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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.
| 1651. |
In are oil drop experiment , the following charges (in arhitrary units )were found an a series of oil droplets .Calculate the magnitude of the charge on the elecron. `3 xx 10^(-19),9xx 10^(-15),12 xx 10^(-15), 18 xx 10^(-15)` |
| Answer» The magnitude of the charge should be smallest and other changes should be integrate multiples of that smaller charge so, in the problem, the smaller charge in `3 xx 10^(-15)` and is also an integral multiple of this charge | |
| 1652. |
In hydrogen atom which enrgy level order si bnot correct ?A. `1s lt 2p`B. ` 2p = 2 s`C. ` 2p gt 2s`D. ` 2plt 3s` |
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Answer» Correct Answer - C In H-atom subshells fo a shell possess same enrgy level. |
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| 1653. |
The wave number of the first line in the balmer series of `Be^(3+)` ? |
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Answer» `Delta E (Be) = Z^(2) Delta E(H)` `bar v (Be) = 16 bar v (H)` `= 16 (15200 cm^(-1)) = 2.43 xx 10^(5) cm^(-1)` |
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| 1654. |
In an oil drop experiment, the charges on oil drops were found as `1.5 xx 10^(-15), 3 xx 10^(-15), 4.5 xx 10^(-15), 6.0 xx 10^(-15)`. Calculate the magnitude of the charge on the electron. |
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Answer» The magnitude of charge shoule be smallest & other charges should be integral multiples of that smallest charge `(q = n e)`. So in the problem, the smallest charge is `1.5 xx 10^(-15)` and is also an integral multiple of all other charges. case 1: `1.5 xx 10^(-15) = 1 xxe` case 2: `3xx 10^(-15) = 2xx e` case 3: `4.5 xx 10^(-15) = 3 xx e` case 4 : `6 xx 10^(-15) = 4 xx e` so charge on the electrons is same in all cases and it will be `1.5 xx 10^(-15)` |
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| 1655. |
The mass numbers of three isotopes of an element are 10,12,14 units. Their percentage abundance is 80,15 and 5 respectively. What is tha tomic weight of the element?A. `10.5`B. `11.5`C. `12.5`D. `13.5` |
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Answer» Correct Answer - A Avg. at. wt `=(sum% "abundance" xx "Atomic weight")/("Totalratio")` |
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| 1656. |
Penetration power of proton isA. More than electronB. Less than electronC. More than neutronD. None |
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Answer» Correct Answer - B Proton is 1837 (approx 1800) times heavier than an electron. Penetration power `prop 1/"mass"` |
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| 1657. |
Calculate the momentum of a moving particle which has a wavelength of `200 nm` |
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Answer» Wavelength of particle `= 200 nm = 200 xx 10^(-9) m` According to de Broglie relationship `lambda = (h)/(P)` `Momentum P = (h)/(lambda)= (4.626 xx 10^(-34)J s)/(200 xx 10^(-9)m)= 3.313 xx 10^(-27) kg m s^(-1)` |
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| 1658. |
In what ratio should `._(17)CI^(37)` and `._(17)CI^(35)` be presents so as to obtain `._(17)CI^(35.5)`?A. `1:2`B. `1:1`C. `1:3`D. `3:1` |
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Answer» Correct Answer - C Let `x%` of `CI^(17) `and `(100 - x)%` of` CI^(15)` be presents in `._(17)CI^(35.5)` `:. 35.5 = (x xx 37 + (100 - x) xx 35)/(100)` Solve for x, `x = 25` `:. % of CI^(37) = 25` `:. % of CI^(o+) = 75` `:. Ratio is 1:3` |
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| 1659. |
The radius of the first Bohr orbit for `H^(o+)` isA. `0.529 Å`B. `0.264 Å`C. `0.132 Å`D. `0.176 Å` |
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Answer» Correct Answer - B ` r_(1) for He^(o+) = (0.529 xx 1^(2))/(2)= 0.264 Å` |
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| 1660. |
The wavelength of the radiations emitted when in a hydrogen atom electron falls from infinity to stationary state is ` : (R_H = 1. 097 xx10^7 m^(-1))` .A. `9.1 xx 10^-8 nm`B. 192 mC. 406 nmD. 91 nm |
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Answer» Correct Answer - D (d) `(1)/(lamda) = R_H[(1)/(oo) -(1)/(l^2)]` :. `lamda =(1)/(R_H) = (1)/(1.097 xx 10^7) m = 9.1 xx 10^-8 m` =`91 xx 10^s m = 91 nm`. |
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| 1661. |
The discovery of neutron becomes very late because.A. it is present in nucleusB. it is a fundamental particle.C. It does moveD. it does not carry any charge. |
| Answer» Correct Answer - D | |
| 1662. |
Proton was discovered byA. ChadwickB. ThomsonC. GoldsteinD. Bohr |
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Answer» Correct Answer - C ( c) Proton is represented by `p` having charge `+ 1` discovered in `1988` by Goldstein. |
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| 1663. |
One would expect proton to have very largeA. Ionization potentialB. RadiusC. ChargeD. Hydration energy |
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Answer» Correct Answer - D `H^+` (proton) will have very large hydration energy due to its very small ionic size Hydration energy `prop 1/"Size"` |
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| 1664. |
The two paricles A and B have de Broglie wavelengths 1 nm and 5 nm respectively if mass of A is four times the mass of B, the ratio of kinetic energies of A and B would be :-A. `5 : 1`B. `25 : 4`C. `20 : 1`D. `5 : 4` |
| Answer» Correct Answer - 2 | |
| 1665. |
A golf has a mass of `40 g ` and a speed of `45 m s^(-1)` .If the speed can be measured an accurary of `2%` calculate the uncertainty in the position |
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Answer» Mass of hall `= 40 g = 40 xx 10^(-3) kg` The uncertyainty in the speed `Delta v = 45 xx (2)/(100) 0.9 m s^(-3) kg` so the uncleretainty in position `Delta v = (h)/(4 pi Delta p) = (h)/(4pi m Delta v) ` `= (6.626 xx 10^(-34) Js)/(4 xx 3.14 xx (40 xx 10^(-3) kg) xx (0.9 m s^(-1)))` `= 1.46 xx 10^(-33) m` |
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| 1666. |
The velocities of two paricles A and B are `0.05` and `0.02m//s` respectively. The mass of B is five times the mass of A. The ratio of their de-Brogile wavelength isA. `2:1`B. `1:4`C. `1:1`D. `4:1` |
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Answer» Correct Answer - A `lambda = (h)/(mv), (lambda_(A))/(lambda_(B)) = (m_(B).v_(B))/(m_(A).v_(A))` |
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| 1667. |
An electron of mass `m` and charge `e` is accelerated from rest through a potential difference `V` in vacuum. The final speed of the electron will beA. `sqrt(V//m)`B. `sqrt(eV//m)`C. `sqrt((2eV//m))`D. None of these |
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Answer» Correct Answer - C `(1)/(2)mv^(2) = ev` |
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| 1668. |
The radius of the second Bohr for `Li^(2+)` isA. `0.529 xx (4)/(3) Å`B. `0.529 xx (2)/(3) Å`C. `0.529 xx (4)/(9) Å`D. `0.529 xx (2)/(9) Å` |
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Answer» Correct Answer - A `r_(n) = (r_(1) for H xx n^(2))/(Z) = (0.529 Å xx n^(2))/(Z)` ` r_(2) for Li^(2+) = (0.529 xx 2^(2))/(3)= (0.529 xx 4)/(3)Å` |
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| 1669. |
The wave leagth of the first line of lyman series of hydrogen is identical to that second line of balmer series for some hydrogen like ion `X` The `IF_(2)` for X isA. `-54.4 eV`B. `-328 eV~`C. `-13.6 eV`D. `-3.8 eV` |
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Answer» Gives `lambda(lyman H_(2) = lambda (Balmer , X)` i.e `R_(H)xx 1^(2)((1)/(1^(2)) - (1)/(2^(2))) = R_(2)X^(2) ((1)/(2^(2)) - (1)/(3^(2)))` Second line Balmer `rArr n = 3` `rArr X = 2` Thus `IE_(2) = - 13.6(z^(2))/(n^(2)) = - 13.6 xx (2^(2))/(n^(2)) = -13.6 eV` |
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| 1670. |
`{:("Given",K,L,M,N,),(,2,8,11,2,):}` The number of electrons present in `l=2` is -(a). `3`(b). `6`(c). `5`(d). `4`A. 3B. 6C. 5D. 4 |
| Answer» Correct Answer - A | |
| 1671. |
Which one of the following pairs of ions have the same electronic configuration?A. `Cr^(3+), Fe^(3+)`B. `Fe^(3+), Mn^(2+)`C. `Fe^(3+),Co^(3+)`D. `Sc^(3+),Cr^(3+)` |
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Answer» Correct Answer - B `{:(._(24)Cr=[Ar]3d^(5)4s^(1),._(24)Cr^(3+)=[Ar]3d^(3)),(._(26)Fe=[Ar]3d^(6)4s^(1),._(26)Fe^(3+)=[Ar]3d^(5)),(._(25)Mn=[Ar]3d^(5)4s^(2),._(25)Mn^(2+)=[Ar]3d^(5)),(._(27)Co=[Ar]3d^(7)4s^(2),._(27)Co^(3+)=[Ar]3d^(6)),(._(21)Sc=[Ar]3d^(1)4s^(2),._(21)Sc^(3+)=[Ar]):}` |
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| 1672. |
An impossible set of four quantum number of an electron isA. `n = 4, l = 2, m =-2, s = +1//2`B. `n =4, l = 0, m = 0, s = +1//2`C. `n = 3, l = 2, m =-3, s = +1//2`D. `n =5, l = 3,m =0, s =- 1//2` |
| Answer» Correct Answer - C | |
| 1673. |
Which of the following pairs of ions have the same electronic configuration ?A. `Cr^(3+),Fe^(3+)`B. `Fe^(3+),Mn^(2+)`C. `Fe^(3+),Co^(3+)`D. `Se^(3+),Cr^(3+)` |
| Answer» `Fe^(3+) and Mn^(3+)` have the same element configuration | |
| 1674. |
In which of the following pairs, the ions are iso electronicA. `Na^(+), Mg^(2+)`B. `Al^(3+),O^(-)`C. `Na^(+),O^(-)`D. `N^(3),Cl^(-)` |
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Answer» Correct Answer - A Isoelectronic have same number of electrons |
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| 1675. |
Which of the following is not according to the panli exclasion principal ?A. B. C. D. a and b both |
| Answer» The set of the four quantum number is not same for the three electron in answer c in answer a both electron have the same set of quantum number white in answer b the first and third electrons have the same set of quantum numbers | |
| 1676. |
If an electron drops from 4th orbit to 2nd orbit in an H-atom, thenA. it gains `2.55eV` of potential energyB. it gains `2.55eV`of total energyC. it emits a `2.55 eV` electronD. it emits a `2,55 eV`photon |
| Answer» Correct Answer - D | |
| 1677. |
Assertion (A) : Limiting line is the balmer series has a wavelength of `364.4 nm` Reason (R ) : Limiting line is obtained for a jump electron from `n = infty`A. If (A) and (R) are both correct and (R) is the correct reason for (A).B. If (A) and (R) are both correct but (R) is not the correct reason for (A).C. If (A) is true but (R) is false.D. If both (A) and (R) false. |
| Answer» Correct Answer - A | |
| 1678. |
The balmer series occurs between the wavelength of `[R = 1.0968 xx 10^7 m^-1]`.A. `4623 Å "to" 6563 Å`B. `1243 Å "to" 6563 Å`C. `3647 Å "to" 6563 Å`D. `3647 Å "to" 7210 Å` |
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Answer» Correct Answer - C ( c) For Balmer series `(1)/(lamda) = R[(1)/(2^2) -(1)/(n^2)]` where `n = 3,4,5,….oo` The obtain the limits for Balmer series `n = 3` and `n = oo` respectively. `lamda_(max) (n=3) = (1)/(R[(1)/(2^2) -(1)/(3^2)]] = (36)/(5 R)` =`(36)/(5 xx 1.0968 xx 10^7) m = 6563 Å` `lamda_(min) (n = oo) = (1)/(R[(1)/(2^2) -(1)/(oo^2)]] = (4)/(R)` =`(4)/(1.0968 xx 10^7) m = 3647 Å`. |
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| 1679. |
The line with smallest wavelength in the Balmer series in the hydrogen spectrum will have the frequency :-A. `8.22xx10^(14)s^(-1)`B. `3.29xx10^(15)s^(-1)`C. `7.22xx10^(16)s^(-1)`D. `8.05xx10^(13)s^(-1)` |
| Answer» Correct Answer - 1 | |
| 1680. |
Assertion (A) : Limiting line is the balmer series has a wavelength of `364.4 nm` Reason (R ) : Limiting line is obtained for a jump electron from `n = infty`A. Statement-I is true, Statement-II is true , Statement-II is correct explanation for Statement-I.B. Statement-I is true, Statement-II is true , Statement-II is NOT a correct explanation for statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true |
| Answer» Correct Answer - A | |
| 1681. |
In Bohr series of lines of hydrogen spectrum, third line from the red end corresponds to which one of the following inner orbit jumps of electron for Bohr orbit in atom in hydrogen :A. `3 to 2`B. `5 to 2`C. `4 to 1`D. `2 to 5` |
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Answer» Correct Answer - B |
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| 1682. |
Calculate the uncertainty in the velocity of a cricket hall (mass `= 0.15 kg)` uncertainty in position is of the order of `1Å` |
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Answer» `Delta u - Delta x = (h)/(4pi m)` `Delta u = (h)/(4pi m - Delta x) = (6.626 xx 10^(-34))/(4 xx (22)/(7) xx 0.15 xx 10^(-10))` `= 3.51 xx 10^(-24) ms^(-1)` |
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| 1683. |
An electron is moving with a kinetic energy of `4.55 xx 10^-25 J`. What will be Broglie wavelength for this electron ? |
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Answer» `KE=(1)/(2)mv^(2)=4.55xx10^(-25)` or `(1)/(2)xx9.1xx10^(-31)xxv^(2)=4.55xx10^(-25)` or `v^(2)=(2xx4.55xx10^(-25))/(9.1xx10^(-31))` Applying de Broglie equation, `lamda=(h)/(mv)=(6.6xx10^(34))/(9.1xx10^(-31)xx10^(3))=0.72xx10^(-6)m`. |
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| 1684. |
Frequancy `=f_(1)`, Time period = T, Energy of `n^(th)` orbit `= E_(n)`, radius of `n^(th)` orbit `=r^(n)`, Atomic number = Z, Orbit number = n : `{:(,"Column-I",,"Column-II",),((A),f,(p),n^(3),),((B),T,(q),Z^(2),),((E),E_(n),(r ),(1)/(n^(2)),),((D),(1)/(r_(n)),(s),Z,):}` |
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Answer» Correct Answer - (a-q);(b-p); (c-q,r) ; (d-r,s) |
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| 1685. |
The possible value of `l` and `m` for the last electron in the `Cl^(- )ion` are :A. `1` and `2`B. `2` and `+1`C. `3` and `-1`D. `1` and `-1` |
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Answer» Correct Answer - D `Cl^(-) = 1s^(2) 2s^(2) 2p^(6)3s^(2)3p^(6)` For last `e^(-) n=3, l=1, m=pm1` |
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| 1686. |
`{:("Given",K,L,M,N,),(,2,8,11,2,):}` The number of electrons present in `l=2` is -A. `3`B. `6`C. `5`D. `4` |
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Answer» Correct Answer - A `K = 2e^(-) = 1s^(2)` `L = 8e^(-) = 2s^(2) 2p^(4)` `M=11e^(-)=3s^(2)3p^(6)3d^(3)` `N = 2e^(-) = 4s^(2)` for `d e^(-)=3, l=2` |
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| 1687. |
Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen. |
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Answer» Correct Answer - `2.725 xx 10^(6)M^(-1)` |
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| 1688. |
Change in angular momentum when an electron makes a transition corresponding to the 3rd line of the Balmer series in `Li^(2+)` ion is :A. `(h)/(2pi)`B. `(2h)/(2 pi)`C. `(3h)/(2pi)`D. `(4h)/(2pi)` |
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Answer» Correct Answer - C |
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| 1689. |
If an electron has spin quantum number of `-(1)/(2) ` and magnetic quantum number of -1 it cannot be present in:A. d-orbitalB. f-orbitalC. s-orbitalD. p-orbital |
| Answer» Correct Answer - C | |
| 1690. |
An electron of mass m and charge -e moves in circular orbit of radius r round the nucleus of charge +Ze in unielectron system. In CGS system the potential energy of electron isA. `(Z^(2)e^(2))/(r)`B. `-(Ze^(2))/(r)`C. `(Ze^(2))/(r)`D. `(mv^(2))/(r)` |
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Answer» Correct Answer - B (b) Potential energy = work done ` = int_(oo)^(r)-(Ze^(2) dr)/(r^(2))=-(Ze^(2))/(r)` |
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| 1691. |
In photoelectric emission, the energy of the emitted electron isA. greater than the incident photonB. same as that of the incident photonC. smaller than the incident photonD. proportional to the intensity of incident photon |
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Answer» Correct Answer - C (c) In the photoelectric effect, the energy of the emitted electron is smaller than that of the incident photon because some energy of photon is used to eject the electron and remaining energy is used to increase the kinetic energy of ejected electron . |
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| 1692. |
Five valence electrons of `.^(15)P` re labelled as If the spin quantum numbers of B and Z is `+(1)/(2)`, the group of electron with three of the quantum number (n,l,m) same are .A. AB,XYZ,BYB. ABC. XYZ,AZD. AB,XYZ |
| Answer» Correct Answer - 2 | |
| 1693. |
The electronic configuration of an element is `1s^(2)2s^(2)2p^(2)3s^(2)3p^(6)3d^(5)4s^(1)` This represents itsA. Excited stateB. Ground stateC. Cationic formD. Anionic form |
| Answer» Ground state , because half -filled d orbit is more stable. | |
| 1694. |
The electronic configuration of an element is `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(1)` This represents itsA. Excited stateB. Ground stateC. Cationic formD. Anionic form |
| Answer» The gives element configuration is the ground state for chromium | |
| 1695. |
The longest ` lambda` for the Lyamn series is …………… (Given ` R_H = 109678 cm^(-1))` :A. ` 1215`B. ` 1315`C. `1415`D. `1515` |
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Answer» Correct Answer - A For longest of Lyman series . ` n_(2) = 2`, `(1)/( lambda) = R_(H) [(1)/(n_(1)^(2)) - (1)/n_(2)^(2)]` because ` Delta E = (hc)/( lambda)` is minimum when `lambda` is longest , Thus ` Delta E = E_(2) - E_(1)` = minimum. |
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| 1696. |
`8g` each of oxygen and hydrogen at `27^(@)C` will have the total kinetic energy in the ratio of ……. |
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Answer» Correct Answer - `1: 16` |
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| 1697. |
Calculate the uncertainty in the velocity of an electron of the uncertainty in its position is of the order of `1 Å` |
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Answer» According to Heisenbenrg uncertainty principle `Delta u - Delta x = (h)/(4pi m)` `Delta u = (h)/(4pi m Delta x) = (6.62 xx 10^(-34))/(4 xx 3.14 xx 9.108 xx 10^(-31) xx 10^(-10))` `= 5.7 xx 10^(5) m s^(-1)` |
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| 1698. |
Calculate the frequency corresponding to the wavelength `4000 Å` |
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Answer» Correct Answer - A::D `v = (c )/(lambda) = (3 xx 10^(10))/(4000 xx 10^(-8)) = 7.5 xx 10^(14) s^(-1)` |
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| 1699. |
Which of the following represents schrodinger equationsA. `(del^2/(delx^2)+del^2/(dely^2)+del^2/(delz^2))Psi=(8pi^2m)/h^2[E-V]Psi=0`B. `hatHPsi=EPsi`C. `(hatT=hatV)Psi=EPsi`D. all of them |
| Answer» Correct Answer - D | |
| 1700. |
The uncertainty principle and the concept of wave nature of matter were proposed by.............and ............respectively. |
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Answer» Correct Answer - Heisenberg, de-Broglie. |
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