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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.
| 51. |
The eyes of a reptille pass a visual signal what is minimum number of photons that must strike the receptor `(h = 6.6 xx 10^(-34))`? |
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Answer» `E = (hc)/(lambda) = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(850 xx 10^(-9)) = 2.3 xx 10^(-19) J` Energy required to trip signal `= 3.15 xx 10^(-14) J` `m xx E = E_(total)` photons that strike the recptor are `n = (3.15 xx 10^(-16))/(2.5 xx 10^(-19)) = 1.37 xx 10^(5)` photrons |
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| 52. |
The highest value of e/m of anode rays has been observed when the discharge tube is filled with:A. nitrogenB. oxygenC. hydrogenD. helium |
| Answer» Correct Answer - C | |
| 53. |
In which of the following species both cation and Anion have same number of electronsA. `CaO`B. `KBr`C. `NaF`D. `MgS` |
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Answer» Correct Answer - C No of electrons = atomic no `+-` charge |
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| 54. |
The eyes of a reptille pass a visual signal to the brain when the visual receptors are struck by photons of wavelength `859` nm. If an energy of `3.15 xx 10^(-14) J` is required to trip the signal, what is minimum number of photons that must strike the receptor `(h = 6.6 xx 10^(-34))`? |
| Answer» `1.37 xx 10^5` photon, | |
| 55. |
It is given that `E=(2.859)/lambda cal//mol` The energy associated with radiation of wavelength `4xx10^(-5)` m will beA. `71.5 kcal//mol`B. `35.75kcal//mol`C. `32.0 kcal//mol`D. `7.15 kcal//mol` |
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Answer» Correct Answer - A Given `E=(2.859)/lambdacal//mol` `lambda=4xx10^(-7)` `E=(2.859 cal//mol)/(4xx10^(-5)m)=71.5kcal//mol` |
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| 56. |
The nuclear size is measured in units ofA. amuB. angstromC. cmD. fermi |
| Answer» Correct Answer - D | |
| 57. |
An atom contains electrons, protons and neutrons. If the mass of each neutrons is halved, and each electron is doubled, then the atomic mass of `._(12)Mg^(24)`A. Gets doubledB. Approximately remain sameC. Approximately get reduced by `5%`D. Approximately get reduced by `25%` |
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Answer» Correct Answer - D `{:(,p,n,"mass"),("Initial",12,12,24),("Final",12,6,18):}` `=(6 xx 100)/(24) = 25%` reduced |
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| 58. |
Atom consist of electrons , protons and neutrons . If the mass attributed to neutron were halved and that attributed to the electrons were doubled , the atomic mass of ` 6C^(12)` would be approximately :A. approximately the sameB. doubledC. reduced approx. 25%D. approx. halved. |
| Answer» Correct Answer - C | |
| 59. |
It is known that an atom contains protons, neutrons and almost electrons. If the mass of neutron is assumed to half of its original value where as that of proton is assumed to be twice of its original value then the atomic mass of `_(7)^(14)N` will be :-A. sameB. `25%`moreC. `14.28%`moreD. `28.5%` less |
| Answer» Correct Answer - 2 | |
| 60. |
Atom consist of electrons , protons and neutrons . If the mass attributed to neutron were halved and that attributed to the electrons were doubled , the atomic mass of ` 6C^(12)` would be approximately :A. Will remain approximately the sameB. Will become approximately two timesC. Will remain approximately halfD. Will be reduced by 25% |
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Answer» Correct Answer - D No change by doubling mass of electrons however by reducing mass of neutron to half total atomic mass becomes 6+3 instead of 6+6 .Thus reduced by 25% |
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| 61. |
It is known that atom contain protons. Neutrons and electrons. If the mass of neutron is assumed to half of its orginal value where as that of proton is assumed to be twice of its original value then the atomic mass of `._(6)^(14)C` will be :-A. sameB. `25%` moreC. `14.28%` moreD. `28.5%` more |
| Answer» Correct Answer - C | |
| 62. |
The uncertainty in momentum of an electron is `1 xx 10^-5 kg - m//s`. The uncertainty in its position will be `(h = 6.62 xx 10^-34 kg = m^2//s)`.A. `1.05 xx 10^(-28)`mB. `1.05 xx 10^(-26)` mC. `5.27 xx 10^(-30)` mD. `5.52 xx 10^(-38)` m |
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Answer» Correct Answer - C |
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| 63. |
The uncertainty in momentum of an electron is `1 xx 10^-5 kg - m//s`. The uncertainty in its position will be `(h = 6.62 xx 10^-34 kg = m^2//s)`.A. `1.05xx10^(-28)m`B. `1.05xx10^(-26)m`C. `5.27xx10^(-30)m`D. `5.25xx10^(-28)m` |
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Answer» Correct Answer - C `Deltax.Deltap=(lambda)/(4pi)` `"put value" " "Deltap=1.0xx10^(-5)kg ms^(-1)` |
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| 64. |
The uncertainty in momentum of an electron is `1 xx 10^-5 kg - m//s`. The uncertainty in its position will be `(h = 6.62 xx 10^-34 kg = m^2//s)`.A. `1.05 xx 10^-28 m`B. `1.05 xx 10^-26 m`C. `5.27 xx 10^-30 m`D. `5.25 xx 10^-28 m` |
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Answer» Correct Answer - C ( c) According to `Delta x xx Delta p = (h)/(4 pi)` `Delta x = (h)/( Delta p xx 4 pi) = (6.62 xx 10^-34)/(1 xx 10^-5 xx 4 xx3.14)` =`5.27 xx 10^-32 m`. |
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| 65. |
Calculate the uncertainty in velocity of a circuit ball of mass `150 g` if the uncertainty in its position is `1 Å (h = 6.6 xx 10^-34 kg m^2 s^-1)`.A. `3.5 xx 10^-24 m s^-1`B. `4.5 xx 10^-24 m s^-1`C. `3.5 xx 10^-24 cm s^-1`D. `4.5 xx 10^-24 cm s^-1` |
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Answer» Correct Answer - A (a) `Delta x xx m. Delta v ge (h)/(4)` `m = 150 g = 0.15 kg` `Delta x = 1 Å = 10^-10 metre` `h = 6.6 xx 10^-34 kg m^2 s^-1` `p = 3.14` `Delta V = (6.6 xx 10^-34)/(0.15 xx 10^-10 xx 4 xx 3.14)` =`3.5 xx 10^-24 ms^-1`. |
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| 66. |
The uncertainty in momentum of an electron is `1 xx 10^-5 kg - m//s`. The uncertainty in its position will be `(h = 6.62 xx 10^-34 kg = m^2//s)`.A. `1.05xx10^(-28)` mB. `1.05xx10^(-26)` mC. `5.27xx10^(-30)` mD. `5.25xx10^(-28)`m |
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Answer» Correct Answer - C According to `Deltax xx Deltap=h/(4pi)` `Deltax=h/(Deltapxx4pi) =(6.62xx10^(-34))/(1xx10^(-5)xx4xx3.14)=5.27xx10^(-30) m` |
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| 67. |
`{:(Column-I,Column-II),((A)7s,(P)"Maximum energy "),((B)4d,(Q)"Maximum number of electrons"),((C)5d,(R)3 "sub shells"),((D)4p,(S)"Minimum number of orbitals"):}` |
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Answer» Correct Answer - A-P,S; B-Q; C-Q; D-R no. of electrons in an orbit `=2n^(2)` no. of orbital in an orbit `= n^(2)` no. of electron in an orbital =2 no. of subshells = n |
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| 68. |
Calculate the uncertainty in the position of a particle when the uncertaintly in momentum it: (a). `1xx10^(-3)g" cm "sec^(-1)` (b) zero. |
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Answer» (a). Given, `DeltaP=1xx10^(-3)g" cm "sec^(-1)` `h=6.62xx10^(-27)erg-sec` `pi=3.142` According to uncertainty principle, `Deltax*Deltap ge (h)/(4pi)` So, `Delta ge (h)/(4pi)*(1)/(Deltap)ge (6.62xx10^(-27))/(4xx3.142)xx(1)/(10^(-3))` `=0.527xx10^(-24)cm` (b). When the value of `Deltap=0`, the value of `Deltax` will be infinity. |
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| 69. |
If uncertainty in momentum of electron is three times the uncertainty in position, then uncertainty in velocity of electron would be:A. `(1)/(4m)sqrt((h)/(3pi))`B. `(1)/(3m)sqrt((h)/(pi))`C. `(1)/(3m)sqrt((4pi)/(h))`D. `(1)/(2m)sqrt((3h)/(pi))` |
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Answer» Correct Answer - D `DeltaxDeltapge()/(4pi)" , "Deltap=3Deltax` `therefore(Deltap)/(3)*Deltap ge (h)/(4pi)" , "Deltap=sqrt((3h)/(4pi))` `mDeltav=sqrt((3h)/(4pi))" , "Deltav=(1)/(2m)sqrt((3h)/(pi))` |
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| 70. |
If uncertainty in position and momentum are equal then uncertainty in velocity is.A. `sqrt((h)/(pi))`B. `sqrt(h)/(2pi)`C. `(1)/(2m)sqrt((h)/(pi))`D. `(1)/(m)sqrt((h)/(pi))`. |
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Answer» Correct Answer - C `Deltax*Deltap ge (h)/(4pi)` `Deltap ge sqrt((h)/(4pi))` . . . When `Deltax=Deltap` `mDeltavgesqrt((h)/(4pi))` ltbgt `Deltavge(1)/(2m)sqrt((h)/(2pi))` |
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| 71. |
The maximum number of sub levels, orbitals and electrons in N shell of an atom are respectivelyA. 4,12,32B. 4,16,30C. 4,16,32D. 4,32,64 |
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Answer» Correct Answer - C no of sub shell = n,no of orbitals `=n^(2)`, no of electrons `= 2n^(2)` |
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| 72. |
All electrons on the 4p sub-shell must be characterized by the quantum numberA. n=4, m=0, `s=pm1/2`B. l=1C. `l=0, s=pm 1/2`D. `s=pm1/2` |
| Answer» Correct Answer - B | |
| 73. |
If uncertainty in position and momentum are equal then uncertainty in velocity is.A. `sqrt((h)/(2 pi))`B. `(1)/(m) sqrt((h)/(pi))`C. `sqrt((h)/(pi))`D. `(1)/(2 m) sqrt((h)/(pi))` |
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Answer» Correct Answer - D (d) `Delta x = Delta p` :. `Delta x. Delta p = (h)/(4 pi)` or `Delta x = sqrt((h)/(4 pi))` Now, `Delta x. Delta u = (h)/(4 pi m)` :. `Delta u = (h)/(4 pi m) xx sqrt((4 pi)/(h)) = (1)/( 2m) xx sqrt((h)/(pi))`. |
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| 74. |
Maximum number of electrons in a sub-shell of an atom is determined by the following.A. `2 n^2`B. `4 l + 2`C. `2 l + 1`D. `4 l - 2` |
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Answer» Correct Answer - B (b) Total number of subshells `= (2l + 1)` `:.` Maximum number of electrons in the subshells `= 2(2 l + 1)` =`4 l + 2`. |
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| 75. |
The measurement of the electron position is associated with an uncertainty in momentum, which is equal to `1 xx 10^-18 g cm s^-1`. The uncertainty in electron velocity is (mass of an electron is `9 xx 10^-28 g`)A. `1 xx 10^6 cm s^-1`B. `1 xx 10^5 cm s^-1`C. `1 xx 10^11 cm s^-1`D. `1.1 xx 10^9 cm s^-1` |
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Answer» Correct Answer - D (d) `Deltax = Deltap therefore Deltax dot Deltap = (h)/(4pi)` or `Deltax= sqrt((h)/(4pi))` `Delta x. Delta u = (h)/(4 pi m)` `Delta p = 1 xx 10^-18g cm sec^-1` `m xx Delta u = 1 xx 10^-18` `:. Delta u = (1 xx 10^-18)/(9 xx 10^-28) = 1.1 xx 10^9 cm sec^-1`. |
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| 76. |
The number of angular and radial nodes of 4d orbital respectively areA. 3,1B. 1,2C. 3,0D. 2,1 |
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Answer» Correct Answer - D No of angular nodes `= l = 2` No of radial nodes `=n -l - 1 = 4 - 2 - 1= ` |
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| 77. |
An obital is found ot contain total nodes= 3 and radial nodes = 1. Obital angular momentum for the electron present in this orbital : |
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Answer» Correct Answer - B |
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| 78. |
The radial probability distribution curve obtined for an orbital wave function `(Phi)` has 3 peaks and 2 radial nodes. The valence electron of which one of the following metals does this wave function `(Phi)` correspond to.A. `Co`B. `Li`C. `K`D. `Na` |
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Answer» Correct Answer - D `Na_(11)to1s^(2),2s^(2),2p^(6),underset("valence electron")ubrace(3s^(1))` Number of radial node `=n-1l-1(n=3)` `=3-0-1=2` |
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| 79. |
The ratio of atomic number of two elements A and B is `1:2` The numebr of electrons present in the valence sheel `(3^rd)` of A is equal to the difference in the number of electrons present in the other two shells. Steps involved for the calculation of ratio of number of electrons present in a penultimate sheel to anti-penultimate sheel of B are given below : Arrange them in the correct sequence : (1) calculation of atomic number of B (2) calcualtion of valence electrons present in A. (3) calacualtion of atomic number of A. (4) calacualtion of number of electrons present in the penultimate and anti-penultimate shells of B. (5) writing electronic configuration of B.A. 23415B. 23154C. 45231D. 42135 |
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Answer» (i) calculation of valence electorns present in A (ii) calculation of atomic number of A (iii) calcualtion of atomic number of B (iv) writing electronic configuration of B (v) calcualtion of number of electrons present in the penultimate and anti-penultimate shells of B. |
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| 80. |
Calculate the wavelength of radiation emitted producing a line in the Lyman series ,when an electron falls from fourth stationary level in hydrogen atom `(R_(H) = 1.1 xx 10^(7)m^(-1))` |
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Answer» `(1)/(lamda)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` `=1.1xx10^(7)((1)/(1^(2))-(1)/(4^(2)))` `=969.6xx10^(-10)` metre `thereforelamda=969.4`Ã… |
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| 81. |
Calculate the wavelength of radiation emitted producing a line in the Lyman series ,when as electron falls dfrom fourth stationary in hydrogen atom `(R_(H) = 1.1 xx 10^(7)m^(-1)` |
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Answer» `(1)/(lambda) = R_(H) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` Given `R_(H) = 1.1 xx 10^(7)` for Lyman series `n_(1) = 1 `and `n_(2) = 4` (gives ) Therefore `(1)/(lambda) =1.1 xx 10^(7)[(1)/(1^(2)) - (1)/(4^(2))]` `lambda= 0.9696 xx 10^(-7) m)` |
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| 82. |
Not considering the electron spin, the degeneracy of second excited state is 9, while the degeneracy of then first excited state of `H^(-)` is : |
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Answer» Correct Answer - 3 |
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| 83. |
Calculate the wavelength of first line of Lyman series of ten times ionised sodium atom ` (Z = 11)` and compare with the wavelength of fire line of Balmer series of (H) atom . |
| Answer» `656 Å, 1.532 xx 10^(-3) xx 10^(-3)` timse of ` lambda_H`, | |
| 84. |
Pick out the isoelectronic structures from the following `underset (I) CH_3^+ underset (II) (H_3) O^+ underset (III) NH_3 underset (IV) CH_3^-`.A. I and IIB. I and IVC. I and IIID. II,III and IV |
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Answer» Correct Answer - D (d) `CH_3^+ = 6 + 3 - 1 = 8e^-`, `H_3 O^+ = 3 + 8 - 1 = 10 e^-` `NH_3 = 7 +3 = 10 e^-, CH_(3)^(-) = 6 + 3 + 1 = 10 e^-`. |
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| 85. |
Energy levels A,B,C of a certain atoms corresponding to increasing values of energy level i.e., `E_(A) lt E_(B) lt E_(C)`. If `lambda_(1), lambda_(2)` and `lambda_(3)` are the wavelengths of radiations corresponding to the transitions C to B,B to A and C to A respectively which of the following statement is correct?A. `lambda_(3) = lambda_(1) +lambda_(2)`B. `lambda_(3) = (lambda_(1)lambda_(2))/(lambda_(1)+lambda_(2))`C. `lambda_(1)+lambda_(2)+ lambda_(3) = 0`D. `lambda_(3)^(2) =lambda_(1)^(2)+lambda_(2)^(2)` |
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Answer» Correct Answer - B `E = (hc)/(lambda)` |
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| 86. |
The azimathal quantum number l of an arbital is `3` what are the possible values of m ? |
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Answer» Correct Answer - A::B::C The posible value of m are `+3,+2 +1,0,-1,-2,-3` |
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| 87. |
Any p arbital can accommodate up toA. 4 electronsB. 2 electrons with parallel spinsC. 6 electronsD. 2 electrons with opposite spins |
| Answer» Correct Answer - C | |
| 88. |
What is the difference in the angular momentum associated with the electron in two successive orbits of a hydrogen atom? |
| Answer» Correct Answer - `h//2pi` | |
| 89. |
The charge cloud of a single electron in a `2p` atomic orbital has two lobes of electron density .This metansA. There is a hight probability of locating the electron in the `2p_(s)` atomic orbital at values of `s gt 0`B. There is a hight probability of locating it at value of `s gt 0` but no probability at all of the locating it any where in the yz palne along which `x = 0`C. There is a greater probability of finding a p - right at the nuclessD. All are correct |
| Answer» Correct Answer - A::B::C | |
| 90. |
Which of the following statement is//are correct ?A. The energy of an electron in a many electron atom generally increases with an increases in value of n but for a given the lower the value of ? The lower the energyB. An electron close to the nucless experiences a large electrostate attractionC. For a given value of n an electron penetrates of the nucless more than n p el,ectron which penetrates more than a d-electron and so onD. None of correct |
| Answer» Correct Answer - A::B::C | |
| 91. |
The electronic configuration of `Ti^(2+)` ion is …………. |
| Answer» `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(1)` | |
| 92. |
The correct electronic configuration of Ti (Z=22) atom isA. `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2`B. `1s^2 2s^2 2p^6 3s^2 3p^6 3d^4`C. `1s^2 2s^2 2p^6 3s^2 3p^6 3d^4`D. `1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^3` |
| Answer» Correct Answer - A | |
| 93. |
Natural chlorine consists of two isotopes : 75% `""_(17)^(35)Cl and 25% ""_(17)^(37)Cl`. Calculate the average atomic mass of chlorine . |
| Answer» Average atomic mass `=(35xx75+37xx225)/(100)=35.5` | |
| 94. |
The electronic configuration of an atom A is a,a+b, a+2b , a and that of B a, a + b , 3a+2b, b. (a) Write the electronic configuration of A and B. (b) Calculate their atomic numbers, (c) If the number of neutrons in A is 5b and that in B is 22.5a, calculate their mass number. |
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Answer» (a) The electronic configuration of A is 2,8 , 14,2 and that of B is 2,8,18,6. (b) Their atomic number are 26 and 34 respectively . (c) The number of neutrons in A is 5b =30 and that in B=22.5 a =45 `therefore` 34+45 =79 respectively. |
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| 95. |
Sunny and Bunny have taken spectra for the following transitions .Spectra taken by sunny consited of bright lines on a dark background .Spectra taken by Bunny consisted of dark lines on a bright background .Identify the electronic transitions corrseponding to the spectra taken by sunny and Bunny .What are those spectra called? (i) n= 5 to n=2 (ii) n=3 to n=6 (iii) n=5 to n=7 (iv) n=3 to n=1 (v) n=1 to n=2 (vi) n=5 to n=3 |
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Answer» When an electron in an atom gains energy it gets excited to a higer energy state fron the ground state,The electron being unstable in the exiced state loses energy and comes back to the ground state either directly or in steps .During this process the bright lines corresponding to the transitions are formed on the dark background .This is called emission spectrum.When the electrons get exicted to higher energy and dark lines corresponding to the transitions are formed on the bright background.This is called absorption specrum . (i) n=5 to n=2 `rarr` emission spectum (ii) n=4 to n =6 `rarr` absorption spectrum (iii) n=5 to n =7 `rarr` absorption spectrum (iv) n=3 to n =1 `rarr` emission spectrum . (v) n= 1 to n=2 `rarr` absorption spectrum (vi) n=5 to n = 3 `rarr` emission spectrum |
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| 96. |
The wavelength associated with an electron moving with a velocity of `10^8 "cm s"^(-1)` isA. 72.7 ÅB. 7.27 ÅC. 0.727 ÅD. 0.277 Å |
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Answer» Correct Answer - B `lambda=h/"mv"=(6.6xx10^(-34))/(9.1xx10^(-31)xx10^8 xx 10^(-2))` `=.725xx10^(-9)m =7.25xx10^(-10)m` =7.25Å`~~`7.27Å |
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| 97. |
A moving electron has `5 xx 10^(-25) J `of kinetic energy .What is the de Broglie wavelength ? |
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Answer» Mass of the electron `(m) = 9.1 xx 10^(-31)kg` `KE = (1)/(2) mv^(2) = 5 xx 10^(-15) J` `lambda = (h)/(sqrt(2m xx KE)) = (6.626 xx 10^(-34))/((2 xx 9.1 xx 10^(-31) xx 5 xx 10^(-25))^(1//2)) = 6.95 xx 10^(-7) m` |
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| 98. |
The number of electrons in one molecule of `CO_2` areA. 22B. 44C. 66D. 88 |
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Answer» Correct Answer - A One molecule of `CO_2` have 22 electrons. |
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| 99. |
The kinetic energy of one electron is `2.8xx10^(-13)` J. What is the de-Broglie wavelengthA. `9.25xx10^(-13)` mB. `9.25xx10^(-16)` mC. `9.25xx10^(-8)` mD. `18.5xx10^(-13)` m |
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Answer» Correct Answer - A `because KE=1/2mv^2` Here, `m=9.1xx10^(-31)`kg, KE=`2.8xx10^(-13)`J `therefore 2.8xx10^(-13)=1/2xx9.1xx10^(-31)xxv^2` `v=sqrt((2xx2.8xx10^(-13))/(9.1xx10^(-31)))` `v=0.784xx10^9` m/s `because lambda=h/"mv"=(6.6xx10^(-34))/(9.1xx10^(-31)xx0.784xx10^9)` `=0.925xx10^(-12)` m `=9.25xx10^(-13) m ` |
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| 100. |
In hydrogen atom, an electron jumps from bigger orbit to smaller orbit, so that radius of smaller orbit is one-fourth of radius of bigger orbit. If speed of electron in bigger orbit was v,then speed in smaller orbit isA. `v/4`B. `v/2`C. `v`D. `2v` |
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Answer» Correct Answer - D Radius of nth orbit,`r_(n)propto n^(2)` `(r_("n big"))/r_("n small")=(n^(2)big)/("n^(2)small)=4/1 ` (given) Rightarrow`n_(big)/n_(small)=2` Rightarrow`n_(small)/n_(big)=1/2` Velocity of electron in nth orbit `v_(n)propto1/n` `v_"(n big)"/(v_"(n small)" )=(n_(small))/n_(big)=1/2` Rightarrow`V_"n small"=2(v_"n big")=2v` |
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