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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.
| 2251. |
The quantum number `+ 1//2` and `-1//2` for the electron spin representA. Rotation of the electron in clockwise and anticlockwise direction respectively.B. Rotation of the electron in anticlockwise and clockwise direction respectively.C. Magnetic moment of the electron pointing up and down respectively.D. Two quantum mechanical spin states which have no classifical analogue. |
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Answer» Correct Answer - D `(Psi)` it is a solution of schrodinger wave equal |
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| 2252. |
`{:(List-I,List-II),((I) psi^(2)"depends upon upon distance",(a)"p-orbitals"),((II)psi^(2)"depends upon distance and on one direction",(b)"d-orbital"),((III)psi^(2)"depends upon distance and on two directions",(c)"f-orbital"),((IV)psi^(2)"depends upon distance and on three directions",(d)"s-orbitals"):}` The correct match isA. `{:(I,II,III,IV),(d,c,b,a):}`B. `{:(I,II,III,IV),(c,b,a,d):}`C. `{:(I,II,III,IV),(d,a,b,c):}`D. `{:(I,II,III,IV),(d,a,c,b):}` |
| Answer» Correct Answer - C | |
| 2253. |
`psi^(2)=0` representA. nodeB. orbitalC. zero amplitude of waveD. wave function |
| Answer» Correct Answer - A | |
| 2254. |
Which is true about `Psi` :-A. `Psi` represented the probability of finding an electron around the nucleusB. `Psi` represent the amplitude of the electron waveC. Both A and BD. None of these |
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Answer» Correct Answer - B `2pir = nlambda` [acc to de-broglie theory] |
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| 2255. |
The ionization enthalpy of hydrogen atom is `1.312xx10^6 "J mol"^(-1)`. The energy required to excite the electron in the atom from n=1 to n=2 isA. `6.56xx10^5 "J mol"^(-1)`B. `7.56xx10^5 "J mol"^(-1)`C. `9.84xx10^5 "J mol"^(-1)`D. `8.51xx10^5 "J mol"^(-1)` |
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Answer» Correct Answer - C `DeltaE=1.312xx10^6[1/1^2 -1/2^2]` `=1.312xx10^6xx3/4=0.984xx10^6 =9.84xx10^5` J/mol |
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| 2256. |
If an electron is present in n =6 level . How many maximum spectral lines would be observed in case of H atom-A. 10B. 15C. 20D. 25 |
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Answer» Correct Answer - B The no.of spectral lines is given by `(n(n-1))/(2)` when n=6 then , the no. of spectral line `=(6xx(6-1))/(2)=(6xx5)/(2)=15` |
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| 2257. |
The energy required for the ionisation of excited hydrogen atom would be (in eV)A. `lt 13.6`B. `gt 13.6`C. `13.6`D. none of these |
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Answer» Correct Answer - A `(n=2) rarr` Energy of first excited state `=-3.4` eV `(n=3) rarr` Energy of second excited state `=-1.51` eV |
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| 2258. |
If the velocity of an electron in the first orbit of hydrogen atom is approximately `2.2xx10^(8) cm//s` is velocity in the fourth orbit would beA. `5.5xx10^(7)cm//s`B. `4.4xx10^(7)cm//s`C. `3.3xx10^(7)cm//s`D. `6.6xx10^(7)cm//s` |
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Answer» Correct Answer - A `mvr=(nh)/(2pi)` `v=(hn)/(2pimr)` `v_(1)/v_(2)=r_(2)/r_(1)rArrv_(2)=(r_(1)/r_(2))V_(1)rArrv_(2)=1/4V_(1)` `V_(2)=1/4xx2.2xx10^(8)` `v_(2)=0.55xx10^(8)cm//s` `=5.5xx10^(7)cm//s` |
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| 2259. |
In hydrogen spectrum, the spectral line of Balmer series having lowest wavelength isA. `H_(alpha)` -lineB. `H_(beta)`-lineC. `H_(gamma)`-lineD. `H_(delta)`-line |
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Answer» Correct Answer - D `lambda prop (1)/(Deltan)` |
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| 2260. |
`{:(,"Column-I",,"Column-II",),((A),"Layman series",(p),"maximum number of spectral line observed" = 6,),((B),"Balmer series",(q),"maximum number of spectral line observed"=2,),((C),"In a sample"5 rarr 2,(r ),2^(nd) "line has wave number"(8R)/(9),),((D),"In a single isolated H-atom for",(s),2^(nd)"line has wave number" (3R)/(4),),(,3rarr1 "transition",,,),(,,(t),"total number of spectral line is" 10,):}` |
| Answer» Correct Answer - `(A) rarr r ; (B) rarr s ; (C ) rarr p ; (D) rarr q` | |
| 2261. |
The atomic nucleus contaitsA. ProtonB. NeutronC. ElectronD. Photons |
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Answer» Correct Answer - A::B The atomic necleus contains protons and neutron |
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| 2262. |
Calculate the uncertinty in the velocity fo a wagon of mass ` 200 kg` whose position is known to an accuracy of ` +- 10 m`. |
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Answer» ` Deltau. Delta x = h/ ( 4 pi m) ("Given " Delta x = 10 m, m = 2000 kg )` ` :. Delta u xx 10 = ( 6.626 xx 10^(-34))/( 4 xx 3.14 xx 2000)` ` Delta u = 2. 64 xx 10^(- 39) ms^(-1)`. |
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| 2263. |
If uncertainty in velocity is`6.62xx10^(-2)m//sec` . For a particle of mass of `(0.05)/pigm`. Its uncertinty in position may beA. `0.5xx10^(-14)`B. `10^(-15)`C. `0.8xx10^(-15)`D. `0.5xx10^(-10)` |
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Answer» Correct Answer - A `Delta"x".(6.662xx10^(-12))ge(6.62xx10^(-34))/(4xxpixx((0.05xx10^(-3))/(pi)))` `Delta"x"ge (10^(-19))/(0.2)" "rArrDelta"x"ge0.5xx10^(-18)` |
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| 2264. |
Which electronic level would allow the hydrogen atom to absorbs a photon but not to emit a photonA. 3sB. 2pC. 2sD. 1s |
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Answer» Correct Answer - D Lowest state is 1 s-state |
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| 2265. |
If uncertainty in position and momentum are equal then uncertainty in velocity is.A. `sqrt(h//2pi)`B. `(1)/2sqrt(h//pi)`C. `sqrt(h//pi)`D. None |
| Answer» Correct Answer - B | |
| 2266. |
Comprehension # 1 Read the following rules and answer the questions at the end of it. Electrons in various suborbits of an filled in increasing order to their energies. Pairing of electrons in various orbitals of a suborbit takes places only after each orbital is half-filled. No two electrons in an atom can have the same set of quantum number. `C (Z=24),Mn^(+)(Z=25), Fe^(2+)(Z=26)` and `Co^(3+) (Z=27)` are isoelectronic each having 24 electrons. Thus,A. all have configuration as `[Ar]4s^(1)3d^(5)`B. Cr and `Mn^(+)` have configuration as `[Ar]4s^(1) 3d^(5)` while `Fe^(2+)` and `Co^(3+)` have configuration as `[Ar]3d^(5)`C. all have configurations as `[Ar]3d^(6)`D. all have configurations as `[Ar] 4s^(2)3d^(6)` |
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Answer» Correct Answer - B `Cr=1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(1)` `Mn^(+)=1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(1)` `Fe^(2+) = 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6)` `Co^(3+)=1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)` |
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| 2267. |
Comprehension # 1 Read the following rules and answer the questions at the end of it. Electrons in various suborbits of an filled in increasing order to their energies. Pairing of electrons in various orbitals of a suborbit takes places only after each orbital is half-filled. No two electrons in an atom can have the same set of quantum number. The sub-shell that arises after f sub-shell is called g sub-shell.A. it contains 18 electrons and 9 orbitalsB. it corresponds to `l=4` and first occurs in `5th` energy levelC. a g-orbital can have maximum of two electronsD. all the above statements are true. |
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Answer» Correct Answer - D for g - sub-shell `n=5` `l = 0, 1,2,3,4` `l=4` (g-subshell) number of electron `=2(2l+1) = 2xx9implies 18` number of orbital `=(2l +1) implies 9` any orbital can have more two electron. |
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| 2268. |
Comprehension # 1 Read the following rules and answer the questions at the end of it. Electrons in various suborbits of an filled in increasing order to their energies. Pairing of electrons in various orbitals of a suborbit takes places only after each orbital is half-filled. No two electrons in an atom can have the same set of quantum number. A compound of vanadium has a magnetic moment of `1.73 BM`. Electronic configuration of the vanadium ion in the compound is :A. `[Ar]4s^(0)3d^(1)`B. `[Ar]4s^(2)3d^(3)`C. `[Ar]4s^(1)3d^(0)`D. `[Ar]4s^(0)3d^(5)` |
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Answer» Correct Answer - A `sqrt(n(n+2))=1.73` `n(n+2)=3` `n+2n=3` `n^(2)+2n-3=0` `(n+3)(n-1)=0` `n=1` Number of unpaired electron =1 `V^(4+)implies [Ar]3s^(1)4s^(0)` |
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| 2269. |
Comprehension # 1 Read the following rules and answer the questions at the end of it. Electrons in various suborbits of an filled in increasing order to their energies. Pairing of electrons in various orbitals of a suborbit takes places only after each orbital is half-filled. No two electrons in an atom can have the same set of quantum number. Which of these ions are expected to be paramagnetic and coloured in aqueous solution ?A. `Fe^(3+),Ti^(3+),Co^(3+)`B. `Cu^(+),Ti^(4+),Sc^(3+)`C. `Fe^(3+),Ni^(2+),V^(5+)`D. `Cu^(+),Cu^(2+),Fe^(2+)` |
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Answer» Correct Answer - A `Fe^(3+)=[Ar]3d^(5)` `Ti^(3+)=[Ar]3d^(1)` `Co^(3+)=[Ar]3d^(6)` all are having unpaired electron hence paramagnetic `&` coloured. |
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| 2270. |
Comprehension # 1 Read the following rules and answer the questions at the end of it. Electrons in various suborbits of an filled in increasing order to their energies. Pairing of electrons in various orbitals of a suborbit takes places only after each orbital is half-filled. No two electrons in an atom can have the same set of quantum number. How many elements would be in the second period of the periodic table if the spin quantum number `(m_(s))` could have the value of -(1)/(2),0,+(1)/(2)`?A. `8`B. `10`C. `12`D. `18` |
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Answer» Correct Answer - C Spin quantum number `(m_(s))=-(1)/(2),0,+(1)/(2)` that is one orbital accomodate maximum `3e^(-)` Number of element in any period `=3r^(2)` `n=(p+2)/(2)` (for even period no.) `n = (2+2)/(2)=2` number of element `implies 3xx4 implies 12` |
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| 2271. |
The number of nodal planes in a `p_(x)` orbital is `:`A. oneB. twoC. threeD. zero |
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Answer» Correct Answer - A For `p_(x),p_(y)`& `p_(z)` only one nodal plane exists. |
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| 2272. |
Statement-I : No two electrons in an atom can have the same values of four quantum number. Because Statement-II : No two electrons in an atom can be simultaneously in the same shell, same subshell, same orbitals and have same spin.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 | |
| 2273. |
Statement-I : Nodal plane of `p_(x)` atmoic orbital is yz plane. Because Statement-II : In `p_(x)` atmoic orbital electron density is zero in the yz plane.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 | |
| 2274. |
The total number of valence electrons in `4. 2g` of ` N_3^-` ion are :A. `1.6 N_A`B. `3.2 N_A`C. `2.1 N_A`D. `4.2 N_A` |
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Answer» Correct Answer - A (a) `42 g` of `N^(-)._3` ion has `16 N_A` valence electrons , `4.2 g` of `N^(-)._3` ion have `= (16 N_A)/(42) xx 4.2 = 1.6 N_A`. |
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| 2275. |
The wavelength associated with a golf ball weighing `200 g` and moving at a speed of `5 m h^(-1)` is of the orderA. `10^(-10)` mB. `10^(-20)` mC. `10^(-30)` mD. `10^(-40)` m |
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Answer» Correct Answer - C `lambda=h/"mv"=(6.626xx10^(-34)xx3600)/(200xx10^(-3)xx5)` `=23.85xx10^(-31)m =2.385xx10^(-30)`m |
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| 2276. |
A resonance hybrid is always more stable than any of its canonical structures. This stability is due to delocalization of electrons.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) We know that a resonance hybrid or the actual molecule is always more stable than any of its canonical structures which is also called hypothetical or imaginary structures. The stability is due delocalization of electrons and is measured in terms of resonance energy or delocalization energy, it is defined as the difference in internal energy of the resonance hybrid and the most stable canonical structure. Therefore, both the assertion and reason are true and the reason is a correct explanation of the assertion. |
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| 2277. |
The atomic nucleus contains :A. Four protonsB. Four neutronsC. Two neutrons and two protonsD. Four protons and two electrons |
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Answer» Correct Answer - C Nucleus of helium is `._2He^4` mean 2 neutrons and 2 protons. |
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| 2278. |
For n = 4 , which of the following is trueA. The total possible values of l are 3B. The highest value of l is 4C. The total number of possible values of m is 7D. The highest value of m is +3 |
| Answer» Correct Answer - 4 | |
| 2279. |
What is the ratio of mass of an electron to the mass of a proton ?A. InfiniteB. `1.8xx10^3`C. 1.8D. None of these |
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Answer» Correct Answer - B `m_p//m_e=1837=1.8xx10^3` |
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| 2280. |
How many electrons in calcium have l = 0 ?A. 6B. 8C. 10D. 12 |
| Answer» Correct Answer - 2 | |
| 2281. |
Assertion: Total number of orbitals associated with principal quantum number n=3 is 6. Reason : Number of orbitals in a shell equals to 2n.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - D Both assertion and reason are false. Total number of orbitals associated with Principal quantum number n=3 is 9.one 3s orbital + three 3p orbital + five 3d orbitals. `therefore` Therefore there are a total number of nine orbitals. Number of orbitals in a shell equals to `n^2`. |
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| 2282. |
A 200 g golf ball is moving with a speed of 5 m per hour. The associated wavelength is (`h=6.625xx10^(-34)` J -sec) of the order ofA. `10^(-10)` mB. `10^(-20)` mC. `10^(-30)` mD. `10^(-40)` m |
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Answer» Correct Answer - C `lambda=h/"mv"=(6.625xx10^(-34))/(0.2 kg xx (5)/(60xx60ms^(-1))) =10^(-30) m xx 2.4` |
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| 2283. |
The correct set of four quantum number for the valence (outermost) electron of radiation `(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 - B `Rb(z = 37)` `[Kt]_(36)5s^(1)` Find the last electron `n = 5,l = 0,m = 0,s= 1//2` |
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| 2284. |
The correct set of four quantum number for the valence (outermost) electron of radiation `(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 (a) Electronic configuration of `Rb_((37))` is `1 s^2 2 s^2 2 p^6 3 s^(3) 3 p_(3)^(6) d^10 4 s^2 4 p^6 5 s^1` So for the valence shell electron `(5 s^1)` `n = 5, l = 0, m = 0, s = + (1)/(2)`. |
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| 2285. |
An electron is present in 4 f sub - shell . The possible values of azimuthal quantum number for this electron are :-A. 0, 1, 2, 3B. 1 ,2, 3, 4C. 3D. 4 |
| Answer» Correct Answer - 3 | |
| 2286. |
An electron has principal quantum number `3`. The number of its (i) sub-shell and (ii) orbitals would be respectively.A. `3 and 5`B. `3 and 7`C. `3 and 9`D. `2 and 5` |
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Answer» Correct Answer - C ( c) The principal quantum number `n = 3`. Then azimuthal quantum number `l = 3` and number of orbitals `= n^2 = 3^2 = 9.3` and `9`. |
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| 2287. |
The wavelength associated with a gold weighing `200g` and moving at a speed of `5m//h` is of the order :-A. `10^(-10)m`B. `10^(-20)m`C. `10^(-30)m`D. `10^(-40)m` |
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Answer» Correct Answer - C Acc to paulis an orbital accomdate maximum two electron, hence paulis exclusion principle voilates. |
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| 2288. |
The wavelength associated with a golf ball weighing 100g and moving at a speed of `5 m//h` is of the orderA. `10^(-10)m`B. `10^(-20)m`C. `10^(-30)m`D. `10^(-40)m` |
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Answer» Correct Answer - C |
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| 2289. |
The correct set of four quantum number for the valence (outermost) electron of radiation `(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 (a) Electronic configuration of `Rb_((37))` is `1 s^2 2 s^2 2 p^6 3 s^3 3 p_3^(6)3 d^10 4 s^2 4 p^6 5 s^1` So for the valence shell electron `(5 s^1)` `n = 5, l = 0, m = 0, s = + (1)/(2)`. |
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| 2290. |
If a shell is having `g` sub-shell, which is correct statement about principal quantum number `n` of this shell.A. `n le 5`B. `n ge 5`C. `n = 5`D. Cannot be determined |
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Answer» Correct Answer - B (b) The maximum number of possible sub-shells associated with a particular shell is equal to principal quantum number i.e. `n ge 5`. |
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| 2291. |
The energy of the electron in the first orbit of `He^+` is `- 871.6 xx 10^-20 J`. The energy of the electron in the first orbit of hydrogen would be.A. `-871.6xx10^(-20)J`B. `-435xx10^(-20)J`C. `-217.9xx10^(-20)J`D. `-108.9xx10^(-20)J` |
| Answer» Correct Answer - C | |
| 2292. |
The energy of the electron in the first orbit of `He^+` is `- 871.6 xx 10^-20 J`. The energy of the electron in the first orbit of hydrogen would be.A. `- 871.6 xx 10^-20 J`B. `- 435.8 xx 10^-20 J`C. `- 217.9 xx 10^-20 J`D. `- 108.9 xx 10^-20 J` |
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Answer» Correct Answer - C ( c) `E_(1 He^+) = E_(1 H) xx z^2` `- 871 xx 10^-20 = E_(1 H) xx 4` `E_(1 H) = - 217.9 xx 10^-20 J`. |
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| 2293. |
The energy of the electron in the first orbit of `He^+` is `- 871.6 xx 10^-20 J`. The energy of the electron in the first orbit of hydrogen would be.A. `-871.6xx10^(-20) J`B. `-435.8xx10^(-20) J`C. `-217.9xx10^(-20) J`D. `-108.9xx10^(-20)` J |
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Answer» Correct Answer - C `E_(1 He^+)=E_(1H)xxz^2` `-871.6xx10^(-20)=E_(1H)xx4` `E_(1H)=-217.9xx10^(-20)J` |
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| 2294. |
The schrodinger wave equation for hydrogen atom is `Psi_(2s)=1/(4sqrt2pi)(1/a_0)^(1//2) [2-r_0/a_0]e^(-2//a_0)` where `a_0` is Bohr radius. If the radial node in 2s be at `r_0`, then find r in terms of `a_0`A. `r_0=2a_0`B. `2r_0=a_0`C. `3//2r_0=a_0`D. `r_0=a_0` |
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Answer» Correct Answer - A `Psi_(2s)=1/(4sqrt2pi) (1/a_0)^(3//2) [2-r_0/a_0] e^(r//a^0)` [uncertainty principle ] `Psi_(2x)^2 =0` at node then , `[(2-r_0)/a_0] =0 rArr r_0 =2a_0` |
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| 2295. |
Why is the source of `alpha`-particles kept inside the lead block ? |
| Answer» `alpha`-particles cannot pentrate lead, but `beta`-and `gamma`-rays can. In order to screen `alpha`-particles from `beta`-and `gamma`-rayws , lead block was used. | |
| 2296. |
Why was a spherical sulphide screen used in `alpha`-ray scattering experiment ? |
| Answer» to observe scintillations even if the `alpha`-rays get deflected at large angles. | |
| 2297. |
A doubly ionised lithium atom is hydrogen like with atomic number `Z=3` Find the wavelength of the radiation required to excite the electron in `Li^(2+)` from the first to the third Bohr orbit. |
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Answer» Correct Answer - `113.74Å` `(1)/(lambda)=RZ^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2)]]=Rxx9[(1)/1^(2)-(1)/3^(2)]` `(1)/(lambda)=8R` `R=(1)/(8R)= 113.74Å` |
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| 2298. |
The ratio of energy of photon of `lambda = 2000 Å` to that of `lambda = 4000 Å` isA. `.^1//_(4)`B. 4C. `.^1//_(2)`D. 2 |
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Answer» Correct Answer - D (d) `E = (hc)/(lamda) , lamda_1 = 2000 Å , 12 = 4000 Å ,` so `(E_1)/(E_2) = (lamda_2)/(lamda_1) = (4000)/(2000) = 2`. |
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| 2299. |
Which of the following graphs is incorrect?A. B. C. D. |
| Answer» Correct Answer - D | |
| 2300. |
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. `2rarr5`B. `3rarr2`C. `5rarr2`D. `4rarr1` |
| Answer» Correct Answer - C | |