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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.
| 1951. |
The mass of a particle is `10^(-10)g` and its radius is `2xx10^(-4)cm`. If its velocity is `10^(-6)cm sec^(-1)` with `0.0001%` uncertainty in measurement, the uncertainty in its position is `:`A. `5.2 xx 10^(-8)m`B. `5.2xx 10^(-7)m`C. `5.2 xx 10^(-6)m`D. `5.2 xx 10^(-9)` |
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Answer» Correct Answer - A `Deltax = (h)/(4pim.Deltav)` where `Deltav = 10^(-12)` |
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| 1952. |
The mass of a particle is `10^(-10)g` and its radius is `2xx10^(-4)cm`. If its velocity is `10^(-6)cm sec^(-1)` with `0.0001%` uncertainty in measurement, the uncertainty in its position is `:`A. `5.2xx10^(-8)m`B. `5.2xx10^(-7)m`C. `5.2xx10^(-6)m`D. `5.2xx10^(-9)J` |
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Answer» Correct Answer - 1 `m=10^(-10)grArr10^(-13)kg,` `Deltav=(0.0001)/(100)xx10^(-6)xx10^(-2)=10^(-14)m sec^(-1)` `=Deltax-Deltap=(h)/(4x)` `rArrDeltax=(h)/(4piDeltap)=(h)/(4pimDeltav),` `Deltax=(6.62xx10^(-34))/(4xx3.14xx10^(-13)xx10^(-14))` `Deltax=(6.62)/(12.56)xx(10^(-34))/(10^(-27)),Deltax=5.2xx10^(-5)m` |
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| 1953. |
Find the wavelenght of the first line of `He^(+)`ion sepectral series whose interval between extreme line is `[(1)/(lambda)-(1)/(lambda)=2.745xx10^(4)cm^(1)]` |
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Answer» Correct Answer - `4689Å` `(1)/(lambda_(1))=RZ^(2)[(1)/(1^(2))-(1)/(oo^(2))]` `(1)/(lambda_(2))=RZ^(2)[(1)/(1^(2))-(1)/(2^(2))]` `(1)/(lambda)=(1)/(lambda_(1))-(1)/(lambda_(2))=((1)/(2^(2))-(1)/(oo^(2)))` `=2.7451xx10^(4) cm^(1)` |
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| 1954. |
A dye absorbs a photon of wavelength `lambda` and re`-` emits the same energy into two phorons of wavelengths `lambda_(1)` and `lambda_(2)` respectively. The wavelength `lambda` is related with `lambda_(1)` and `lambda_(2)` as `:`A. `lambda=(lambda_(1)+lambda_(2))/(lambda_(1)lambda_(2))`B. `lambda=(lambda_(1)lambda_(2))/(lambda_(1)+lambda_(2))`C. `lambda=(lambda_(1)^(2)lambda_(2)^(2))/(lambda_(1)+lambda_(2))`D. `lambda=(lambda_(1)lambda_(2))/((lambda_(1)+lambda_(2))^(2))` |
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Answer» Correct Answer - B `E=E_(1)+E_(2),(hc)/(lambda)=(hc)/(lambda_(1))+(hc)/(lambda_(2))" "rArr" "(hc)/(lambda)=hc((lambda_(2)+lambda_(1))/(lambda_(1)lambda_(2))),lambda=(lambda_(1)lambda_(2))/(lambda_(1)+lambda_(2))` |
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| 1955. |
What is the shortest wavelength line in the Paschen series of `Li^(2+)` ion ?A. `(R)/(9)`B. `(9)/(R)`C. `(1)/(R)`D. `(9R)/(4)` |
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Answer» Correct Answer - 3 `(1)/(lambda)=RZ^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]=Rxx3^(2)[(1)/(3^(2))-(1)/(oo^(2))]" "rArr" "R or lambda=(1)/(R)` |
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| 1956. |
Humphry series discovered in `H-"atomic"` spectra has lowest energy radiations among all series. Lowest state for this series is `n_1 = 6`.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - A (a) For Humphry series, `(n_2 = 7,8,9….)` and `n_1 = 6`. |
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| 1957. |
Statement : Monochromatic X-rays fall on lighter elements such as carbon and show scattering and effect is known as Compton effect . Explanation : `lambda` scattered light is always lower than ` lambda` incident light .A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - B (b) In compton effect `lamda_("scattered") gt lamda_("incident")`. |
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| 1958. |
If shortest wavelength of H-atom in Balmer series is X then (i) What is the shortest wave length in Lyman series. (ii) What is the longest wave length in Paschen series. |
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Answer» (i) `(1)/(lambda)=Z^(2)R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` `lambda=[(1)/(Z^(2)R_(H))][(n_(1)^(2)n_(1)^(2))/(n_(2)^(2)-n_(1)^(2))]` Shortest wavelength correspond to maximum energy `(E=(hc)/(lambda))` hence transition must be `n_(2)=oo, n_(1)=2` `=[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]=((1)/(4))=(n_(1)^(2)n_(2)^(2))/(n_(2)^(2)-n_(1)^(2))=4` `x=((1)/(Z^(2)R_(H)))4=(4)/(R_(H))implies R_(H)=(4)/(x)` (a) For shortest wavelength Lyman series `n_(1)=1,n_(2)=oo` `lambda=((1)/(R_(H)))implieslambda=(x)/(4)` (b) For longest wavelength in Paschen series `n_(1)=3,n_(2)=4` `implies (n_(2)^(2)n_(1)^(2))/(n_(2)^(2)-n_(1)^(2))=(144)/(16-9)=[(144)/(7)]=lambda=((1)/(R_(H)))xx(144)/(7)=(x x 144)/(4xx7)=(36x)/(7)` |
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| 1959. |
The shortest wavelength of H-atom in Lyman series is x, then longest wavelength in Balmer series of `He^(+)` isA. `(4)/(3)`B. `(36)/(5)`C. `(1)/(4)`D. `(5)/(9)` |
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Answer» Correct Answer - A (a) Longest wavelength in Lyman Series of hydrogen atom arisis from transition between `n = 2` to `n = 1` whose number is given by `overline v_(H(2rarr1)) = R xx 1^2 ((1)/(1^2) - (1)/(2^2)) = (3)/(4) R` Shortest wavelength in Balmer series of `He^+` arises from transition between `n = a` to `n = 2` whose wave number is given by `overline v_(He(alpha -2)) = R xx 2^2 ((1)/(2^2) -(1)/(alpha^2)) = R` =`(overlinev_(He))/(overlinev_H) = (lamda_H)/(lamda_(He))` `:. (lamda_H)/(lamda_(He)) = (4R)/(3R) = (4)/(3)`. |
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| 1960. |
If the shortest wavelength of H-atom in Lyman series is `x`, then longest wavelnght in Balmer series of `HE^(2+)` is :A. ` (9x)/5`B. `(36x)/5`C. `x/4`D. `(5x)/9` |
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Answer» Correct Answer - A `1/(lambda _L) = RH [1/1^2 - 1/(prop)]` for shortest ` lambda` of lyman series of H `1/(lambda_B) = Z^2 R_H [1/2^2 -1/3^2]` for longest `lambda` of Bolmer series of ` He^+` ` :. (lambda_B )/(lambda_L) = (1 xx 36)/( 5 xx 4) :. Lambda _B = 9/5 . lambda_L = 9/5 x`. |
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| 1961. |
The French physicist Louis de Broglie in 1924 postulated that matter like radiation , should exhibit a dual behaviour. He proposed the following relationship between the wavelength `.lambda` of a material particle,its linear momentum P and Planck constant h. `lambda=(h)/(p)=(h)/(mv)` The de Broglie relaion that the wavelength of a particle should decrease as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles.The waves or de Broglie waves. These waves differ from the electromagnetic waves as they: (i) have lower velocities (ii) have no electrical and magnetic fields and (iii) are not emitted by the particle under consideration. The expermental confirmation of the de Broglie relation was obtained when Davission and Germer in 1927, observed. As diffraction is a characteristic property of waves, hence the beam of electrons behave as a wave as proposed by de Broglie. Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron . He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we can determine its position. The converse is also true. This is summed up in what we now call the 'Heisenberg uncertainty principle' : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of uncertainty in the position, `Deltax` and the uncertainty in the momentum `Delta(mv)` must be greater than or equal to `(h)/(4pi), i.e.,` `Deltax Delta(mv)ge(h)/(4pi)` The transition so that the de Broglie wavelength of electron becomes 3 times of its initial value in `He^(+)` ion will be :A. `2 to 5`B. `3 to 2`C. `2 to 6`D. `1 to 2` |
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Answer» Correct Answer - c |
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| 1962. |
The French physicist Louis de Broglie in 1924 postulated that matter like radiation , should exhibit a dual behaviour. He proposed the following relationship between the wavelength `.lambda` of a material particle,its linear momentum P and Planck constant h. `lambda=(h)/(p)=(h)/(mv)` The de Broglie relaion that the wavelength of a particle should decrease as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles.The waves or de Broglie waves. These waves differ from the electromagnetic waves as they: (i) have lower velocities (ii) have no electrical and magnetic fields and (iii) are not emitted by the particle under consideration. The expermental confirmation of the de Broglie relation was obtained when Davission and Germer in 1927, observed. As diffraction is a characteristic property of waves, hence the beam of electrons behave as a wave as proposed by de Broglie. Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron . He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we can determine its position. The converse is also true. This is summed up in what we now call the 'Heisenberg uncertainty principle' : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of uncertainty in the position, `Deltax` and the uncertainty in the momentum `Delta(mv)` must be greater than or equal to `(h)/(4pi), i.e.,` `Deltax Delta(mv)ge(h)/(4pi)` The correct order of wavelength of Hydrogen `(._(1)H^(1))` Deuterium `(._(1)H^(2))` and Tritium `(._(1)H^(3))` moving with same kinetic energy is :A. `lambda_(H)gtlambda_(D)gtlambda_(T)`B. `lambda_(H)=lambda_(D)=lambda_(T)`C. `lambda_(H)ltlambda_(D)ltlambda_(T)`D. `lambda_(H)ltlambda_(D)gtlambda_(T)` |
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Answer» Correct Answer - a |
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| 1963. |
Calculate energy of electron which is moving in the orbit that has its radius , sixteen times the radius of first Bohr orbit for H-atom. |
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Answer» Correct Answer - `-136xx10^(-19)` Joules Radius `=16(RH)=16xx0.0529` `16xx0529=0.0529xx(n^(2))/(Z)` `16=(n^(2))/(1)" "n=4` `T.E.=-13.6xx(n^(2))/(Z^(2))l.v=0.85l.v=-1.36xx10^(-19)J` |
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| 1964. |
The ionisation energy of a H-like atom is `4R_(h)` a. Calculate the wavelength of radiation emitted when an electron jumps from the first excited state to the ground state b. What is the radius of first orbit of this atom?Given `1R_h= 2.18 xx 10^(-19) J` |
| Answer» `(i) 303.89 Å` (ii) `2.645 xx 10^(-9) cm,` | |
| 1965. |
The shortest wavelength of `He^(+)` in Balmer series is `x`. Then longest wavelength in the Paschene series of `Li^(+2)` is :-A. `(36x)/(5)`B. `(16x)/(7)`C. `(9x)/(5)`D. `(5x)/(9)` |
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Answer» Correct Answer - B `(IE)_(Li^(2+))=(IE)_(H)xxƶ^(2)` `21.8xx10^(-19)xx9J//"atom"` `lambda = (h)/sqrt(2ME)` `lambda=(6.62xx10^(-34))/sqrt(2xx9.1xx10^(-31)xx2.18xx10^(-9)xx9)` `lambda=1.17A^(@)` |
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| 1966. |
The French physicist Louis de Broglie in 1924 postulated that matter , like radiation , should exhibit dual behaviour. He proposed the following relationship between the wavelenght lambda of a material particle , its linear momentum p and planck cosntant h. `lambda= (h)/(p)=(h)/(mv)` The de Broglie relation implies that the wavelength of a partices should decreases as its velocity increases. It also implies that the for a given velocity heavir particule should have shorter wavelenght than lighter particles. The waves associated with particles in motin are called matter waves or de Broglie waves. These waves differ from the electromagnetic waves as they (i) have lower velocities have no electrical and magnetic fields and are not emitted by the particle under consideration. The experimental confirmation of the de Broglie relation was obtained when Davisson and Germer. in 1927, observed that a beam of electrons is diffracted by a nickel arystal. As diffarceted by a nickel . As diffraction is a characteristic property of waves, hence the beam of electron dehaves as a wave, as proposed by de Broglie. Werner Heisenberg cobnsiderd the imits of how precisely we can measure propoerties of an electron or other microscopic particle like electron. he determined that there is accurately we measure the momentum of a particle, the less accurately we can determine its position . The converse is laso true . The is summed up in what we now call the "Hesienberg uncertainty princple: It isimpossibble to determine simultameously ltbr. and percisely both the momentum ans position of particle . The product of uncertainly in the position, Deltax and the ncertainly in the momentum Delta(mv) mudt be greater than or equal to `(h)/(4pi)` i.e. `etaDeltax Delta(mv)ge(h)/(4pi)` The correct order of wavelenght of Hydrogen `(._(1)H^(1))` Deuterium `(._(1)H^(2))` and Tritium `(._(1)H^(3))` moving withsame kinetic energy is.A. `lambda_(H)gtlambda_(D)gtlambda_(T)`B. `lambda_(H)=lambda_(D)=lambda_(T)`C. `lambda_(H)ltlambda_(D)ltlambda_(T)`D. `lambda_(H)ltlambda_(D)gtlambda_(T)` |
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Answer» Correct Answer - A `lambda=(h)/sqrt(2m.KE)` for same KE `lambdapropto(1)/sqrt(m)` so`lambda_(H)gtlambda_(0)gtlambda_(T)` |
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| 1967. |
Which is correct paramagnetic or der ?A. Mn Cr ZnB. Fe Zn CoC. Cr Fe ZnD. Hg Mn Fe |
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Answer» Correct Answer - C Cr, Mn, Fe, Zn, Hg, Co, have 6,5,4, ,0,03 upaired elecrons repectively . |
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| 1968. |
The electron energy in hydrogen atom is given by `E_(n) =(-21.7xx10^(12))/n^(2)ergs`. Calculate the energy requried to remove an `e^(-)` completely from n=2 orbit . What is the largest wavelenght in cm of light that can be used to cause this transition . |
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Answer» For complete removal from n=2 `DeltaE_(n)`=`+2.1xx10^(-12)[(1)/2^(2)-(1)/oo^(2)]` =`5.425xx10^(-12)`erg `(hc)/(lambda)`=`DeltaE_(n) " " rArr lambda=(hc)/(DeltaE_(n) )`=`3.7xx10^(-5)`cm |
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| 1969. |
The electron energy in hydrogen atom is given by `E_(n)=(-2.18xx10^(-18))//n^(2)J`. Calculate the energy required to remove an electron completely from the `n=2` orbit. What is the longest wavelength of light in cm that can be used to cause this transition? |
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Answer» Correct Answer - `5.425xx10^(-12)ergs, 3.7xx10^(-5)cm` `E_(n)=(-21.7xx10^(-12))/(n^(2)) 1 erg =10^(-7)` Joule `E_(n)=(-21.7xx10^(-12))/(4)` `J.E.=0-[(-21.7xx10^(-12))/(4)]=(21.7xx10^(-12))/(4)` `=5.425xx10^(-12)` ergs (b) `5.425xx10^(-12)=(6.624xx10^(-34)xx10^(8))/(lambda)` `lambda=(6.624xx3xx10^(8)xx10^(12))/(5.425xx10^(34))=3.7xx10^(-14)(nm)` `3.7xx10^(-14)xx10^(9)cm=3.7xx10^(-5)cm` |
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| 1970. |
Degenrate orbitals means :A. Orbitals having same enrgyB. Orbitals having same wave functionC. Orbitials having different enrgy but different wave functionD. Orbitlas having different enrgy and same wave function . |
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Answer» Correct Answer - A Degenergat orbitals have no link with wave function . Degenergat orbitals means for same energy . |
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| 1971. |
In a malti-electrons atom which of the following orbitals deseribed by the three quantum number will have the same energy in the absence of megnetic and electric field ?I.`n = 1, l = 0, m = 0`II. `n = 2, l = 0, m = 0`III.`n = 2, l = 1, m = 1` IVgt`n = 3, l = 2, m = 1` V`n = 3, l = 2, m = 0`A. I and IIB. II and IIIC. III and IVD. IV and V |
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Answer» IV `3d_(2), v:3d_(2) ` Degemerate orbital II `2s,II2p` (Assumed to have same energies in the absence of magnitic and electric field) |
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| 1972. |
What are the possible value of `m_(1)` for the different orbital of a. p sub -shell b. d sub-shell |
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Answer» Correct Answer - A::B a. `+ 1,0, - 1 : `b . `+ 2, + 1 0, -1,-2` |
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| 1973. |
Which of the following sets of quantum number for orbitals in hydrogen atom has a greater energy of electrons ? a. `n = 3,l = 2,m = +1` b. `n = 3,l = 2,` and `m = -1` |
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Answer» Correct Answer - A Bohr sets represent `3d` sub-levels whose each d orbital has the same energy i.e. d orbital are degenerate Therefore both a and b are equal in energy |
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| 1974. |
What is the maximum number of electron that can be accommodated In the sub-with` l = 3`? In the shell-with` n = 3`? In the orbital with` m_(1) = +3`? |
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Answer» For a sub-shell `1= 3`,number of orbitals `= 2l + 1 = 2 xx 3 + 1 = 7` Number of electron` 2 xx 7 = 14` b. For a sub-shell `n = 3`,The number of orbitals is `= n^(2) = 3^(2)= 9` Number of electron` = 9 xx 2 = 18` c. Orbit with `n_(1) = +5` correspomding to only one orbital and bence it can accommodate only two electrons |
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| 1975. |
how many electrons can be accommodated in a sub-shell for which n=3, l=1A. 8B. 6C. 18D. 32 |
| Answer» Correct Answer - B | |
| 1976. |
Correct set of four quantum numbers for the valence (outermost) electron of rubidium `(Z = 37)` isA. `5,0,0,+(1)/(2)`B. `5,1,0,+(1)/(2)`C. `5,1,1,+(1)/(2)`D. `6,0,0,+(1)/(2)` |
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Answer» Correct Answer - A |
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| 1977. |
If x is the number of electron is an atom the configuration should be expresent asA. `I_(x)`B. `nl^(x)`C. `mn^(x)`D. None of these |
| Answer» The electronic configuration of an state is exprecessed by first writing the principle quantum number (n) and followed by the Azomaithal quantum number (1) qand then writing the number of electron (x) as reperscript | |
| 1978. |
Which one of the following sets correctly represents the increase in the paramagnetic property of ionsA. `Cu^(+2) gt V^(+2) gt Cr^(+2) gt Mn^(+2)`B. `Cu^(+2) lt Cr^(+2) lt V^(+2) lt Mn^(+2)`C. `Cu^(+2) lt V^(+2) lt Cr^(+2) lt Mn^(+2)`D. `V^(+2) lt Cu^(+2) lt Cr^(+2) lt Mn^(+2)` |
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Answer» Correct Answer - C paramagnetic property depends upon the number of unpaired electrons, higher the no of unpaired electron, higher the paramagnetic property. |
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| 1979. |
The kinetic energy of an electron is `4.55 xx 10^(-25)J`. Calculate the wavelength . `[h = 6.6 xx 10^(-34)Js`, mass of electron `= 9.1 xx 10^(-31)kg]` |
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Answer» `K.E = (1)/(2)mv^(2) = 4.55 xx 10^(-25)` `=(1)/(2) xx 9.1 xx 10^(-31) xx v^(2) = 4.55 xx 10^(-25)` `rArr v^(2) = (2 xx 4.55 xx 10^(-25))/(9.1 xx 10^(-31)) = 10^(6) rArr v = 10^(3) ms^(-1)` Applying de Broglie equation `lambda = (h)/(mv) = (6.6 xx 10^(-34))/(9.1 xx 10^(-31)xx 10^(3)) = 0.72 xx 10^(-6)m` |
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| 1980. |
The kinetic energy of an electron is `4.55 xx 10^(-25) J` .The mass of electron is `9.1 xx 10^(-34) kg` Calculate velocity of the electron |
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Answer» Calculate of the electron Kinetic energy `= (1)/(2) mv^(2) = 4.55 xx 10^(-25) J = 4.55 xx 10^(-25) kg m^(2)s^(-2)` or `v^(2) = (2 xx KE)/( m) = (2 xx (4.55 xx 10^(-25) kg m^(2)s^(-2) J))/((9.1 xx 10^(-31) kg))` `= 10^(6) m^(2)s^(-2)` or Velocity (v) = `(10^(4)m^(2)s^(-2))^((1)/(2)) = 10^(3) m s^(-1)` |
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| 1981. |
Assertion (A) : In an atom, the velocity of electron in the higher orbits keeps on decreasing Reason (R ) : Velocity of electron in inversely proportional to the radius of the orbitA. If both (A) and (R ) correct and (R ) is the correct explanation for (A)B. If both (A) and (R ) correct and (R ) is the correct explanation for (A)C. If (A) is correct but (R ) is incorrectD. If (A) is incorrect but (R ) is correct |
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Answer» Correct Answer - A Velocity `= 2.18 xx 10^(6) xx (Z)/(n)` |
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| 1982. |
Choose the incorrect statement (s) :A. Increasing order of wavelength is : `"Microwaves" gt "Radio waves"gt "IR waves"gt "Visible waves" gt "UV waves"`B. The order of Bohr radius is `(r_(n) :` where n is orbit number for a given atom) `r_(1)lt r_(2) lt r_(3) lt r_(4)`C. The order or total energy is `(E_(n) :` where n is orbit number for a given atom ) `E_(1) gt E_(2) gt E_(3) gt E_(4) `D. The order of velocity of electron in `H, He^(+), Li^(+), Be^(3+)` species in second Bohr orbit is `Be^(3+) gt Li ^(+2)gt He^(+)gt H` |
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Answer» Correct Answer - a c |
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| 1983. |
Which of the following `is//are` correct statement (s).A. The difference in angular momentum associated with the electron present in consecutive orbits of H-atom is `(n-1)(h)/(2pi).`B. Energy difference between energy levels will be changed if PE at infinity assigned value other than zero.C. Frequency of spectral line in a H-atom is in the order of `(2 to 1) lt(3 to 1) lt(4 to 1).`D. On moving away from the nucleus, kinetic energy of electron decreases. |
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Answer» Correct Answer - c d |
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| 1984. |
Last line of Lyman series for `H-` atom has wavelength `lambda_(1) A,2^(nd)` line of Balmer series has wavelength `lambda_(2)A` thenA. `(16)/(lambda_(1)) = (9)/(lambda_(2))`B. `(16)/(lambda_(2)) = (3)/(lambda_(1))`C. `(4)/(lambda_(1)) = (1)/(lambda_(2))`D. `(16)/(lambda_(1)) = (3)/(lambda_(2))` |
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Answer» Correct Answer - B |
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| 1985. |
Which of the following series in H specits accure IR regionA. LymanB. PashenC. BracketD. Balmer |
| Answer» Correct Answer - B::C | |
| 1986. |
The number fo bodal planes is greatest for the orbital :A. `1s`B. `2p`C. ` 3d`D. `2s` |
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Answer» Correct Answer - C Total node `=n-1` |
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| 1987. |
Atomic weight of ` Ne` is ` 20.2 Ne` is a mixuture of ` Ne^(20)` and ` Ne^(22)`. Relative abunduance fo heavior isotope is :A. ` 90`B. `20`C. `40`D. `10` |
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Answer» Correct Answer - D Average isotopc wt `= sum % x ` isopopic wt . ` (% xx "wt. of isopope "+ % xx "wt . Of other isopope")/( 100)` `therefore 20. 2 = ( a xx 20 + ( 100 -a) xx 22)/( 100)` `therefore A = 90`, per cent fo lighter isopope ` = 100 - 90 =10`. |
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| 1988. |
Last line of Lyman series for `H-` atom has wavelength `lambda_(1) A,2^(nd)` line of Balmer series has wavelength `lambda_(2)A` thenA. `(16)/(lambda_(1))=(9)/(lambda_(2))`B. `(16)/(lambda_(2))=(3)/(lambda_(1))`C. `(4)/(lambda_(1))=(1)/(lambda_(2))`D. `(16)/(lambda_(1))=(3)/(lambda_(2))` |
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Answer» Correct Answer - 2 `(1)/(lambda_(1))=R(1)^(2)[(1)/(1^(2))-(1)/(oo^(2))]" "` and `" "(1)/(lambda_(2))=R(1)^(2)[(1)/(2^(2))-(1)/(4^(2))]` `:. " "lambda_(1)=(1)/(R)" "`and `" "lambda_(2)=(16)/(3R)" ":." "(16)/(lambda_(2))=(3)/(lambda_(1))` |
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| 1989. |
The shortest wavelength in Lyman series of `Li^(2+)` ion is :A. `10.13 "Å"`B. `135 "Å"`C. `13.5 "Å"`D. `101.3 "Å"` |
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Answer» Correct Answer - D |
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| 1990. |
What is de-Broglie wavelength of a He-atom in a container at room temperature. `("use"U_(avg)=sqrt((8kT)/(pim)))` |
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Answer» Correct Answer - `0.79A^(@)` `U_(avg)=sqrt((8kJ)/(pim))` `U_(avg)=sqrt((8xx1.38xx10^(-23)xx298)/(3.14xx4xx1.67xx10^(-27)))` `U_(avg)=1.25xx10^(3)` `lambda=(h)/(mV)implies (6.62xx10^(-34))/(4xx1.67xx10^(-27)xx1.25xx10^(3))` `lambda=0.79Å` |
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| 1991. |
Assertion (A) : If the potential difference applied to an electron is made `4` time , the de Broglie wavelength associated is halved Reason (R ) : On making potential difference `4` times , velocity is doubled and hence `lambda` is halvedA. If both Statement -I & Statement -II are True & the Statement -II is a correct explanation of the Statement I.B. If both Statement -I & Statement -II are True but the Statement -II is not a correct explanation of the Statement- I.C. If Statement -I is True but the Statement -II is False.D. If Statement -I is True but the Statement -II is True . |
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Answer» Correct Answer - A `lambda_(1)=sqrt(150/(V))Å`,`lambda_(2)sqrt((150)/(4V))Å`=`(1)/(2)sqrt((150)/(V))Å`=`lambda_(1)/(2)` `(1)/(2)m_(e)u_(1)^(2)=eV" "(1)/(2)m_(e)u_(2)^(2)= 4 eV` `rArru_(1)=sqrt((2ev)/(m_(e)))` `lambda_(1)=(h)/("mu"_(1))" "=2mu_(1)` `lambda_(2)=(h)/("mu"_(2))=(h)/(2"mu"_(1))=(lambda_(1))/(2)` |
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| 1992. |
de Broglie proposed dual nature for electron by putting his famous equation `lambda = (h)/(mv).` Later on, Heisenberg proposed uncertainty principle as `deltapDeltax ge (h)/(4pi).` On the contrary, Particle nature of electron was established on the basis of photoelectric effect. When a photon strikes the metal surface it gives up its energy to the electron. Part of this energy (say W) is used by the electrons to escape from the metal and the remaining energy imparts kinetic energy `(1//2 mv^(2))` to the ejected photoelectron. The potential applied on the surface to reduce the velocity of photoelectron to zero is known as stopping potential . The circumference of third orbit of a single electron species is 3 nm. What may be the approximate wavelength of the photon required to just ionize electron from this orbit?A. 91.1 nmB. 364.7 nmC. 821 nmD. 205 nm |
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Answer» Correct Answer - c |
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| 1993. |
de Broglie proposed dual nature for electron by putting his famous equation `lambda = (h)/(mv).` Later on, Heisenberg proposed uncertainty principle as `deltapDeltax ge (h)/(4pi).` On the contrary, Particle nature of electron was established on the basis of photoelectric effect. When a photon strikes the metal surface it gives up its energy to the electron. Part of this energy (say W) is used by the electrons to escape from the metal and the remaining energy imparts kinetic energy `(1//2 mv^(2))` to the ejected photoelectron. The potential applied on the surface to reduce the velocity of photoelectron to zero is known as stopping potential . When a beam of photons of a particular energy was incident on a surface of a particular pure metal having work function =(40 eV), some emitted photoelectrons had stopping potential equal to 22 V. some had 12V and rest had lower values. Calculate the wavelength of incident photons assuming that at least one photoelectron is ejected with maximum possible kinetic energy :A. `310 "Å"`B. `298 "Å"`C. `238 "Å"`D. `200 "Å"` |
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Answer» Correct Answer - d |
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| 1994. |
Calculate the momentum of a particle which has a de Broglie wavelength of `2 Å,(h = 6.6 xx 10^(-34) kg m^(2) s^(-1))` |
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Answer» Correct Answer - A::B::C::D `3.3 xx 10^(-24) kg m s^(-1)` Hint: `P = (h)/(lambda)` |
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| 1995. |
What must be the velocity of a beam of electron if they are to display a de Broglie wavelength of `1 Å` |
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Answer» Correct Answer - A::C `7.3 xx 10^(6) kg m s^(-1)` Hint :`v = (h)/(lambda m)` |
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| 1996. |
Assertion (A) : If the potential difference applied to an electron is made `4` time , the de Broglie wavelength associated is halved Reason (R ) : On making potential difference `4` times , velocity is doubled and hence `lambda` is halvedA. If both (A) and (R ) correct and (R ) is the correct explanation for (A)B. If both (A) and (R ) correct and (R ) is the correct explanation for (A)C. If (A) is correct but (R ) is incorrectD. If (A) is incorrect but (R ) is correct |
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Answer» Correct Answer - A Both are correct |
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| 1997. |
What hydrogen-like ion has the wavelength difference between the first lines of Balmer and Lyman series equal to `59.3 nm`? `R_(H) = 109678 cm^(-1)`A. 2B. 3C. 4D. 1 |
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Answer» Correct Answer - B |
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| 1998. |
A particle X moving with a certain velocity has a de Broglie wavelength of `1 "Å"` . If particle Y has a mass of `25%` that of X and velocity `75%` that of X ,de Broglie wavelength of Y will be :A. `3 "Å"`B. `5.33 "Å"`C. `6.88 "Å"`D. `48 "Å"` |
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Answer» Correct Answer - B |
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| 1999. |
An electron in a hydrogen like atom makes transition from a state in which itd de Broglie wavelength is `lambda_(1)` to a state where its de Broglie wavelength is `lambda_(2)` then wavelength of photon (`lambda`) generated will be :A. `lambda = lambda_(1) - lambda_(2)`B. `lambda= (4mc)/(h) {(lambda_(1)^(2) lambda_(2)^(2))/(lambda_(1)^(2) - lambda_(2)^(2))}`C. `lambda = sqrt((lambda_(1)^(2) lambda_(2)^(2))/(lambda_(1)^(2)- lambda_(2)^(2)))`D. `lambda= (2mc)/(h) {(lambda_(1)^(2) lambda_(2)^(2))/(lambda_(1)^(2) - lambda_(2)^(2))}` |
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Answer» Correct Answer - D |
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| 2000. |
Which of the following elements are isotopesA. `C^(12)`B. `C^(13)`C. `C^(14)`D. `N^(14)` |
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Answer» Correct Answer - A::B::C Isotopes: Electrons that contain same atomic number |
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