresponds toA. Infinite energyB. The maximum energyC. The minimum energyD. Zero energy |
|
Answer» Correct Answer - C |
|
| 7. |
Which of the following statements about the Bohr model of the hydrogen atom is false ?A. Acceleration of electron in `n = 2` orbit is less than that in `n = 1` orbiB. Angular momentum of electron in `n = 2` orbit is more than that in `n = 1` orbitC. Kinetic energy of electron in `n = 2` orbit is less than that in `n = 1` orbiD. Potential energy of electron in `n = 2` orbit is less than that in `n = 1` orbit |
|
Answer» Correct Answer - D |
|
| 8. |
An electron in the `n = 1` orbit of hydrogen atom is bound by `13.6 eV`. If a hydrogen atom I sin the `n = 3` state, how much energy is required to ionize itA. 13.6 eVB. 4.53 eVC. 3.4 eVD. 1.51 eV |
|
Answer» Correct Answer - D |
|
| 9. |
The wavelength of the first line of Balmer series is `6563 Å`. The Rydbergs constant fro hydrogen is aboutA. `1.09 xx 10^(7)` per mB. `1.09 xx 10^(8) per m`C. `1.09 xx 10^(9) per m`D. `1.09 xx 10^(5) per m` |
|
Answer» Correct Answer - A |
|
| 10. |
An electron in the `n = 1` orbit of hydrogen atom is bound by `13.6 eV`energy is required to ionize it isA. 13.6 eVB. 6.53 eVC. 5.4 eVD. 1.51 eV |
|
Answer» Correct Answer - A |
|
| 11. |
If the series limit of Lyman series for Hydrogen atom is equal to the series limit Balmer series for a hydorgen like atom, then atomic number of this hydrogen-like atom will beA. 1B. 2C. 3D. 4 |
|
Answer» Correct Answer - B |
|
| 12. |
The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of `108.5mm`. The ground state energy of an electron of this ion will eA. 3.4 eVB. 13.6 eVC. 54.4 eVD. 122.4 eV |
|
Answer» Correct Answer - C |
|
| 13. |
What hydrogen-like ion has the wavelength difference between the first lines of the Balmer Lyman series equal to `59.3nm`? |
|
Answer» From the formula of the previous problem `Delta lambda=(176 pi c)/(15 Z^(2)R)` or `Z=sqrt((176pi c)/(15R Delta lambda))` Substitution of `Delta lambda = 59.3nm` and `R` and the previous gives `Z=3` This identifies the ion as `Li^(++)` |
|
| 14. |
An ionic atom equivalent to hydrogen atom has wavelength equal to `1//4` of the wavelengths of hydrogen lines. The ion will beA. `He^(+)`B. `Li^(++)`C. `Ne^(9+)`D. `Na^(10+)` |
|
Answer» Correct Answer - A |
|
| 15. |
An electron has a mass of `9.1 xx 10^(-31)kg`. It revolves round the nucleus in a circular orbit of radius `0.529 xx 10^(-10)` metre at a speed of `2.2 xx 10^(6) m//s`. The magnitude of its linear momentum in this motion isA. `1.1 xx 10^(-34) kg - m//s`B. `2.0 xx 10^(-24) kg - m//s`C. `4.0 xx 10^(-24) kg - m//s`D. `4.0 xx 10^(-31) kg - m//s` |
|
Answer» Correct Answer - B |
|
| 16. |
The figure shows a graph between `1n |(A_(n))/(A_(1))|` and `1n|n|`, where `A_(n)` is the area enclosed by the `n^(th)` orbit in a hydrogen like atom. The correct curve is |
|
Answer» Correct Answer - A |
|
| 17. |
An electron with kinetic energy `T~~4eV` is confirned within a region whose linear dimension is `l=1 mu m`. Using the uncertianty principle, evaluate the relative uncertainty of its velocity. |
|
Answer» The momentum the electron is `Deltap_(x)=sqrt(2mT)` Uncertainty in its momentum is `Delta p_(x)geħ//Deltax=ħ//l` Hence relative nucertainty `(Delta P_(x))/(p_(x))=(ħ)/(lsqrt(2mT))=sqrt((ħ^(2))/(2ml^(2)))//T=(Deltav)/(v)` Substitution gives `(Delta v)/(v)=(Delta p)/(p)= 9.75xx10^(-5)~~10^(-4)` |
|
| 18. |
The ratio of the wavelengths for `2 rarr 1` transition in `Li, He`, and H isA. `1 : 2 : 3`B. `1 : 4 : 9`C. `4 : 9 : 36`D. `3 : 2 : 1` |
|
Answer» Correct Answer - C |
|
| 19. |
In a beryllium atom, if `a_(0)` be the radius of the first orbit, then the radius of the second orbit will be in generalA. `na_(0)`B. `a_(0)`C. `n^(2) a_(0)`D. `(a_(0))/(n^(2))` |
|
Answer» Correct Answer - C |
|
| 20. |
Energy of an electron in an excited hydrogen atom is `-3.4eV`. Its angualr momentum will be: `h = 6.626 xx 10^(-34) J-s`.A. `1.11 xx 10^(34) J sec`B. `1.51 xx 10^(-31) J sec`C. `2.11 xx 10^(-34) J sec`D. |
|
Answer» Correct Answer - C |
|
| 21. |
The ionization potential for second `H`e electron isA. 13.6 eVB. 1.36 eVC. 54.4 eVD. 100 eV |
|
Answer» Correct Answer - C |
|
| 22. |
The ratio of the energies of the hydrogen atom in its first to second excited state isA. `1//4`B. `4//9`C. `9//4`D. 4 |
|
Answer» Correct Answer - C |
|
| 23. |
The energy required to remove an electron in a hydrogen atom from `n = 10` state isA. 13.6 eVB. 1.36 eVC. 0.136 eVD. 0.0136 eV |
|
Answer» Correct Answer - C |
|
| 24. |
The ionisation potential of hydrogen atom is `13.6` volt. The energy required to remove an electron in the `n = 2` state of the hydrogen atom isA. 27.2 eVB. 13.6 eVC. 6.8 eVD. 3.4 eV |
|
Answer» Correct Answer - D |
|
| 25. |
When a sample of solid lithium is placed in a flask of hydrogen gas then following reaction happened `._1^1 H + ._3 Li^7 rarr ._2 He^4 + ._2He^4`. This statement is. .A. 1B.C. May be true at a particular pressureD. None of these |
|
Answer» Correct Answer - B |
|
| 26. |
Which state of triply ionised Beryllium `(Be^(+++))` the same orbital radius as that of the ground state hydrogen ?A. `n = 4`B. `n = 3`C. `n = 2`D. `n = 1` |
|
Answer» Correct Answer - C |
|
| 27. |
The irradiation of lithium and beryllium targets by a monoergic stram of protons reveals that the reaction `Li^(7)(p,n) 1.65 MeV` is initialted whereas the reaction `Be^(9)(p,n)B^(9)-1.85MeV` does not take place. Find the possible values of kinetic energy of the protons. |
|
Answer» Since the reaction `Li^(7)(p,n)Be^(7)(Q= -1.65MeV)` is initiated, the incident proton energy must be `ge(1+(M_(p))/(M_(d)))xx1.65= 1.89MeV` since the reaction `Be^(9)(p,n)B^(9)(Q= -1.85MeV)` is not initiated `Tle(1+(M_(p))/(M_(ve)))xx1.85= 2.06MeV` Thus `1.89MeVleT_(p)le2.06MeV` |
|
| 28. |
Find the number of free electrons per one sodium atom at `T=0` if the Fermi level is equal to `E_(F)= 3.07eV` and the density of sodium is `0.97 g//cm^(3)`. |
|
Answer» We calculate the concentration `n` of electron in the `Na` metal from `E_(max)=E_(F)=( ħ^(2))/(2m)(3 pi^(2)n)^(2//3)` we get from `E_(F)= 3.07eV` `n= 2.447xx10^(22) per c.c` From this we get the number of electrons per one `Na` atom as `(n)/(rho).(M)/(N_(A))` where `rho=` density of `Na, M=` molar weight in gm of `Na,N_(A)=` Avagadro number we get `0.963` electrons per one `Na` atom. |
|
| 29. |
Write missing symbols, denoted by `x`, in the following nuclear reactions: (a) `B^(10)(x,alpha)Be^(8)`, (b) `O^(17)(d,n)x`, (c )`Na^(23)(p,x)Ne^(20)`, (d)`x(p,n)Ar^(37)`. |
|
Answer» (a) The particle `x` must carry two nucleous and a unit of positive charge. The reaction is `B^(10)(d, alpha)B_(e )^(8)` (b) The particle `x` must contain a proton in addition to the constituents of `O^(17)`. Thus the reaction is `O^(17)(d,n)F^(18)` (c ) The particle `x` must carry nucleon number `4` and two units of `+ve` charge. Thus the particle must be `x-alpha` and the reaction is `Na^(23)(p, alpha)Ne^(20)` (d) The particle `x` must carry mass number `37` and have one unit less of positive charge. Thus `x=Cl^(37)` and the reaction is `Cl^(37)(p,n)Ar^(37)` |
|
| 30. |
Nuclear reactions are given as (i) `square (n, p)_(15) p^(32)` (ii) `square (p, alpha)_(8) O^(16)` (iii) `._(7)square^(4) (p) ._(6)C^(14)` missing particle or nuclide (in box `square` ) in these reactions are respectivelyA. `S^(32), F^(19), ._(0)n^(1)`B. `F^(19), S^(32), ._(0)n^(1)`C. `Be, F^(19), ._(0)n^(1)`D. None of these |
|
Answer» Correct Answer - A |
|
| 31. |
To activate the reaction`(n,alpha)` with stationary `B^(11)` nuclei, neutrons must have the threshold kinetic energy `T_(th)=4.0MeV`. Find the energy of this reaction. |
|
Answer» We have `4.0=(1+(m_(n))/(M_(B^(11))))|Q|` or `Q=(11)/(12)xx4MeV= -3.67MeV` |
|
| 32. |
The yeild of a nuclear reaction producing radionuclides may be described in two ways: either by the ratio `omega` of the numbers of nuclear reactions to the number of bombarding particles, or by the quantity `k`, the ratio of the activity of the formed radionuclide to the number of bombarding particles, Find: (a) the half-life of the formed radionuclide, assuming `omega` and `k` to be known: (b) the yield `omega` of the reaction `Li^(7)(p,n)Be^(7)` if after irradiation of a lithum target by a beam or protons(over `t=2.0`) hours and with `=1.35.10^(8)dis//s` and its half-life to `T=53 days`. |
|
Answer» (a) Assuming of course, that each reaction produces a radio nuclide of the same type, the decay constant `alpha` of the radionuclide is `k//w`. Hence `T=(In 2)/(lambda)=(w)/(l)In2` (b) number of bombarding particles is : `(It)/(e )` (e = charge on proton.) Then the number of `Be^(7)` produced is : `(It)/( e)w`. If `lambda=` decay constant of `Be^(7)=(In2)/(T)`, then the acitvity is `A=(It)/(e )omega. (In2)/(T)` Hence `w=(eAT)/(It In 2)= 1.98xx10^(-3)` |
|
| 33. |
An alpha-particle with kinetic energy `T_(alpha)= 7.0MeV` is scattered elastically by an initially stationary `Li^(6)` nucleus. Find the kinetic energy of the recoil nucleus if the angle of divergence of the two particle is `Theta= 60^(@)`. |
|
Answer» Initial momentum of the `alpha` particle is `sqrt(2mT_(alpha)) hat(i)` (where `hat(i)` is a unit vector in the incident direction.) Final momenta are respectively `vec(p)_(alpha)` and `vec(p_(Li)`. Conservation of momentum reads `vec(P)_(a)+vecP_(Li)= sqrt(2mT_(alpha))hat(i)` Squarting `p_(alpha)^(2)+p_(Li)+2p_(alpha)p_(Li)cos Theta= 2mT_(alpha)` (1) where `Theta` is the angle between `vec(p)_(alpha)` and `vec(p)_(Li)`. Also by energy conservation `(p_(alpha)^(2))/(2m)+(p_(L)^(2))/(2M)=T_(alpha)` (`m & M` are respectively the masses of `alpha` particle and `Li^(6)`.)So `p_(alpha)^(2)+(m)/(M)p_(Li)^(2)= 2mT_(alpha)` (2) Substituting (2) from (1) we see that `P_(Li)[(1-(m)/(M))p_(Li)+2p_(alpha)cos Theta]=0` Thus if `p_(Li) ~~ 0` `p_(alpha)= -(1)/(2)(1-(m)/(M))p_(Li)sec Theta`. Since `p_(alpha), p_(Li)` are both positive number (being magnitudes of vectors) we must have `-le cos Theta lt 0 if m lt M` This being understood, we write `(P_(Li)^(2))/(2M)[1+(M)/(4m)(1-(m)/(M))^(2) sec^(2)Theta]=T_(alpha)` Hence the recoil energy of the `L_(i)` nucleus is `(P_(Li)^(2))/(2M)=(T_(alpha))/(1+((M-m)^(2))/(4mM)sec^(2)Theta)` As we pointed out above `Theta ~~ 60^(@)`. If we take `Theta= 120^(@)`, we get recoil energy of `Li= 6MeV`. |
|
| 34. |
A stationary `Pb^(200)` nucleus emits an alpha-particle with kinetic energy `T_(alpha)= 5.77MeV`. Find the recoil velocityof a daughter nucleus. What friction of the total energy liberated in this decay is accountd for by the recoil energy of the daughter nucleus? |
|
Answer» The momentum of the `alpha`-particle is `sqrt(2M_(alpha)T)`. This is also the recoil momentum of the daughter nuclear in opposite direction. The recoil velocity of the daughter nucleus is `(sqrt(2M_(alpha)T))/(M_(d))=(2)/(196)sqrt((2T)/(M_(p)))= 3.39xx10^(5)m//s` The energy of the daughter nucleus is `(M_(alpha))/(M_(p))T` and this represents a fraction `(M_(alpha)//M_(d))/(1+(M_(alpha))/(M_(d)))=(M_(alpha))/(M_(alpha+M_(d)))=(4)/(200)=(1)/(50)=0.02` of total energy. Here `M_(d)`is the mass of the daughter nucleus. |
|
| 35. |
The yeild of the nuclear reaction `C^(13)(d,n)N^(14)` has maximum magnitudes at the following values of kinetic energy `T_(i)` of bombarding deuterons: `0,60,0.90, 1.55,`and `1.80MeV`. Making use of the tabel of atomic masses, find the corresponding energy levels of the transitional nucleus through which this reaction proceeds. |
|
Answer» The reaction is `d+C^(13rarrN^(15))rarrn+N^(14)` Maxima of yields determine the energy levels of `N^(15)`. As in the previous problem the excitation enrgy is `E_(exc)=Q+E_(K)` where `E_(k)=` available kinetic enrgy. This is found is as in the previous problem. The velocity of the centre of mass is `(sqrt(2m_(d)T_(i)))/(m_(d)+m_(c ))=(m_(d))/(m_(d)+m_(c ))sqrt((2Ti)/(m_(d)))` So `E_(k)=(1)/(2)m_(d)(1-(m_(d))/(m_(d)+m_(c ))^(2)(2Ti)/(m_(d)))+(1)/(2)m_(c )((m_(d))/(m_(d)+m_(c )))^(2)(2Ti)/(m_(d))=(m_(c ))/(m_(d)+m_(c ))Ti` `Q` is the `Q` value for the ground state of `N^(15)`: we have `Q=c^(2)xx(Delta_(d)+Delta_(c )^(13)-Delta_(N)^(15))` `1=c^(2)xx(0.01410+0.00335-0.00011)am u` `=16.14MeV` The excitaiton energies then are `16.66MeV,16.92MeV` `17.49MeV` and `17.70MeV`. |
|
| 36. |
Minimum energy required to takeout the only one electron from ground state of `He^(+)` isA. 13.6 eVB. 54.4 eVC. 27.2 eVD. 6.8 eV |
|
Answer» Correct Answer - B |
|
| 37. |
What kinetic energy must a proton possess to split a deuteron `H^(2)` whose binding energy is `E_(b)=2.2 MeV`? |
|
Answer» The result of the previous problem applies and we find that energy required to split a deuteron is `Tge(1+(M_(p))/(M_(d)))E_(b)= 3.3MeV` |
|
| 38. |
A `P^(32)` radionuclide with half-life `T= 14.3` days is prodiced in a reactor at a constant rate `q= 2.7.10^(9)` nuclei per second. How soon after the beiginning of production of that radionuclide will its activity be equal to `A= 1.0.10^(9) dis//s`? |
|
Answer» `(dN)/(dt)= g(dN)/(dt)=underset("supply")underset (uarr) g-lambda underset(decay)underset(darr)N` We see that `N` will approach a constant value `(g)/(lambda)`. This can also be proved directly. Multiply by `e^(lambda t)` and write `(dN)/(dt)e^(lambda t)+lambdae^(lambda t)N="ge"^(lambda t)` Then `(d)/(dt) (Ne^(lambda t))= "ge"^(lambda t)` or `Ne^(lambda t)=(g)/(lambda)e^(lambda t)+ const` At `t=0` when the production is started, `N=0` `0= (g)/(lambda)+constant` Hence `N=(g)/(lambda)(1-e^(-lambda t))` Now the activity is `A= lambdaN= g(1-e^(-lambda t))` From the problem `(1)/(2.7)= 1-e^(-lambda t)` This gives `lambda t= 0.463` so `t=(4.463)/(lambda)=(0.463xxT)/(0.693)=9.5 days` Algebraically `t=-(T)/(In 2) In(1-(A)/(g))` |
|
| 39. |
What function of the rodioactive cobalt nuclei whose half-life is `71.3` days decays during a month? |
|
Answer» We calculate `lambda` first `lambda=(In2)/(T_(1//2))=9.722xx10^(-3)` per day Hence fraction decaying in a month `=1-e^(-lambda t)= 0.253` |
|
| 40. |
Knowing the decay constant `lambda` of a nucleus, find: (a) the probability of decay of the nucleus during the time from `0` to `t`, (b) the mean lifetime `tau` of the nucleus. |
|
Answer» (a) The probability of survival(i.e., not decaying) in time `t` is `e^(-lambda t)`. Hence the probability of decay is `1-e^(- lambdat)` (b) The probaility that the particle decays in time `dt` around time `t` is the difference `e^(-lambda t)-e^(-lambda(t+dt))= e^(-lambdat)[1-e^(e lambda dt)]= lambda e^(-lambda t)dt` Therefore the mean life time is `T= int_(0)^(oo)t lambdae^(-lambdar)dt// int_(0)^(oo) lambdae^(-lambdat)dt=(1)/(lambda) int_(0)^(oo)xe^(-x)dx//int_(0)^(oo)e^(-x)dx=(1)/(lambda)` |
|
| 41. |
Hydrogen atom emits blue light when it changes from `n = 4` energy level to the `n = 2` level. Which colour of light would te atom emit when it changes from the `n = 5` level to the `n = 2` level ?A. RedB. YellowC. GreenD. Violet |
|
Answer» Correct Answer - D |
|
| 42. |
A gold foil of mass `m= 0.20 g` was irradiated during `t=6.0` hours by a thermal neutron flux falling normally on its suface. Following `tau=12 hours` after the completion of irradiation the activity of the foil became equal to `A= 1.9.10^(7)dis//s`. Find the neutron flux density if the effective cross-section of formation of a radioactive nucleus is `sigma= 96b`, and the half-life is equal to `T=2.7 days` |
|
Answer» We apply the formula of the previous problem except that have `N=no`. Of radio nuclide and no. of host nuclei originally. Here `n=(0.2)/(197)xx6.023xx10^(23)=6.115xx10^(20)` Then after time `t N=(n.J.sigma.T)/(In 2)(1-e^(-(tIn 2)/(T)))` T= half life of the radionuclide After the source of neutrons is cut off the acitivity after time `T` will be `A=(n.J.sigma.T)/(In 2)(1-e^(-tIn2//T))e^(tauIn2//Txx(In 2)/(T))=n.J.sigma(1-e^(-tIn2//T))e^(-tauIn2//T)` Thus `J=Ae^(tau In 2//T)//n sigma(1-e^(-tIn 2//T))= 5.92xx10^(9)part//cm^(2).s` |
|
| 43. |
Find the strangeness `S` and the hypercharge `Y` of a neutral elementry particle whose isotopic spin projection is `T_(z)= +1//2` and baryon charge `B= +1`. What particle is this. |
|
Answer» From the Gell-Mann Nishijama formula `Q=T_(Ƶ)+(Y)/(2)` we get `O=(1)/(2)+(Y)/(2) or Y=-1` Also `Y=B+Simplies S-2`. Thus the particle is `=^(o)0` |
|
| 44. |
Which of the following processes are forbidden by the law of conservation of lepton charge: `(1)nrarrp+e^(_)+v`, (2) `pi^(+)rarrmu^(+)+e^(-)+e^(++)`, (3) `pi^(-)rarrmu^(-)+v`, (4)`p+e^(-)rarrn+v`, (5)`mu^(+)rarre^(+)v+overset(~)v`, (6) `K^(-)rarr mu^(-)+overset(~)v` ? |
|
Answer» (1) The process `nrarrp+e^(-)+v_(e )` cannot occur as there are `2` more leptons `(e^(-),v_(e ))` on the right compared to zero on the left. (2) The process `pi^(+)rarr mu^(+)e^(-)+e^(+)` is forbidden because this corresponds to a change of lepton number by, (`0` on the left -1 on the right) (3) The process `pi^(-)rarr mu^(-)+v_(mu)` is forbidden `mu^(-), v_(mu)` being both leptons `DeltaL=2hre`. (4),(5),(6) are allowed (except that one must distinguish between muon neutrinoes and electron neutrinoes). The correct names would be (4) `P+e^(-)rarrn+v_(e )` (5) `mu^(+)rarre^(+)+v_(e )+overset(~)v_(mu)` (6) `K^(-)rarr mu^(-)+overset(~)v_(mu)`. |
|
| 45. |
Which of the following process processes are forbidden by the law of conservation of strangeness : (1)`pi^(-)+prarr Sigma^(-)+K^(+)`, (2) `pi^(-)+prarrSigma^(+)+K^(-)`, (3)`pi^(-)+prarrK^(+)+K^(-)+n`, (4) `n+prarr ^^ ^(0)+Sigma^(+)`, (5) `pi^(-)+nrarr Xi^(-)+K^(+)+K^(-)`, (6) `K^(-)+prarr Omega+K^(+)+K^(0)` ? |
|
Answer» (1) `pi^(-)+prarrSigma^(-)+K^(+)` `0 0 -1 1` so `DeltaS=0` allowd. (2) `pi^(-)+prarrSigma^(+)+K^(-)` `0 0 -1 1` so `DeltaS= -2` forbiddeb (3) `pi^(-)+prarrK^(-)+K^(+)+n` `0 0 rarr-1 1 0` so `DeltaS=0`, allowed (4)`n+prarr^^ ^(0)+Sigma^(+)` `0 0 -1 -1` so `DeltaS= -2`. forbidden (5) `pi^(-)+nrarr=^(-)+K^(+)+K^(-)` `0 0rarr -2 1 -1` so `DeltaS= -2`. forbidden. (6) `K^(-)+prarrSigma^(-)+K^(+)K^(o)` `-1 0 3+1+1` so `DeltaS=0` allowed. |
|
| 46. |
Indicate the reasons why the following processes are forbidden: (1) `Sigma^(-)rarr^^ ^(0)+pi^(-)`, (2) `pi^(-)+prarrK^(+)+K^(-)`, `K^(-)+n rarr Omega^(-)+K^(+)+K^(0)`, (6) `murarr e^(-)+v_(e )+v_(mu)`. |
|
Answer» (1)` Sigma^(-)Sigma^(-)rarr^^ ^(0)+pi^(-)`, (2) `pi^(-)+prarr K^(+)+K^(-)`, `K^(-)+n rarr Omega^(-)+K^(+)+K^(0)`, (6) `murarr e^(-)+v_(e )+v_(mu)`. is forbidden by energy conservation. The mass difference `M_(Sigma)-M_(^^ ^(0))= 82(MeV)/(c^(2)) lt m_(pi)^(-)` ( The process `1rarr 2+3` will be allowed only if `m_(1)gt m_(2)+m_(3)`.) (2) `pi^(-)+prarrK^(+)K^(-)` is disallowed by conservation of baryon number. (3) `K^(-)+n rarr Omega^(-)+K^(+)+K^(0)` is forbidden by conservation of charge (4) `n+prarr Sigma^(+)^^ ^(0)` is forbidden by strangerness conservation. (5) `pi^(-)rarrmu^(-)+e^(-)+e^(+)` is forbidden by conservation of muon number (or lepton number). (6) `mu^(-)rarre^(-)+v_(e )+ overset(~)v_(mu)` is forbidden by the seperate conservation of muon number as well as lepton number. |
|
| 47. |
A narrow stream of electrons with kinetic enrgy `T= 10keV` passes through a polycrystalline aluminium foil, forming a system of diffraction fringes on a screen. Calculate tha interplanar distacne corresponding to the reflection of third order from a certain system of crystal planes if it is responsible for a diffraction ring of diameter `D= 3.20cm`. The distacnce between the foil and the screen is `l=10.0cm`. |
|
Answer» See the analogous problem with `X`- rays (5.156) The glancing angle is obtained from `tan 2 theta=(D)/(2l)` where `D =` diameter of the ring, `l=` distance from the foil to the screen. Then for the third order Bragg reflection `2 d sin theta=k lambda=k(2pi ħ)/(sqrt(2mT)),(k=3)` Thus `d=(pi ħk)/(sqrt(2mTsin theta))=0.232nm` |
|
| 48. |
A stream of electrons accelerated by a potential difference `V` falls on the surface of a metal whose inner potential is `V_(i)= 15V`. Find: (a) the refractive index of the metal for the electrons accelerated by a potential difference `V= 150V`, (b) the values of the raito `V//V_(i)` at which the refracticve index differs from unity by not more than `eta=1.0%`. |
|
Answer» Inside the metal, there is a negative potential energy of `-eV_(i)`. (This potential energy prevents electron from leaking out and can be measured in photoelectric effect ets.)An electron whose `K.E` is `eV` outside the metal will find its `K.E` increased to `e(V+V_(i))` in the metal. Then (a) de Broglie wavelength in the metal `=lambda_(m)=(2piħ)/(sqrt(2m e (V+V_(i))))` Also de Broglie wavelength in Vacuum `=lambda_(0)=(2pi ħ)/(sqrt(2 mVe))` Hence refractive index `n=(lambda_(0))/(lambda_(m))=sqrt(1+(V_(i))/(V))` Substituting we get `n=sqrt(1+(1)/(10))~=1.05` (b) `n-1= sqrt(1+(V_(i))/(V))-1leeta` then `1+(V_(i))/(V)le(1+eta)^(2)` or `V_(i)leeta(2+eta)V` or `(V)/(V_(i))ge(1)/(eta(2+eta))` For `eta= 1% = 0.01` we get `(V)/(V_(i))ge50` |
|
| 49. |
A certain atom has three electrons `(s,p and d)` in addition to filled shells, and is in a state with the greatest possible total mechanical moment for a given configuraion. In the corresponding vector model of the atom find the angle between the spin momentum and the total anular momentum of the given atom |
|
Answer» The total angular momentum is greatest when `L,S` are both greatest and add to form `J`. Now for triplet of `s,p,d` electrons Maximum spin `rarr S=(3)/(2)` corresponding to `M_(s)ħsqrt((3)/(2).(5)/(2))=(ħsqrt(15))/(2)` Maximum orbital angular momentum `rarr` `L=3` corresponding to `M_(L)=ħsqrt((3)/(2).(5)/(2))=(ħsqrt(15))/(2)` Maximum total angular momentum `J=(9)/(2)` corresponding to `M=(ħ)/(2)sqrt(99)` In vector model `vec(L)=vec(J)-vec(S)` or in magnitude squared `L(L+1)ħ^(2)=J(J+1)ħ^(2)+S(S+1)ħ^(2)-2vec(J).vec(S)` Thus `cos( lt vec(J),vec(S))=(J(J+1)+S(S+1)-L(L+1))/(2sqrt(J(J+1)sqrt(S(S+1))))` Substitution gives `lt (vec(J),vec(S))=31.1^(@)`. |
|
| 50. |
The mean path length of alpha-particles in air under standard conditions is defined by the formula `R=0.98.10^(27)v_(0)^(3) cm`, this formula, find for an alpha-particle with initial kinetic enrgy `7.0 MeV`: (a) its mean path length, (b) the average number of ion pairs formed by the given alpha-particle over the whole path `R` as well as over its first half, assuning the ion pair formation energy to be equql to `34 eV`. |
|
Answer» (a) For an alpha particle with initial `K.E 7.0 MeV`, the initial velocity is `V_(0)=sqrt((2T)/(M alpha))` `=sqrt(2xx7xx1.602xx10^(-6))/(4xx1.672xx10^(-24))` `= 1.83xx10^(9)cm//sec` Thus `R=6.02 cm` (b) Over the whole path the number of ion pairs is `(7xx10^(6))/(34)= 2.06xx10^(5)` Over the first half of the path:- We write the formula for the mean path as `R alphaE^(3//2)` where `E` is the initial energy. Thus if the energy of the `alpha` particle after traversing the first half of the path is `E_(1)` then `R_(0)E_(1)^(3//2)=(1)/(2)R_(0)E_(0)^(3//2) or E_(1)= 2^(-2//3)E_(0)` Hence number of ion pairs formed in the first half of the path length is `(E_(0)-E_(1))/(34eV)=(1-2^(-2//3))xx2.06xx10^(5)= 0.76xx10^(5)` |
|