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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. |
Consider the ground state `Cr` atom `(Z = 24)`. The number of electron with the azimuthal number `l = 1` and `2` ,respectively areA. 16, 5B. 16, 4C. 12, 4D. 12, 5 |
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Answer» Correct Answer - D Cr(Z=24)`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)3d^(5)` No. of p-electrons (l=1)=12 electrons No. of d-electrons (l=2)=5 electrons |
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| 52. |
An electron in an atom can be completely designated with the help of four quantum numbers. Out of these, the first three i.e., principal (n), azimuthal (l) and magnetic (m) quantum number are obtained from the solution of Shrodinger wave equation while the spin(s) quantum number arises from the spin of the electron around its axis clockwise or antiaclockwise. Ot of these principal quantum number tells about the size, azimuthal quantum number about the shape and magnetic quantum signifies the orientation of the electron orbital. How many electrons in a given atom have the following set of quantium numbers? `n = 3, l =2, m = +2, s = -1//2`A. 1B. 18C. 14D. not possible |
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Answer» Correct Answer - A Only one electron can have this value. |
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| 53. |
An electron in an atom can be completely designated with the help of four quantum numbers. Out of these, the first three i.e., principal (n), azimuthal (l) and magnetic (m) quantum number are obtained from the solution of Shrodinger wave equation while the spin(s) quantum number arises from the spin of the electron around its axis clockwise or antiaclockwise. Ot of these principal quantum number tells about the size, azimuthal quantum number about the shape and magnetic quantum signifies the orientation of the electron orbital. The electronic configuration of P in `H_(3)PO_(4)` isA. `1s^(2)2s^(2)2p^(6)3s^(2)3p^(3)`B. `1s^(2)2s^(2)2p^(6)3s^(2)`C. `1s^(2)2s^(2)2p^(6)`D. `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)` |
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Answer» Correct Answer - C Oxidation of P in `H_(3)PO_(4)` `x+3(+1)+4(-2)=0` `x=8-3=+5` The configuration of `P^(5+)` ion is : `1s^(2)2s^(2)2p^(6)`. |
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| 54. |
`| Psi^(2)|` can haveA. Any value from zero to 1B. Any value from `-1` to `+1`C. A positive non-zero valueD. A non -zero value |
| Answer» Correct Answer - 1 | |
| 55. |
Choose the following regarding Bohr modelA. It introduce the concept of quantisation of energyB. Radius of shellis given by ` r= ( 0.529n^(2))/( z) Å`C. K.E. of electron is given by `( 2pi^(2) Me^(4) z^(2)k^(2))/( n^(2)h^(2))`D. Two isotopes have same ionisation energy |
| Answer» Correct Answer - 1,2 | |
| 56. |
Energy of level 1,2,3 of a certain atom corresponds to increasing value of energy `E_(1) lt E_(2) lt E_(3)` . If `lambda_(1) , lambda_(2)` and `lambda_(3)` are the wavelength of radiation corresponding to transition `3 rarr 2, 2 rarr1` and `3 rarr1` respectively. Which ofte following statement is `//` are correct ?A. `(1)/(lambda_(3))= ( 1)/(lambda_(1))+(1)/(lambda_(2))`B. `lambda_(3)= (lambda_(1) lambda_(2))/(lambda_(1)+lambda_(2))`C. `(1)/(lambda_(2))= ( 1)/(lambda_(1))+(1)/(lambda_(3))`D. `lambda_(2)= (lambda_(1) lambda_(3))/(lambda_(1)+lambda_(2))` |
| Answer» Correct Answer - 1,2 | |
| 57. |
The potential energy of an electron in hydrogen atom is `-3.02 eV`, its kinteic energy will beA. `1.51 eV`B. `15.10 eV`C. ` 13.6 eV`D. ` 1.36 eV` |
| Answer» Correct Answer - 1 | |
| 58. |
Find out the incorrect matchA. Electron - Dual natureB. Rainbow - Discontinuous spectraC. Atoms - Line spectraD. Particle nature - Photoelectric effect |
| Answer» Correct Answer - 2 | |
| 59. |
How many degenerate orbitals are presentin a subshell if electron associated with that subshell possesses orbital angular momentum `=2 sqrt (3) h` ?A. No degenerate orbitalsB. Seven degenerates orbitalsC. Three degenerate orbitalsD. No of degenerate orbitals are same as in subshell which possesses `mu_(L) = 3.46 ` B.M. |
| Answer» Correct Answer - 2,4 | |
| 60. |
If radius of `2^(nd)` orbit is x then di-Broglie wavelength in`4^(th)`orbit is given gyA. `8 pi x `B. `2pi x `C. `4 pi x`D. `6 pi x ` |
| Answer» Correct Answer - 2 | |
| 61. |
An electron is moving in `3^(rd)` orbit of Hydrogen series and radius of first orbit is x thenA. de-Broglie wavelength is `6 pi x`B. de-Broglie wavelength is`2pi x`C. Velocity of electron is `(h)/( 6 pi x m)`D. Velocity of electron is `(h)/( 2pixm)` |
| Answer» Correct Answer - 1,3 | |
| 62. |
How many nodal planes are present in `4d_(z^(2)) ` ?A. 2B. 1C. ZeroD. 3 |
| Answer» Correct Answer - 3 | |
| 63. |
What is the energy associated with `3^(rd)` energy shell of hydrgoen atom ?A. `- 2.18 xx 10^(-18) J`B. ` - 0.342 xx 10^(-19) J`C. ` -0. 726 xx 10^(-18) JD. ` - 2.42 xx 10^(-19) J` |
| Answer» Correct Answer - 4 | |
| 64. |
An electron is moving in `3^(rd)` orbit of `Li^(+2)` and its separation energy is y.The separation energy of an electron moving in `2^(nd)` orbit of `He^(+)` isA. `(4y)/( 9)`B. `(y)/(9)`C. `-(y)/(9)`D. y |
| Answer» Correct Answer - 4 | |
| 65. |
(i) Calculate the total number of electrons present in one mole of methane (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of `.^(14)C` (Assume that the mass of neutron `= 1.675 xx 10^(-27) kg`) (iii) Find (a) the total number and (b) the total mass of protons in 34 mg of `NH_(3)` at S.T.P. (Assume the mass of proton `= 1.6726 xx 10^(-27) kg`) Will the answer change if temperature and pressure are changed ? |
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Answer» (i) 1 molecule of `CH_(4)` contains electrons `= 6 + 4 = 10` `:.` 1 mole, i.e., `6.022 xx 10^(23)` molecules will contain electrons `= 6.022 xx 10^(24)` (ii) (a) 1 g atom of `.^(14)C = 14g = 6.022 xx 10^(23)` atom `= (6.022 xx 10^(23)) xx 8` neutrons (as each `.^(14)C` atom has `14 - 6 = 8` neutrons) Thus, 14g or 14000 mg have `8 xx 6.022 xx 10^(23)` neutrons `:.` 7 mg will have neutrons `= (8 xx 6.022 xx 10^(23))/(14000) xx 7 = 2.4088 xx 10^(21)` (b) Mass of 1 neutrons `= 1.67 xx 10^(-27) kg` `:.` Mass of `2.4088 xx 10^(-21)` neutrons `= (2.4088 xx 10^(21)) (1.67 xx 10^(-27) kg) = 4.0347 xx 10^(-6) kg` (iii) (a) 1 mol of `NH_(3)` = 17 g `NH_(3) = 6.022 xx 10^(23)` molecules of `NH_(3)` `= (6.022 xx 10^(23)) xx (7 +3)` protons `= 6.022 xx 10^(24)` protons `:.` 34 mg i.e., `0.034 g NH_(3) = (6.022 xx 10^(24))/(17) xx 0.034 = 1.2044 xx 10^(22)` protons (b) Mass of one proton `= 1.6726 xx 10^(-27) kg` `:.` Mass of `1.2044 xx 10^(22)` protons `= (1.6726 xx 10^(-27)) xx (1.2044 xx 10^(22)) kg = 2.0145 xx 10^(-5) kg` There is no effect of temperture and pressure. |
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| 66. |
Identify the `Mg^(2+)` ion from the figure where, n and p represent the number of neutrons and protons respectively.A. B. C. D. |
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Answer» Correct Answer - d Correct answer d since the ion has 10 electrons it is `Mg^(2+)` ion |
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| 67. |
(i) Calculate the total number of electrons present in 1 mole of methane . (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of `.^(14) C`. (Assume that mass of a neutron ` = 1 . 6 75 xx 10 ^(-27) g)` (iii) Find (a) the total number of protons and (b) the total mass of protons in ` 32 mg` of ` NH_3` at ` STP`. ( mass of proton ` = 1 . 6 72 xx 10^(-27) g)` Will the answer change if the temperature and pressure are changed ? |
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Answer» number of electrons prsent in 1 molecule of methane `(CH_(4))` `{1(6) +4(1)}=10` number of electrons present in 1 mole i.e `6.023 xx 10^(23) ` molecules of methane `=6.022 xx10^(23) xx10 =6.022 xx10^(24)` (ii) (a) number of atoms of `.^(14)C` in 1 mole = `6.023 xx10^(23)` since 1 atom `.^(14)C` contains (14 – 6) i.e., 8 neutrons, the number of neutrons in ` 14 g` of `^.(14) C` is `( 6.023 xx10^(23) ) xx 8 `, Or `14 g` of `.^(14)C` contains ` (6.022 xx 10^(23) xx8)` neutrons. Number of neutrons in 7 mg `(6.022xx10^(23)xx8xx7mg)/(1400 mg)` ` 2.4092 xx10^(21) ` (b) Mass of one neutron `=1.67493 xx10^(-27)kg` Mass of total neutrons in` 7g` of `.^(14)C` `(2.4092 xx 10^(21)) (1.67493 xx10^(-27) kg) ` `4.0352 xx 10^(-6)kg ` (iii) (a) 1 mole of ` NH_(3) ={1(14) +3(1) } g of NH_(3) ` ` 17 g` of `NH_(3) ` `6.022 xx10^(23) "molecules" of NH_(3)` Total number of protons present in 1 molecules of `NH_(3)` `{1(7) +3(1)}` 10 Number of protons in ` 6.023xx10^(23)` molecules of `NH_(3)` `(6.023xx10^(23) ) (10)` `6.023xx10^(24)` `Rightarrow 17 g of NH_(3) "contains" (6.023 xx10^(24)) ` protons. Number of protons in 34 mg of `NH_(3)` `( 6.022 xx10^(24)xx34 mg)/(17000 mg)` `1.2046 xx 10^(22)` (b) Mass of one proton `= 1.67493 xx 10^(-27) kg` Total mass of protons in 34 mg of `NH_(3)` `( 1.67493 xx 10^(27) kg) (1.2046 xx 10^(22)` `2.0176xx10^(-5) kg` The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed. |
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| 68. |
In a hydrogen atom, an electron jumps from third orbit to the orbit. Find out the spectral line. |
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Answer» Step I. Calculation of the wave length of the spectral line According to Rydbergy formula : `(1)/(lambda)=bar(v)=R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))] , R_(H)=109678 cm^(_1) , n_(1)=1, n_(2)=3`. `:. " " (1)/(lambda)=109678 [(1)/(1^(2))-(1)/(3^(2))]=109678 [(9-1)/(9)]cm^(-1)`. `"or" " " lambda=(9)/(109678 xx8) cm =(9xx10^(-2)m)/(109678 xx 8)=1.0257 xx 10^(-7) m`. Step II. Calculation of the frequency of the spectral line `c=v lambda "or" v=c//lambda , c=3.0 xx 10^(8)ms^(-1), lambda =1.0257 xx 10^(-7) m`. `:. " " v=((3.0 xx 10^(8)ms^(-1)))/((1.0257 xx 10^(-7)m))=2.9248 xx 10^(15) s^(-1)`(Hz). |
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| 69. |
An electron in H -atoms jumps from some higher level to 3rd energy level.If three spectral lines are possible for the transition, find the initial position of electron. |
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Answer» Number of spectral lines `=((n_(2)- n_(1))(n_(2)-n_(1)+1))/(2)` ` implies 3 = ((n_(2)-3)(n_(2)-3+1))/(2)` `implies 6 = n_(2)^(2) - 5n_(2) + 6` ,brgt `implies n_(2)(n_(2)-5)=0` or `n_(2)= 0,n_(2)= 5` Since`n_(2)= 0`is not possible , hence `n_(2) = 5` |
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| 70. |
Which of the following is correct? (i) A particle of light, a photon, has a definite enegry `E = hv`. (ii) The photon has momentum, `mc`. (iii) The momentum of photon is related to the wavelength of the light.A. `(i),(ii)`B. `(i),(ii),(iii)`C. `(i),(iii)`D. `(ii),(iii)` |
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Answer» Correct Answer - B According to the de Broglie equation, `mc = h//lambda`. |
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| 71. |
An element E loses one `alpha` and `beta`-particles in three successive stages. The resulting element will beA. an isobar of EB. an isotone of EC. an isotope of ED. E itself |
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Answer» Correct Answer - C `._(Z)^(A)Eoverset(-alpha)rarr _(Z-2)^(A-4)X overset(-beta)rarr _(Z-1)^(A-4)Y overset(-beta)rarr_(" "Z)^(A-4)W` Both elements E and W have same atomic number but different mass numbers. Hence, both are isotopes. So, the resulting element will be an isotope of E. |
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| 72. |
Calculate the range of frequencies of visible light from `3800 - 7600 Å` |
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Answer» Correct Answer - `3.95 xx 10^(14) " to " 7.89 xx 10^(14) s^(-1)` `lambda = 3800 Å = 3800 xx 10^(-10) m = 3.8 xx 10^(-7) m, " " v = (c)/(lamda) = (3 xx 10^(8) ms^(-1))/(3.8 xx 10^(-7) m) = 7.89 xx 10^(14) s^(-1)` `lamda = 7600 Å = 7600 xx 10^(-10) m = 7.6 xx 10^(-7) m, " " v = (c)/(lamda) = (3 xx 10^(8) ms^(-1))/(7.6 xx 10^(-7) m) = 3.95 xx 10^(14) s^(-1)` |
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| 73. |
Which of the following orbitals has maximum enegry?A. `2s` orbitals of `Li`B. `2s`orbital of `H`C. `2s` orbital of `Na`D. All have the same enegry |
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Answer» Correct Answer - B Energies of the orbitals in the same subshell decrease with increase in the atomic number `(Z)`, thus, the enegry of the `2s`orbital of `H` atom is gerater than that of the `2s` orbital of `Li` and that of `Li` is greater than that of `Na`, and so on, that is, `E_(2s)(H) gt E_(2s)(Li) gt E_(2s)(Na) gt...` |
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| 74. |
How many times is the velocity of the electron in the first shell of `He^(+)` ion as compared to that in the first shell of hydrogen atom ? |
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Answer» Correct Answer - 2 `v_(n) = v_(0) Z//n`. For H, Z = 1 and for 1st shell, n = 1 Hence, `v_(1) = v_(0)` For `He^(+), Z = 2` and for 1st shell, n = 1. Hence `v_(1) = 2v_(0)` |
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| 75. |
An electron is moving in 3rd orbit of Hydrogen atom . The frequency of moving electron isA. `2.19 xx 10^(14)rps`B. `7.3 xx 10^(14)` rpsC. `2.44 xx 10^(14)` rpsD. `7.3 xx 10^(10) ` rps |
| Answer» Correct Answer - 3 | |
| 76. |
The number of orbitals present in the `3rd` shell isA. `3`B. `1`C. `9`D. `18` |
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Answer» Correct Answer - C Each subshell of quantum number `n` contains `n^(2)` orbitals. Thusm, for `n - 3`, there are `(3)^(2) = 9` orbitals. |
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| 77. |
Which of the following symbols are not acceptable ? (i) `._(35)^(79)Br` (ii) `.^(79)Br` (iii) `._(79)^(35)Br` (iv) `._(35)Br`A. `(ii),(iv)`B. `(iii),(iv)`C. `(ii),(iii),(iv)`D. `(ii),(iii)` |
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Answer» Correct Answer - B `(3)` is not acceptable because shorthand notation for any nuclide consists of the symbol of the element with the atomic number written as a subscript on the left and mass number as a superscript on the left. `(4)` is not acceptable because if we know the name of the element `(Br)`, we always know its atomic number `(35)`, but we may not know its mass number. hence, it is essential to indicate its mass number. |
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| 78. |
Symbols `._(35)^(79)Br` and `.^(79)Br` can be written whereas symbols `._(79)^(35) Br` and `._(35)Br` are not accepted. Answer in brief. |
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Answer» In the symbol `._(A)^(B)X` of an element : A denotes the atomic number of the element B denotes the mass number of the element. The atomic number of the element can be identified from its symbol because no two elements can have the atomic number. However, the mass numbers have to be mentioned in order to identify the elements. Thus, Symbols `._(35)^(79)Br` and `.^(79) Br` are accepted because atomic number of Br will remain 35 even if not mentioned. Symbol`._(79)^(35) Br` is not accepted because atomic number of Br cannot be 79 (more than the mass number = 35). Similarly, Symbol `._(35)Br` cannot be accepted because mass number has to be mentioned. This is needed to differentiate the isotopes of an element. |
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| 79. |
Number of angular nodes for 4d orbtial is.........A. 4B. 3C. 2D. 1 |
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Answer» Correct Answer - C No. of angular nodes = l For 4d orbital, n = 4, l = 2 `therefore " "` No. of angular nodes for 4d orbital = 2 |
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| 80. |
The orbital angular momentum of an electron in `2s`-orbital isA. ZeroB. OneC. TwoD. Three |
| Answer» Correct Answer - A | |
| 81. |
A new electron enters the orbital when:A. value of n is minimumB. value of l is minimumC. value of (n + 1) is minimumD. value of (n + m) is minimum |
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Answer» Correct Answer - C The electron will enter into an orbital with minimum value of n + l. |
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| 82. |
Calculate the energy of an electron in the first Bohr orbit of`He^(+)`. |
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Answer» `E_(n) = -2.18 xx 10^(-18) ((Z^(2))/(n^(2)))J = - 13.6 (Z^(2))/(n^(2)) eV = 54.4 eV` `Z=2 ` `n=1` `E_(1)= (- 2.18 xx 10^(-18 ) xx ( 2)^(2))/( (1)^(2))` `E_(1) =- 8.72 xx 10^(-18) J` |
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| 83. |
The electrons identified by quantum numbers n and l : (1) `n = 4, l = 1` (2) `n = 4, l = 0` (3) `n = 3, l = 2` (4) `n = 3, l = 1` can be placed in order of increasing energy as :A. `(1) gt (3) gt (2) gt (4)`B. `(3) gt (4) gt (2) gt (1)`C. `(4) gt (2) gt (3) gt (1)`D. `(2) gt (4) gt (1) gt (3)` |
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Answer» Correct Answer - C `(1) = 4p, (2) = 4s, (3) = 3d, (4) = 3p` The order of increasing energy is `3p lt 4s lt 3d lt 4p " i.e., " (4) lt (2) lt (3) lt (1)` |
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| 84. |
How many orbitals in total are associated with `4^(th)` energy level?A. 4B. 9C. 16D. 7 |
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Answer» Correct Answer - C No. of orbitals = `n^(2)`, if n = 4 No. of orbitals = 16 |
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| 85. |
The correct set of quantum numbers for the unpaired electron of chlorine atom isA. `n=2,l=1,m=0`B. `n=2,l=1,m=1`C. `n=3,l=1,m=1`D. `n=3,l=0,m=0` |
| Answer» Correct Answer - 3 | |
| 86. |
Which of the following represent(s) the correct set of four quantum numbers of a 5d electron ?A. `4,3,2, +(1)/(2)`B. ` 4,2,1, +(1)/(2)`C. ` 4,3,-2,+(1)/(2)`D. ` 4,1,1,+(1)/(2)` |
| Answer» Correct Answer - 2 | |
| 87. |
The electrons identified by the following quantum numbers `n` and `l: (i) n = 4, l = 1, (ii) n = 4, l = 0, (iii) n = 3, l = 2`, and (iv) `n = 3, l = 1` can be placed in the order of increasing enegry from the lowest to the highest asA. `(iii) lt (i) lt (iv) lt (ii)`B. `(i) lt (iii) lt (ii) lt (iv)`C. `(ii) lt (iv) lt (i) lt (iii)`D. `(iv) lt (ii) lt (iii) lt (i)` |
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Answer» Correct Answer - D According to `(n +l)` rule, the electroc with lower `(n +l)` has lower enegry. Thus, (ii) and (iv) `(n +l = 4)` have lower enegry than (i) and (iii) `(n +l = 5)`. If two `e^(-)s` have same `(n +l)` value, then `e^(-)` of lower `n` has lower enegry. Thus, (iv) has lower enegry than (ii), while (iii) has lower enegry than (i)`. |
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| 88. |
Maximum number of orbitals is given shell identified by the principal quantum number `n` is equak toA. `n`B. `n^(2)`C. `2n^(2)`D. `2n` |
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Answer» Correct Answer - B All the orbitals of a given value of `n` consitute a single shell of atom. With the increase in the value of `n`, the number of allowed orbitals increases and is given by `n^(2)`. |
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| 89. |
The elctron identified by quantum numbers n and l (i) `n=4,l=1` (ii) `n=4,l=0` (iii) `n=3,l=0` (iv) `n=3,l=1` can be placed in order of increasing energyA. `(iii) lt (iv) lt (i) lt (ii)`B. `(iii) lt (iv) lt (ii) lt (i)`C. `(i) lt (iii) lt (ii) lt (iv)`D. `(iii) lt (i) lt (iv) lt (ii)` |
| Answer» Correct Answer - 2 | |
| 90. |
The orbital diagram in which the Aufbau principle is violated isA. B. C. D. |
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Answer» Correct Answer - C Aufbau principle is violated in option( c) as electrons switch to p orbitals without completely filling the s orbital which has comparatively lower energy. |
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| 91. |
Which of the sequences given below shows the correct increasing order of energy?A. 3s, 3p, 4s, 4p, 3d, 5s, 5p, 4dB. 3s, 3p, 3d, 4s, 4p, 4d, 5s, 5pC. 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5pD. 3s, 3p, 4s, 4p, 5s, 3d, 4d, 5p |
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Answer» Correct Answer - C 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p is the correct order of orbitals with increasing energy. |
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| 92. |
Which of the following set of quantum number belongs to highest energy.A. n=4, l=0, m=0, s=+1/2B. n=3, l=0, m=0, s=+1/2C. n=3, l=1, m=+1, s=+1/2D. n=3, l=2, m=+1, s=+1/2 |
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Answer» Correct Answer - D 3d orbital has the highest energy. |
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| 93. |
Give one example of each of the following: (i) Isotope of `._(17)^(35)Cl` (ii) Isobar of `._(18)^(40)Ar` (iii) Isotone of `._(6)^(14)C` |
| Answer» (i) `._(17)^(37)Cl` (ii) `._(19)^(40)K` (iii) `._(8)^(16)O` | |
| 94. |
The most abundant isotope of hydrogen is :A. Hydrogen ionB. ProtiumC. DeuteriumD. Tritium. |
| Answer» Correct Answer - D | |
| 95. |
If a body weighing 1.0g is travelling along x axis at 100 cm `s^(-1)` within 1 cm `s^(-1)`, what is the theoretical uncertainty in its position ? |
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Answer» According to uncertainty principle, `Delta xmDeltaupsilon =(h)/(4 pi) or Deltax=(h)/(4 pi m Delta upsilon)` Since there is uncertainty of 1 cm `s^(-1)` in measuring velocity, the means that velocity varies between 99 xm `s^(-1)` to 101 cm ` s^(-1) or Delta upsilon = (upsilon_(2)-upsilon_(1))=(101-99)=2 cm s^(-1) or 0.02 m s^(-1)`. `Deltax=6.626 xx 10^(-34) kg m^(2)s^(-1), m=1.0g =10^(-3)kg` `Delta x=((6.626xx10^(-34)kgm^(2)s^(-1)))/(4xx3.143 xx (10^(-3)kg)xx (0.02ms^(-1)))=2.636 xx 10^(-30)m` |
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| 96. |
Energy of the state `S_(1)` in units of the hydrogen atom ground state energy isA. 0.75B. 1.5C. 2.25D. 4.5 |
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Answer» Correct Answer - C For `S_(1)` state of `Li^(2+)`, n = 2 and Z = 3. `therefore" "` Energy of the state `S_(1)` in the units of hydrogen atom ground state energy is : `E=E_(H)xx(Z^(2))/(n^(2))=E_(H)xx(3^(2))/(2^(2))=9/4E_(H)=2.25xxE_(H)` |
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| 97. |
The orbital angular momentum quantum number of the state `S_(2)` is |
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Answer» Correct Answer - B The state `S_(2)` has one radial node and its energy is equal to the ground state energy of the hydrogen atom. This is possible only for 3p orbital. 3p orbital has one radial node, `n=3,l=1rArr(n-l-1)=(3-1-1)=1 E=E_(H)xx(Z^(2))/(n^(2))=E_(H)xx(3^(2))/(3^(2))=E_(H)` Azimuthal quantum number for the state `S_(2)`, l = 1. |
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| 98. |
On solving Schrodinger wave equation for hydrogen atom, the values of the enery obtained are called......and the corresponding values of the wavefunction `(Psi)` are called....... |
| Answer» eigen values, eigen functions | |
| 99. |
What is the physical significance of `Psi and Psi^(2)` ? |
| Answer» `Psi` as such has no physical significance. `Psi^(2)` gives the probability of finding the electron at any point around the nucleus. | |
| 100. |
What is the significance of the sign `psi^(2)` ? |
| Answer» `psi^(2)` gives the measure of the probable electron density in different regions in space. | |