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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.
| 151. |
A narrow beam of protons with velocity `v=6.10^(6)m//s` falls normally on a silver foil of thickness `d= 1.0 mu m`. Find the probability of the protons to be scattered into the rear hemisphere `(theta gt 90^(@))`. |
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Answer» The requisite probability can be written easily by analogy with (b) of the presious problem. It is `P=(N(pi//2))/(I_(0)tau)=nd((Ze^(2))/((4piepsilon_(0))2mv^(2)))^(2)4piint_(pi//2)^(x)(cos theta//2d theta)/("sin"^(3)(theta)/(2))` The intergal is unity. Thus `P= pind((Ze^(2))/((4pi epsilon_(0))mv^(2)))^(2)` Substitution gives using `n=(rho_(Ag)N_(A))/(A_(Ag))=(10.5xx10^(3)xx6.023xx10^(26))/(108),P= .006` |
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| 152. |
Detemine how many times the intensity of a narrow beam of thermal neutrons will decrease after passing through the heavy water layer of thickness `d=5.0 cm`. The effective cross-section are of neutrons quitting the beam due to scattering if the thickness of the plate is `d= 0.50 cm`. The effective cross-section of interaction of deuterium and oxygen nuclei with thermal neutrons are equal to `sigma_(1_= 7.0 b ` and `sigma_(s)= 11 b` respectivley. |
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Answer» Here `(1)/(eta)=e^(-n_(2)sigma_(2)+n_(1)sigma_(1)d)` where `1` refers to `O^(1)` and 2 o r `D` nuclei Using `n_(2)=2n,n_(1)=n=` concentration of `O` nuclei in heavy water we get `(1)/(eta)= e^(-2sigma_(2)+sigma_(1)nd)` Now using the data for heavy water `n=(1.1xx6.023xx10^(23))/(20)= 3.313xx10^(22)per c c` Thus substituting the values `eta= 20.4=(I_(0))/(I)` |
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| 153. |
The shortest wavelength in the Lyman series of hydrogen spectrum is `912 Å` correcponding to a photon energy of `13.6eV`. The shortest wavelength in the Balmer series is aboutA. `3648 Å`B. `8208 Å`C. `1228 Å`D. `6566 Å` |
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Answer» Correct Answer - A |
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| 154. |
Radius of the first orbit of the electron in a hydrogen atom is `0.53 Å` . So, the radius of the third orbit will beA. `2.12 Å`B. `4.77 Å`C. `1.06 Å`D. `1.59 Å` |
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Answer» Correct Answer - B |
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| 155. |
The kinetic energy of electron in the first Bohr orbit of the hydrogen atom isA. `-6.5 eV`B. `-27.2 eV`C. `13.6 eV`D. `-13.6 eV` |
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Answer» Correct Answer - C |
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| 156. |
As the electron in the Bohr orbit is hydrogen atom passes from state `n = 2` to `n = 1` , the `KE (K) and PE (U)` change asA. K two-fold, U four-foldB. K four-fold, U two-foldC. K four-fold, U also four-foldD. K two-fold, U also two-fold |
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Answer» Correct Answer - C |
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| 157. |
The radius of the Bohr orbit in the ground state of hydrogen atom is `0.5 Å`. The radius o fthe orbit of the electron in the third excited state of `He^(+)` will beA. `8 Å`B. `4 Å`C. `0.5 Å`D. `0.25 Å` |
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Answer» Correct Answer - B |
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| 158. |
The fraction `f` of radioactive material that has decayed in time `t`, varies with time `t`. The correct variation id given by the curve. .A. AB. BC. CD. D |
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Answer» Correct Answer - B |
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| 159. |
If in hydrogen atom, radius of `n^(th)` Bohr orbit is , `n_(r)` frequency of revolution of electron in `n^(th)` orbit is `f_(n)` choose the correct optionA. B. C. D. Both (a) and (b) |
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Answer» Correct Answer - D |
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| 160. |
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. `Y rarr 2Z`B. `W rarr X + Z`C. `W rarr 2Y`D. `X rarr Y + Z` |
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Answer» Correct Answer - C |
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| 161. |
From the condition of the foregoing problem find: (a) normalized eigenfunctions of the particle in the states for which `Psi(r )` depends only on `r` , (b) the most probable value `r_(pr)` for the ground state of the particle and probability of the particle to be in in the region `r lt r_(pr)` |
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Answer» (a) The nomalized wave functions are obtained from the normalization `1=int|Psi|^(2)dV=int|Psi|^(2) 4pir^(2)dr` `=int_(0)^(ro)A^(2)4pichi^(2)dr=4piA^(2)int_(0)^(ro)"sin"^(2)(npi r)/(r_(o))dr` `=4piA^(2)(r_(0))/(npi)int_(0)^(bar(npi))sin^(2)xdr=4piA^(2)(r_(0))/(npi).(npi)/(2)=r_(0).2piA^(2)` Hence `A=(1)/(sqrt(2pir_(0)))` and`Psi=(1)/(sqrt(2pi.r_(0)))("sin"(npir)/(r_(0)))/(r)` (b) The radial probability distribution function is `P_(n)(r )=4pir^(2)(Psi)^(2)=(2)/(r_(0)) "sin"^(2)(npir)/(r_(0))` For the ground state `n=1` so `P_(1)(r )=(2)/(r_(0))"sin"^(2)(pi r)/(r_(0))"sin"^(2)(pi r)/(r_(0))` By inspection this is maximum for `r=(r_(0))/(2)`. Thus `r_(pr)=(r_(0))/(2)` The probability for the particle to be found in the region `r lt r_(pr)` is clearly `50%` as one can immediately see from a graph of `sin^(2)x`. |
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| 162. |
Making use of the solution of the foregoing problem, determine the probability of the particle with energy `E=U_(0)//2` to be located in the region `xgtl`, if `l^(2)U_(0)=((3)/(4)pi)^(2)( ħ^(2))/(m)`. |
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Answer» `U_(0)l^(2)=((3)/(4)pi)U_(0)l^(2)=((3)/(4)pi)^(2)( ħ^(2))/(m)` and `El^(2)=((3)/(4)pi)^(2)( ħ^(2))/(2m)` or `kl=(3)/(4)pi` It is easy to check that the condition of the bound state is satisfied. Also `alphal=sqrt((2m)/ħ^(2)(U_(0-E))l^(2))= sqrt((mU_(0))/ħ^(2)l^(2))=(3)/(4)pi` Then from the previous problem `D=Ae^(alphal)sin kl=A(e^(3//4))/(sqrt(2))` By normalization `I=A^(2)[ int_(0)^(l)sin^(2)kxdx+int_(l)^(oo)(3^(3pi//2))/(2)e^(-(3pi//2)x//l)dx]` `A^(2)[(1)/(2)int_(0)^(1)(1-cos 2kx)dx+1int_(0)^(oo)(1)/(2)e^(-3x)/(2)ydy]` `=A^(2)[(1)/(2)[-(sin 2kl)/(2k)]+(1)/(2).((l)/(3pi))/(2)]=A^(2)l[(1)/(2)[1+((1)/(3pi))/(2)]+(1)/(2)((1)/(3pi))/(2)]` `A^(2)l[(1)/(2)+(2)/(3pi)]=A^(2)(l)/(2)(1+(4)/(3pi))` or `A =sqrt((2)/(l))(1+(4)/(3pi))^(-1//2)` The probability of the particle to be located in the region `x gt l` is `P=int_(l)^(oo)Psi^(2)dx=(2)/(l)(1+(4)/(3pi))^(-1)int_(l)^(oo)(e^(3)pi//2)/(2) e^(-(3 x)/(2)(x)/(l))dx` `=(1+(4)/(3pi))^(-1)int_(l)^(oo)e^(3pi//2)e^(-(3pi//2)y)dy=(2)/(3pi)xx(3pi)/(3pixx4)=14.9%`. |
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| 163. |
Using the Schrodinger equation, demonstrate that at the point where the potential energy `U(x)` of a particle has a infinte discountinuity, the wave function renains smooth, I.e., its first derivaltive with respect to the coordinate is continuous. |
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Answer» We can for definiteness assume that the discontinuity occurs at the point `x=0`. Now the schordinger equation is `( ħ^(2))/(2m)(d^(2)Psi)/(dx^(2))+U(x)Psi(x)=EPsi(x)` We intergate this equation around `x=0`i.e., from `x=-epsilon_(1) to x=epsilon_(2)` where `epsilon_(1),epsilon_(2)` are small positives numbers. Then `( ħ^(2))/(2m)int_(-epsilon_(1))^(+epsilon_(2))(d^(2)Psi)/(dx^(2))dx=int_(-epsilon_(1))^(+epsilon_(2))(E-U(x)Psi(x)dx)` ltbrltgt or `((dPsi)/(dx))_(+epsilon_(2))=((d Psi)/(dx))_(-epsilon_(1))=-(2m)/ (ħ^(2)) int_(-epsilon_(1))^(epsilon_(2))(E-U(x))_(dx)Psi(x)` SInce the potential and the energy `E` are finite and `Psi(x)` is bounded by assumption, the intergaral on the right exists and `rarr0 asepsilon_(1),epsilon_(2)rarr0` Thus `((d Psi)/(dx))_(epsilon_(2))=((dPsi)/(dx))_(-e_(1)) as epsilon_(1),epsilon_(2)rarr0` So `((dPsi)/(dx))` is continuous at `x=0` (the point where `U(x)` has a infinite jump discontuinuityI `((d Psi)/(dx))_(epsilon_(2))=((dPsi)/(dx))_(-e_(1)) as epsilon_(1),epsilon_(2)rarr0` So `((dPsi)/(dx))` is continuous at `x=0` (the point where `U(x)` has a infinite jump discontuinuity. |
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| 164. |
For electron moving in `n^(th)` orbit of the atom , the angular velocity is proportional to:A. nB. `1//n`C. `n`D. `1//n` |
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Answer» Correct Answer - D |
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| 165. |
Which one of the relation is correct between time period and number of orbits while an electron is revolving in a orbitA. `n^(2)`B. `(1)/(n^(2))`C. `n^(3)`D. `(1)/(n)` |
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Answer» Correct Answer - C |
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| 166. |
Energy of an electron in n th orbit of hydrogen atom is `(k = (1)/(4 pi epsilon_(0)))`A. `- (2pi^(2)k^(2)me^(4))/(n^(2)h^(2))`B. `- (4 pi^(2)m ke^(2))/(n^(2)h^(2))`C. `- (n^(2)h^(2))/(2 pi k me^(4))`D. `- (n^(2)h^(2))/(4pi^(2) k me^(2))` |
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Answer» Correct Answer - A |
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| 167. |
The first member of the paschen series in hydrogen spectrum is of wavelength `18,800 Å`. The short wavelength limit of Paschen series isA. `1215 Å`B. `6560 Å`C. `8225 Å`D. `12850 Å` |
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Answer» Correct Answer - C |
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| 168. |
Find the binding energy of a valence electrons in the ground state of a `Li` atom if the wavelength of the first line of the sharp series is known to be equal to `lambda_(1)=813nm` and the short-wave cut-off wavelength of that series to `lambda_(2)=350nm`. |
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Answer» For the first line of the sharp series `(3Srarr2P)` in a `Li` atom `(2piħC)/(lambda_(1))=-(ħR)/((3+alpha_(0))^(2))+(ħR)/((2+alpha_(1))^(2))` For the short wave cut-off wave-length of the same series `(2piħc)/(lambda_(2))=(ħR)/((2+alpha_(1))^(2))` From these two equations we get on substraction `3+alpha_(0)=sqrt(ħR//(2piħc(lambda_(1)lambda_(2)))/(lambda_(1)lambda_(2)))` `=sqrt((Rlambda_(1)lambda_(2))/(2piCDeltalambda)),Delta lambda=lambda_(1)-lambda_(2)` Thus in the ground state, the binding energy of the electron is `E_(b)=(ħR)/((2+alpha_(0))^(2))=ħR//(sqrt(Rlambda_(1)lambda_(2)/(2picDeltalambda)-1))^(2)=5.32eV` |
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| 169. |
Calculate in the atomic mass units the mass of (a) a `Li^(8)` atom whose nucleus has the binding energy `41.3MeV`, (b) a `C^(10)` nucleus whose binding energy per uncleon is equal to `6.04 MeV`. |
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Answer» (a) We have for `Li^(8)` `41.3MeV= 0.044361 am u=3Delta_(H)+5Delta_(n)-Delta` Hence `Delta=3xx0.00738+5xx0.00867-0.09436= 0.02248` amu (b) For `C^(10) 10xx6.04= 60.4MeV` `=0.06488` `=6 Delta_(H)+4Delta_(n)-Delta` Hence `Delta= 6xx0.00783+4xx0.00867-0.01678` amu Hene the mass of `C^(10) is 10.01678` |
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| 170. |
A stationary neutral particle disintegrated into a proton with kinetic energy `T= 5.3MeV` and a negative pion. Find the mass of that particle. What is its name. |
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Answer» By conservation of energy-momentum `Mc^(2)=E_(p)+E_(pi)` `O=vec(p_(p))+vec(p_(pi))` Then `m_(pi)^(2)c^(4)=E_(pi)^(2)-vec(p_(pi))c^(2)=(Mc^(2)-E_(p))^(2)-c^(2)vec(P_(p))` `=M^(2)c^(4)-2MC^(2)E_(p)+m_(p)^(2)c^(4)` This is a quadratic equation in `M` `M^(2)-2(E_(p))/(c^(2))M+m_(p)^(2)+m_(pi)^(2)=0` or using `E_(p)=m_(p)c^(2)+T` and solving `(M-(E_(p))/(c^(2)))^(2)=(E_(p)^(2))/(c^(4))-m_(p)^(2)+m_(pi)^(2)` Hence, `M=(E_(p))/(c^(2))+sqrt(m_(pi)^(2)+(T)/(c^(2))(2m_(p)+(T)/(c^(2))))` From the table of masses we identifiy the particle as a `^^` particle |
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| 171. |
Find the number of neutrons generated per unit time in a uranium reactor whose thermal power is `p=100MW` if the average number of neutrons liberated in each nuclear splitting is `v=2.5`. Each splitting is assumed to release an energy `E=200MeV`. |
| Answer» `N_(0)` of fissions per unit time is clearly `P//E`. Hence no. of neutrons produced per unit time to `(vP)/(E )`. Substitution gives `7.80xx10^(18) n eutrons//sec` | |
| 172. |
When the wave of hydrogen atom comes from infinity into the first then the value of wave number isA. 109700 cmB. 1097 cmC. 109 cmD. None of these |
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Answer» Correct Answer - A |
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| 173. |
In the following transitions, which one has higher frequency ?A. `3 - 2`B. `4 - 3`C. `4 - 2`D. `3 - 1` |
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Answer» Correct Answer - D |
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| 174. |
Which of the following transitions will have highest emission wavelength ?A. `n = 2 " to " n = 1`B. `n = 1 " to " n = 2`C. `n = 2 " to " n = 5`D. `n = 5 " to " n = 2` |
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Answer» Correct Answer - D |
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| 175. |
Which of the following transitions in a hydrogen atom emits photon of the highest frequency ?A. `n = 1 " to " n = 2`B. `n = 2 " to " n = 1`C. `n = 2 " to " n = 6`D. `n = 6 " to " n = 2` |
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Answer» Correct Answer - A |
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| 176. |
In a hydrogen atom, which of the following electronic transitions would involve the maximum energy changeA. From `n = 2 " to " n = 1`B. From `n = 3 " to " n = 1`C. From `n = 4 " to " n = 2`D. From `n = 3 " to " n = 2` |
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Answer» Correct Answer - B |
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| 177. |
The absorpotion transitions between the first and the fourth energy states of hydrogen atom are `3`. The emission transitions between these states will beA. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - D |
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| 178. |
The energy levels of the hydrogen spectrum is shown in figure. There are some transitions `A, B, C, D` and `E`. Transition `A, B`, and `C` respectively represent A. First member of Lyman series, third spectral line of Balmer series and the second spectral line of Paschen seriesB. Ionization potential of hydrogen, second spectral line of Balmer series and third spectral line of Paschen seriesC. Series limit of Lyman series, third spectral line of Balmer series and second spectral line of Paschen seriesD. Series limit of Lyman series, second spectral line of Balmer series and third spectral line of Paschen series |
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Answer» Correct Answer - C |
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| 179. |
At a certain voltage applied to an `X`-rays tuve with aluminium anticathode the short-wave cut-off wavelength of the continuous `X`-rays spectum is equal to `0.50nm`. Will the `K` series of the characteristic specturm whose excitation potential is equal to `1.56kV` be also observed in this case? |
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Answer» Since the short wavelength cut off of the continuous specturm is `lambda_(0)=0.50nm` the voltage applied must be `V=(2 pi ħ )/(2 lambda_(0))=2.48kV` since this is greater than the excitation potential of the `K` series of the characteristic specturm (which is only `1.56kV`) the latter will be observed. |
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| 180. |
Find the de Broglie wavelength of relaativistic electrons reaching the anticathode of an `X`-ray tube if the short wavelength limit of the continous `X`-rays spectrum is equal to `lambda_(sh)= 10.0p m`? |
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Answer» For relativistic elelctrons, the formula for the short wavelength limit is `X`-rays will be `(2pi ħc)/(lambda_(sh))=m_(0)c^(2)((1)/(sqrt(1-beta^(2)))-1)=csqrt(P^(2)+m^(2)c^(2))-mc^(2)` or `((2pi ħ)/(lambda_(sh))+mc)^(2)=P^(2)+m^(2)c^(2)` or `((2piħ)/(lamda_(sh)))((2piħ)/(lambda_(sh))+2mc)=P^(2)` or `p=(2piħ)/(lambda_(sh))sqrt(1+(mc lambda_(sh))/(piħ))` Hence `lambda_(sh)=lambda_(sh)//sqrt(1+(mc lambda_(s h))/(piħ))-3.29p m` |
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| 181. |
Find the posible multiplicities `x` of the terms of the types (a)`.^(x)D_(2), (b).^(x)P_(a//2), (c ).^(x)F_(1)`. |
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Answer» (a) Here `J=2,L=2`. Then `S=0,1,2,3,4` and the multiplicities `(2S-1)` are `1,3,5,7,9` (b) Here `J=3//2,L=1` Then `S=(5)/(2),(3)/(2),(1)/(2)` and the multiplicities are `6,4,2` (c )Here `J=1,L=3`. Then `S=2,3,4` and the multiplities are `5,7,9` |
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| 182. |
Into what number of sublevels are the following terms split in a weak magnetic field: (a)`.^(3)P_(0)`, (b) `.^(2)F_(5//2)`, (c ) `.^(4)D_(1//2)` ? |
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Answer» (a) The term `3P_(0)` does not split in weak magnetic firld as it has zero total angular momentum. (b) The term `.^(2)F_(5//2)` will split into `2xx(5)/(2)+1=6` sublevels. The shift in each sublevel is given by `DeltaE= -g mu_(B)M_(Ƶ)B` where `M_(j)= -J(J-1),....,J` and `g` is the Landi factor `g=1+((5xx7)/(4)+(1xx3)/(4)-3xx4)/(2xx(5xx7)/(4))=1+(38-48)/(70)=(6)/(7)` (c ) In this case for the `.^(4)D_(1//2)` term `g=1+((1xx)/(4)+(3xx5)/(4)-2xx3)/(2xx(1xx3)/(4))=1+(3+15-24)/(6)=1-1=0` Thus the energy difference vanish and the level does not split. |
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| 183. |
A system comprises an atom in `.^(2)P_(3//2)` state and a `d` ekectron. Find the possible spectral terms of that system. |
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Answer» The atom has `s_(1)=1//2, l_(1)=1,j=(3)/(2)` The electron has `S_(2)=(1)/(2), l_(2)=2` so the total angular momentum quantum number must be `j_(2)=(3)/(2)` or `(5)/(2)` ltbtgt In `L-S` compling we get `S=0,1.L=1,2,3` and the terms that can be formed are the same as written in the problem above. The possible values of angular momentum are consistant with the addition `j_(1)=(3)/(2) to j_(2)=(3)/(2)` or `(5)/(2)`. The latter gives us `J=0,1,2,3, 1,2,3,4` All these values are reached above. |
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| 184. |
An atom in the state `.^(2)P_(3//2)` is located on the axis of a loop of radius `r=5 cm` carrying a current, `I=10 A`. The distance between the aotm and the centre of the loop is equal to the radius of the latter. How great may be the maximum force that the magnetic field of that current exerts on the atom? |
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Answer» The force on an atom with magnetic moment `vec(mu)` in a magnetic field of induction `vec(B)` is given by `vec(F)=(vec(mu).vec(grad))vec(B)` In the present case, the maximum force arise when `vec(mu)` is along the axis of close to it. Then`F_(Ƶ)=(mu_(Ƶ))_(max)(del B)/(delƵ)` Here `(mu_(Ƶ))_(max)=gmu_(B)J`. The Lande factor `g` is for `^(2)P_(1//2)` `g=1+((1)/(2)xx(3)/(2)+(1)/(2)xx(3)/(2)-1xx2)/(2(1)/(2)xx(3)/(2))=1-(1//2)/(3//2)=(2)/(3)` and `J=(1)/(2) so (mu_(Z))_(max)=(1)/(3)mu_(B)`. The magnetic field is given by `B_(Ƶ)=(mu_(0))/(4 pi).(2Ipi r^(2))/((r^(2)+Ƶ^(2))^(3//2))` or `(del B_(Ƶ))/(delƵ)=-(mu_(0))/(4pi)6I pir^(2)(Ƶ)/((r^(2)+Ƶ^(2))^(5//2))` Thus `((del B_(Ƶ))/(delƵ=r))=(mu_(0))/(4 pi)(3I pi)/(sqrt(8)r^(2))` Thus maximum force is `F=(1)/(3)mu_(B)(mu_(0))/(4 pi)(3 pi)/(sqrt(8))(I)/(r^(2))` Substitution gives (using data in `MKS` units) `F=4.1xx10^(-27)N` |
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| 185. |
Energy of the electron in `nth` orbit of hydrogen atom is given by `E_(n) =-(13.6)/(n^(2))eV`. The amount of energy needed to transfer electron from first orbit to third orbit isA. 13.6 eVB. 3.4 eVC. 12.09 eVD. 1.51 eV |
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Answer» Correct Answer - C |
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| 186. |
A hydrogen atom in the normal state is locatred at a distance `r=2. 5cm` from a long straigh condctor carrying a current `I= 10A`. Find the force acting on the atom. |
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Answer» The magnetic field at a distacne `r` from a long current carrying wire is mostly tangential and given by `B_(varphi)=(mu_(0)I)/(2 pi r)=(mu_(0))/(4pi)(2I)/(r )` The force on a magnetic dipole of mement `vec(mu)` due to this magnetic iels is also tangential and has magnitude `(vec(mu).grad)B_(varphi)` This force is nonvanishing only when the component of `vec(mu)` along `vec(r )` non zero. Then `F=mu_(r )(del)/(del r)B_(varphi)= -mu_(r )(mu_(0))/(4pi)(2I)/(r^(2))` Now the maximum value of `mu_(r )=+-mu_(B)` Thus the force is `F_(max)=mu_(B)(mu_(0))/(4pi)(2I)/(r^(2))= 2.97xx10^(-26)N` |
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| 187. |
A particle in the ground state is located in a unidimesional square potential well of length `l` with absolutely impenetrable walls `(0 lt x lt l)`. Find the probability of the particle staying within a region `(1)/(3)l le x le (2)/(3)l`. |
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Answer» We look for the solution of Schrodinger eqn. with `-(ħ^(2))/(2m)(d^(2)Psi)/(dx^(2))=E Psi,0lexlel`….(1) The boundary condition of impenetrable walls means `Psi(x)=0` for `x=0` and `x=l` (as `Psi(x)=0` for `xlt0` and `xgtl`,) Then solution of (1) is `Psi(x)=A "sin"sqrt(2mE)/(ħ)x+B "cos"sqrt(2mE)/(ħ)x` Then `Psi(0)=0implies B=0` `Psi(l)=0implies A "sin"sqrt(mE)/(ħ)l-0` `A~~ 0` so `sqrt(2mE)/(ħ)l=n pii` Hence `E_(n)=(n^(2)pi^(2)ħ^(2))/(2ml^(2)),n= 1,2,3`.... Thus the ground state wave fucnction is `Psi(x)=A "sin"(pi x)/(l)` We evaluate `A` by nomalization `1=A^(2)int_(0)^(i)"sin"^(2)(pi x)/(l)dx=A^(2)(l)/(pi)int_(0)^(x)sin^(2) theta d theta=A^(2)(l)/(pi).(pi)/(2)` Thus `A+(sqrt(2)/(l))` Finally, the probability `P` for the particle to lie in `(l)/(2) le (2l)/(3)` is `P=P((l)/(3)lexle(2l)/(3))=(2)/(l)int_(l/3)"sin"^(2)(pix)/(l)dx` `=(2)/(pi)int_(pi//3) sin^(2) theta d theta=(1)/(pi)int_(pi//3)^(2pi//3)(1- cos 2 theta)d theta` `(1)/(pi)(theta-(1)/(2) sin 2 theta)^(2pi//3)=(1)/(pi)((2pi)/(3)-(pi)/(3)-(1)/(2)"sin"(4pi)/(3)+(1)/(2)`sin`(2pi)/(3))` `=(1)/(pi)((pi)/(3)+(1)/(2)(sqrt(3))/(2)+(1)/(2)(sqrt(3))/(2))=(1)/(3)+(sqrt(3))/(2pi)= 0.609` |
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| 188. |
Negative points with kinetic energy `T=100 MeV` travel an average distance `l=11m` from origin to decay. Find the proper lifetime of these poins. |
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Answer» Energy of pions is `(1+eta)m_(0)c^(2)` so `(1+eta)m_(0)c^(2)=(m_(0)c^(2))/(sqrt(1-beta^(2)))` Hence `(1)/(sqrt(1-beta^(2)))=1+eta or beta=(sqrt(eta(2+eta)))/(1+eta)` Here `beta=(v)/(c )` of pion. Hence time dilation factor is `1+eta` and the distance traversed by the pion in its lifetime will be `(c betatau_(0))/(sqrt(1-beta^(2)))=ctau_(0)sqrt(eta(2+eta))= 15.0 meters` on substituting the values of various quantities. (Note. The factor `(1)/(sqrt(1-beta^(2))` can be looked at as a time dilation effect in the laboratory frame or as length contraction factor brought to the other side in the proper frame of the pion). |
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| 189. |
Determine the overall degeneracy of a `3D` state of a `Li` atom. What is the physical meaning of that value? |
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Answer» For a `3d` state of `Li` atom, `S=(1)/(2)` because there is only one electron and `L=2`. The total degeneracy is `g=(2L+1)(2S+1)=5xx2=10`. The states are `.^(2)D_(3/2)` and `.^(2)D_(5//2)` and we check that `g= 4+6=(2xx(3)/(2)+1)+(2xx(5)/(2)+1)` |
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| 190. |
Write the spectral designation of the term whose degeneracy is equal to seven and the quantum numbers `L` and `S` are interrelated as `L=3S`. |
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Answer» The degeneracy is `2J+1`. So we must have `J=3` From `L=3S`, we see that `S` must be an integer since `L` is intergal and `S` can be either intergal or half intergal. If `S=0` then `L=0` but this is consistent with `J=3`. For `S ge2,L ge6` and then `J=3`. Thus the state is `.^(3)F_(3)` |
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| 191. |
In a hydrogen atom, the distance between the electron and proton is `2.5 xx 10^(-11)m`. The electricl force of attraction between then will beA. `2.8 xx 10^(-7) N`B. `3.7 xx 10^(-7) N`C. `6.2 xx 10^(-7) N`D. `9.1 xx 10^(-7) N` |
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Answer» Correct Answer - B |
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