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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.
| 1551. |
The wave function `Phi_(n,l,m_(1))` is a methematical function whose value depends upon shperical upon spherical polar coordinates `(r,theta,phi)` of the electron and characterized by the quantum numbers, n,l and `m_(1)`. Here r is distance from nucleus, `theta` is colatitude and `phi` is azimuth. in the mathematical functions given in the table, Z is atomic number, `a_(0)` is Bohr radius. . Q. For hydrogen atom the only CORRECT combination is :A. (I) (iv) (R)B. (I) (I) (P)C. (II) (i) (Q)D. (I) (i) (S) |
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Answer» Correct Answer - D 1s (orbita) has wave function `Phi_("nlm")prop((Z)/(a_(0)))^((3)/(2))E^(-Zr//a_(0))` `E=-(Z^(2))/(n^(2))xx13.6eV` `DeltaE_(2-4)=(1)/(16)xx13.6-(-(1)/(4)xx13.6)` `=(3)/(16)xx13.6eV` `DeltaE_(2-6)=-(1)/(36)xx13.6-(-(1)/(4)xx13.6)` `=(8)/(36)xx13.6eV` `(DeltaE_(2-4))/(DeltaE_(2-6))=(3//16xx13.6eV)/(3//36xx13.6eV)` `=(3xx36)/(16xx8)=(27)/(32)` `thereforeDeltaE_(2-4)=(27)/(32)DeltaE_(2-6)` |
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| 1552. |
The probability of finding the electron in the orbital isA. `100 %`B. `90-95%`C. `70-80%`D. `50-60%` |
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Answer» Correct Answer - B The probablity of finding the electron in the orbital is `90-95%` |
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| 1553. |
The wave function, `psi_(n), l, m_(l)` is a mathematical function whose value depends upon spherical polar coordinates `(r,theta,phi)` of the electron and characterised by the quantum number `n,l` and `m_(l)`. Here `r` is distance from nucleus, `theta` is colatitude and `phi` is azimuth. In the mathematical functions given in the Table, `Z` is atomic number and `a_(0)` is Bohr radius. For hydrogen atom, the only CORRECT combination isA. (I ) (i)(P)B. (I)(iv)(R)C. (II)(i)(Q)D. (I)(i)(S) |
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Answer» Correct Answer - D |
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| 1554. |
Which have the same number of s-electrons as the d-electrons in ` Fe^(2+)` ?A. `Li`B. `Na`C. `N`D. `P` |
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Answer» Correct Answer - D `Fe^(2+)` has `6e^(-)` in 3d orbitals |
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| 1555. |
Through what potential difference must an electron pass to have a wavelength of `500A^(@)`. |
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Answer» Correct Answer - `6.03xx10^(-4)"volt"` `500=sqrt((150)/(V))` `therefore (150)/(250000)=V" "thereforeV=6xx10^(-4)`volt |
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| 1556. |
The maximum number of electrons in p-orbital with n=5, m=1 isA. 6B. 2C. 14D. 10 |
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Answer» Correct Answer - B The maximum number of electron in any orbital is 2 |
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| 1557. |
The number of waves in an orbit areA. `n^(2)`B. nC. `n-1`D. `n-2` |
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Answer» Correct Answer - B The number of waves in an orbit is equal to n. (principal quantum number) |
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| 1558. |
Which of the following statement about proton is correct?A. Proton is the nucleus of deuteriumB. Proton is an `alpha`- particleC. Proton is an ionised hydrogen moleculeD. Proton is ionised hydrogen. |
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Answer» Correct Answer - Proton is ionised hydrogen Proton is ionised hydrogen |
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| 1559. |
The ion that is most stableA. `Fe^(+)`B. `Fe^(2+)`C. `Fe^(3+)`D. `Fe^(4+)` |
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Answer» Correct Answer - C Half filled d-orbital is stable |
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| 1560. |
Which one is the ground stateA. B. C. D. |
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Answer» Correct Answer - B Each orbital has almost two electron |
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| 1561. |
Cr has electronic configuration asA. `3s^(2)3p^(2)3d^(4) 4s^(1)`B. `3s^(2)3p^(6)3d^(5) 4s^(1)`C. `3s^(2)3p^(6)3d^(6)`D. none of these |
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Answer» Correct Answer - B `Cr (24)=1s^(2), 1s^(2),2s^(2)2p^(6),3s^(2)3p^(6)3d^(5)4s^(1)` |
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| 1562. |
Which of the following electron configurations is correct for iron,(atomic number26)?A. `1s^(2), 2s^(2)2p^(6), 3s^(2)3p^(6) 3d^(5)`B. `1s^(2), 2s^(2)2p^(6), 3s^(2)3p^(6), 4s^(2), 3d^(5)`C. `1s^(2), 2s^(2)2p^(6), 3s^(2)3p^(6), 4s^(2), 3d^(7)`D. `1s^(2), 2s^(2) 2p^(6), 3s^(2)3p^(6)3d^(6), 4s^(2)` |
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Answer» Correct Answer - D (d) Firstly the electrons are filled in increasing order of energy and then rearrange the subshells in increasing order as `._(26)Fe=1s^(2), 2s^(2) 2p^(6), 3s^(2) 3p^(6)3d^(6),4s^(2)` |
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| 1563. |
Which of the following shows an increasing value of `e//m`?A. `n lt alpha lt p lt e`B. `n lt p lt alpha lt e`C. `n lt p lt e lt alpha`D. ` p lt n lt alpha lt e` |
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Answer» Correct Answer - `n lt alpha lt p lt e` `n lt alpha lt p lt e` |
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| 1564. |
Which of the following statements is incorrect?A. Extra stability of half filled and completely filled orbitals among s and p block elements is reflected in trends of IE across a periodB. Extra stability of half filled and completely filled orbitals among s and p block elements is reflected in E.A trends across a period.C. Aufbau principle is incorrect for cases where energy difference between ns and `(n-1)d` subshell is largerD. Extra stability to half filled subshell is due to higher exchange energies |
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Answer» Correct Answer - C Aufbau rule |
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| 1565. |
The wave number of a spectral line is `5xx10^(5)m^(-1)` The energy corresponding to this line isA. `3.49xx10^(-23)` kjB. `4.45xx10^(-24)` jC. `5.50xx10^(-22)` JD. `9.93xx10^(-23)` kJ |
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Answer» Correct Answer - D Energy `E=(hc)/lambda = hc barv` `=6.62xx10^(-34)xx3xx10^(8)xx5xx10^(5)(barv=1/lambda)` `=9.93xx10^(-23)` kj |
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| 1566. |
The electronic configuration of gadolinium (Atomic number 64) isA. `6s^(2)5d^(1) 4f^(7)`B. `6s^(2)5d^(0)4f^(8)`C. `6s^(1)5d^(0)4f^(7)`D. `6s^(1)5d^(2)4f^(7)` |
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Answer» Correct Answer - A follow hunds rule, aufbau rule, paulis rule |
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| 1567. |
An electronic transition in hydrogen atom result in the formation of `Halpha` line of Hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in `kcal mol^(-1)`) areA. `-313.6-34.84`B. `-313.6-78.4`C. `-78.4-34.84`D. `-78.4-19.6` |
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Answer» Correct Answer - B Energy of an electron in bth orbit `E_(n)=(2pi^(2)Z^(2)e^(4))/(n^(2)h^(2))` On substituting the values of k m e and h we get `E_(n)=(2.172xx10^(-18)Z^(2))/n^(2)J atom^(-1)` `=(1311.8Z^(2))/n^(2)kJ mol^(-1)=-(313.52Z^(2))/n^(2)kcal mol^(-1)` [.: 1 kacal =4.184kJ] For H - atom Z=1 For Lyman series `n_(1)=1 n_(2)=2` `=-(313.52xx (1)^(2))/(1)^(2)kcalmol^(-1)` `=-313.52kcal mol^(-1)` `=-313.6 kcal mol^(-1)` Energy of electron in `n_(2)` orbit `=-(313.52xx(1)^(2))/(2)^(2)kcalmol^(-1)` `-(313.52)/4 kcal mol^(-1)` `=-78.38 kcal mol^(-1)` |
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| 1568. |
One energy difference between the states `n = 2` and `n= 3` is `EeV`, in hydrogen atom. The ionisation potential of `H` atom is -A. 3.2 EB. 5.6 EC. 7.2 ED. 13.2 E |
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Answer» Correct Answer - C ( c) Ionization energy `= E_prop - E_n` =`0 - [-(Z^2)/(n^2) xx 13.6 eV]` =`(Z^2)/(n^2) xx 13.6 eV` `13.6 Z_2 ((1)/(n_1^2) -(1)/(n_2^2))` `n_1 = 2, n_2 = 3` =`13.6 xx 1((9 - 4)/(4 xx 9))` =`(5)/(9) xx 13.6 = 7.2 E`. |
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| 1569. |
What is the ratio of wavelength of (II) line fo Balmer series and (I) line fo Lyman seies ? |
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Answer» Correct Answer - 4 For (II) line of Balmer seres ` N_1 = 2 , n_2 =4` ` :. 1/( lambda_(2B)) = R_H xx [1/2^2- 1/4^2]` ` = R_H xx 3/(16) `…(1) For (I) line fo Lyman series ` n_1 = 1, n-2 =2` ` :. 1/( lambda_(1l) = R_H xx [1/1^2 -1/2^2]= R_H xx 3/4` …..(2) by eqs , (1) and (2) ltbRgt ` ( lambda_(2B))/( lambda 1L) = (R_H xx 3 xx 16)/( 4 xx 3xx R_H) = 4`. |
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| 1570. |
One energy difference between the states `n = 2` and `n= 3` is `"E eV"`, in hydrogen atom. The ionisation potential of `H` atom is -A. 3.2 EB. 5.6 EC. 7.2 ED. 13.2 E |
| Answer» Correct Answer - 3 | |
| 1571. |
Find the ratio ofenergy of a photon of ` 2000Å` wavelength radiation to that of ` 400Å` wavelength radiation to that of `4 000Å` radiation . |
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Answer» Correct Answer - 2 `E = (hc)/( lambda) i.e., E prop 1/( lambda)` `:.E_1/E_2 = ( lambda_2)/( lambda_1) = (4000)/(2000)=2`. |
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| 1572. |
The electrons in a poly -electronic atom are filled one by one in roder fo increasing energy level . The enrgy fo subsgells and orientation depeds upon the values of three quantum numbers 9 i.e., n. l and m respectively ) derived from Schrodinger wave equation . The different orbitals fo a subsshells however possess same energy level and are called degenerater orbitals but their enrgy level charges in presenct fo magentic field and the orbitals are non-degenerate . (A) spectral line is noticed it an electron jups form one level to tohere . the paramgnetic nature of eelment sis due to the presence of unpaired electron . The element which has as many as `s` electrons as (p) electrons by belong to (III) period is ?A. `O`B. `Mg`C. `Al`D. `C` |
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Answer» Correct Answer - B `._(12) Mg:1s^2 , 2s^2 2p^6 , 3s^2`. |
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| 1573. |
The electrons in a poly -electronic atom are filled one by one in roder fo increasing energy level . The enrgy fo subsgells and orientation depeds upon the values of three quantum numbers 9 i.e., n. l and m respectively derived from Schrodinger wave equation . The different orbitals fo a subsshells however possess same energy level and are called degenerater orbitals but their enrgy level charges in presenct of magentic field and the orbitals are non-degenerate . (A) spectral line is noticed it an electron jups form one level to tohere . the paramgnetic nature of eelment sis due to the presence of unpaired electron . The total mangetic moment fo ` Ni^(2)` ion is :A. ` sqrt(2) BM`B. `sqrt(8) BM`C. `sqrt(15) BM`D. ` sqrt(12)BM` |
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Answer» Correct Answer - B `Ni^(2+) , 3d^8 i.e, n=2 ` unpaired eelctron s. Magnetic moment `= sqrt(n(n +2)) BM` ltbRgt ` = sqrt(2(2 +2)) = sqrt( 8)`. |
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| 1574. |
The radiation having maximum wave length isA. Ultraviolet raysB. Radio wavesC. X-raysD. Infra-red rays |
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Answer» Correct Answer - B |X-rays|U.V rays|I.R.|Ratio `rarr lambda uarr` |
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| 1575. |
One energy difference between the states `n = 2` and `n= 3` is `"E eV"`, in hydrogen atom. The ionisation potential of `H` atom is -A. `3.2E`B. `7.2E`C. `5.6E`D. `13.2E` |
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Answer» Correct Answer - B `(E_(1))/(3^(2)) -(E_(1))/(2^(2)) = E rArr (-5E_(1))/(36) = E, E_(1) = - 7.2E` `I.E. +E_(1) = +7.2E` |
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| 1576. |
The ionisation potential of hydrogen atom is `13.6 eV` The energy required to remve as electron in the `n = 2` state of the hydrogen atom isA. `3.4 eV`B. `6.8 eV`C. `13.6 eV`D. `27.7 eV` |
| Answer» Correct Answer - A | |
| 1577. |
Ionisation potential of hydrogen atom is ` 13 . 6 eV`. Hydrogen atom in the ground state is excited by monochromatic light fo energy ` 12. 1 eV` . The spectral lines emitted by hydrogen according to Bohr`s theory will be.A. OneB. TwoC. ThreeD. four |
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Answer» Correct Answer - C ( c) The electron in `H` atom is excited to `III` snell after absorbing `12.1 eV`. The possible transition state `= (3 - 1) = 3`. |
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| 1578. |
Ionisation potential of hydrogen atom is ` 13 . 6 eV`. Hydrogen atom in the ground state is excited by monochromatic light fo energy ` 12. 1 eV` . The spectral lines emitted by hydrogen according to Bohr`s theory will be. |
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Answer» Correct Answer - 3 `E_n = -13. 6 + 12. 1 =- 1.5 eV` ` :. N^2 = E_1 /E_n = ( 13.6)/( 1.5) =9` ` n=3` Thus transiton occurs from ` 3rd` to `1st` orbit ltbRgt ` :.` Spectral lines emitted ` = sum Delta n = sum (3 - 1)` ` = sum 2 = 1 + 2 = 3` . |
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| 1579. |
The Speeds of the fiat and ferrari racing cars are recorded to `+-4.5xx10^(-4)msec^(-1)`. Assuming the track distance to be known within `+-16m`, is the uncertainty principle violated for a 3500 kg car? |
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Answer» `DeltaxDeltav=4.5xx10^(-4)xx16` `=7.2xx10^(-3)m^(2)sec^(-1)` . . . .(i) `(h)/(4pim)=(6.626xx10^(-34))/(4xx3.14xx3500)` . . .(ii) `=1.507xx10^(-38)` Since, `Deltax Deltav ge h//4pim`. Hence, Heisenberg uncertainty principle is not violated. |
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| 1580. |
The average atimic mass of two isopoes with mass numbers A and `A+2` is `A+0.25`. Calculate the percentage abundance of the isotopes. |
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Answer» (i) `(n_(1)xxA)+(n_(2)xx(A+2)=(n_1+n_2)(A+0.25)` from (ii) `87.5% and 12.5%` |
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| 1581. |
The m value not possible for a double dumbell shaped orbital is |
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Answer» Correct Answer - C For d-orbital `l = 2`, so `m=-2, -1,0,+1,+2` are possible |
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| 1582. |
Calculate the momentum of radiation of wavelength `0.33 nm` |
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Answer» We have `lambda = (h)/(mv)` `:. Mv = (h)/(lambda) = (6.625 xx 10^(-34))/(0.33 xx 10^(-9)) = 2.01 xx 10^(-24)kg m s^(-1)` |
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| 1583. |
Calculate the deBroglie wavelength of an electron travilling at `1%` of the speed of the light |
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Answer» The mass of electron is `9.11 xx 10^(-31) kg ` One percent of the speed of light is `s = (1.100)(3.00 xx 10^(8) m s^(-1))` `= 3.00 xx 10^(6) m s^(-1)` The momentum of the electron is given by `p = mv = (9.11 xx 10^(-31) kg) (3.00 xx 10^(6) m s^(-1))` ` = 2.733 xx 10^(-24) kg m s^(-1)` The de Broglie wavelength of this electron is `lambda = (h)/(p) = (6.626 xx 10^(-34) Js)/(2.733 xx 10^(-24) kg m s^(-1))` `= 2.424 xx 10^(-10) m = 242 .4 "pm" ` The wavelength is of atomic dimenssion |
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| 1584. |
The ratio of velocity of the electron in the third and fifth orbit of `Li^(2+)` would be :A. `3:5`B. `5:3`C. `25:9`D. `9:25` |
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Answer» Correct Answer - B `T_(n,z) alpha (n^(3))/(z^(2))` |
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| 1585. |
The circumference of `n^(th)` orbit in H-atom can be expressed in terms of deBroglie wavelength ` lambda` as :A. `(0.529)nlambda`B. `sqrt(nlambda)`C. `(13.6)lambda`D. `nlambda` |
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Answer» Correct Answer - D `m_(y) = 0.25 m_(x), v_(y) = 0.75 v_(x)` ` lambda = (h)/(mv) " "lambda_(x) = (h)/(m_(x)v_(x)),lambda_(y)=(h)/(m_(y)v_(y))` `lambda_(y)=(h)/(0.25M_(x)xx0.75v_(x)) " "lambda_(y)=5.33 A` |
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| 1586. |
`A` particle X moving with a certain velocity has a debroglie wave length of `1A^(@)`. If particle Y has a mass of `25%` that of X and velocity `75%` that of X, debroglies wave length of Y will be :-A. `3A^(@)`B. `5.33A^(@)`C. `6.88A^(@)`D. `48A^(@)` |
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Answer» Correct Answer - B Orbital angular momentum `=sqrt(l(l+1))h` `{:(,,s,p,d,f),(,l=,0,1,2,3):}` |
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| 1587. |
Which of the followin g orbits of hydrogen atom should have the value of their radius in the radius `1:4`?A. K and LB. L and NC. m and ND. a and b are correct |
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Answer» a.Ratio of radil of arbit K and L `(n_(1))/(n_(2)) = (n_(1)^(2))/(n_(2)^(2)) = (1^(2))/(2^(2)) = 1:4` b..Ratio of radil of arbit L and N `(n_(1))/(n_(2)) = (n_(1)^(2))/(n_(2)^(2)) = (2^(2))/(4^(2)) = 4:16 or 1:4` |
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| 1588. |
The ratio of the fifth orbit of `He^("Theta") and Li^("Theta")`will beA. `2:3`B. `3:2`C. `4:1`D. `5:3` |
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Answer» Correct Answer - `3:2` Here n is a constant therefore `r_(1)/(r_(2)) = (Z_(2))/Z_(1) = (3)/(2) = 3: 2` |
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| 1589. |
The work function for a metal si ` 4 eV`. To emit a photoelectron of zero velocity from the surface fo the metal the wavelength of incident light showld be :A. `2700Å`B. `1700Å`C. `5900Å`D. `3100Å` |
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Answer» Correct Answer - D ` hv=` "work function" `+ KE ` Given `KE =0` , `Thus ` hv = 4e V` ` or ` 4 = ( 12375)/( lambda) ` where `lambda` "is in" `Å` `thereforelambda = 3100Å`. |
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| 1590. |
Give the electronic of the following a `H^(o+)`b.`Li^(o+)` c. `F^(o+)` d. `N^(o+)` |
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Answer» Correct Answer - A a.`H^(Theta) = 1s^(2)` b. `Li^(o+a)` = ls^(2)` c. `F^(Theta) = 1s^(2)2s^(2)2p^(6)` `d. N^(2+) = 1s^(2)2s^(2)2p^(1)` |
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| 1591. |
What is the lowest value of n that allow g orbitals |
| Answer» For g suub-shell `l= 4` The minimum value of n for which l can be `4` is `4 +1 `or `5` | |
| 1592. |
Which of the following atoms and ions are isoelectronic (i.e. Have dor the same number of electrons) with a neon atom a C b. `O^(2-)` c.` n^(Theta)` d. ` F^(o+)` e. `Na^(o+)` f. `AI^(3+)` |
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Answer» Correct Answer - A `O^(2-),Ne^(Theta),AI^(3+)` |
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| 1593. |
Which of the following atoms and ions are isoelectronic i.e. have the same numbe of electrons with the neon atomA. `F^-`B. oxygen atomC. mgD. `N^-` |
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Answer» Correct Answer - A `F^-` have the same number of electrons with the neon atom |
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| 1594. |
The subshell that arises after f is called the g subshell.How many electrons may occupy the g subshell?A. 9B. 7C. 5D. 18 |
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Answer» Correct Answer - D g subshells has 9 orbitals |
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| 1595. |
Calculate the energy of one moles of quanta of radiation whose frequency is `5 xx 10^(10)sec^(-1)` |
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Answer» Energy of 1 mol quantum `= Nhv` `= 6.023 xx 10^(23) xx 6.626 xx 10^(-34) xx 5xx 10^(10)` `= 19.95 J mol^(-1)` |
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| 1596. |
The subshell that rises after f subshell is called g subshell What is the total number of orbitals in the shell in which the g subshell first occur?A. 9B. 16C. 25D. 36 |
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Answer» Correct Answer - C For g-subshell, l=4 |
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| 1597. |
What is the wavelength of a photon emitted during a transition from `n = 5` state to the `n = 2` state in the hydrogen atomA. 434 nmB. 234nmC. 476nmD. 244nm |
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Answer» Correct Answer - A `bar(v) = (1)/(lambda) = R [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` |
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| 1598. |
What is the number of photons of light with a wave length 4000 pm that provide 1J energy ? |
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Answer» wavelength, `lambda = 4000 p m = 4 xx 10^(-19)m` `h = 6.625 xx 10^(-34)Js, c = 3.0 xx 10^(8)ms^(-1)` Energy of one photon, `E = (hc)/(lambda)` `= (6.625 xx 10^(-34)Js xx 3 xx 10^(8)ms^(-1))/(4xx 10^(-9)J) = 4.969 xx 10^(-17)J` Number of photons providing 1 joule of energy `= (1)/(4.969 xx 10^(-17)) = 2.01 xx 10^(16)` |
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| 1599. |
An electron having the quantum numbers n=4, l=3 , m=0 , `s=-1/2` would be in the orbitalA. 3sB. 3pC. 4dD. 4f |
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Answer» Correct Answer - D For f orbital l =3 |
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| 1600. |
Calculate the energy associated with photon of light having a wavelength 6000Ã…. [`h=6.624xx10^(-27)erg-sec`.] |
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Answer» We know that `E=hv=h.(c)/(lamda)` ltBrgt `h=6.624xx10^(-27)erg-sec,c=3xx10^(10)cm//sec` So, `E=((6.624xx10^(-27))xx(3xx10^(10)))/(6xx10^(-5))=3.312xx10^(-12)erg`. |
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