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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.
| 301. |
Assertion: It is essential that all the lines available in the emission spectrum will also be available in the absorption spectrum. Reason: The spectrum of hydrogen atom is only absorption spectrum.A. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true but (R) is falseD. (A) is false but (R) is true |
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Answer» Correct Answer - C The minimum frequency required to eject an electron from the surface of metal is called threshold frequency. |
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| 302. |
Assertion(A): Threshold frequency is a characteristic for a metal Reason(R): Threshold frequency is a maximum frequency required for the ejection of electron from the metal surface.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 - C Assertion is true but Reason is false . Threshold frequency is a minimum frequency required for the emission of electrons from the metal surface |
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| 303. |
Assertion :Absorption spectrum conists of some bright lines separated by dark spaces . Reason : Emission spectrum consists of dark linesA. 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 We know that Absorption spectrum is produced when white light is passed through a substance and transmitted light is analysed by a spectrograph. The dark spaces corresponds to the light radiation absorbed by the substance. And emission spectrum is produced by analysing the radiant energy emitted by an excited substance by a spectrograph. Thus discontinuous spectra consisting of a series of sharp lines and separated by dark bands are obtained . Therefore both the Assertion and Reason are false. |
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| 304. |
The energy of a hydrogen atom in its ground state is `-13.6 eV`. The energy of the level corresponding to the quantum number n=5 isA. `-0.54eV`B. `-5.40eV`C. `-0.85eV`D. `-2.72eV` |
| Answer» Correct Answer - A | |
| 305. |
The quantum number of four electrons (el to e4) are given below :- `{:(,n,,l,,m,,s,,),(e1,3,,0,,0,,+1//2,,),(e2,4,,0,,1,,1//2,,),(e3,3,,2,,2,,-1//2,,),(e4,3,,1,,-1,,1//2,,):}` The correct order of decreasing energy of these electrons is :A. `e4gte3gte2gte1`B. `e2gte3gte4gte1`C. `e3gte2gte4gte1`D. none |
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Answer» Correct Answer - C `r_(1)=(r_(2))/(n^(2))" "r_(1)=(r )/(4)" "," "r_(3)=r_(1)xxn^(2)` `r_(3)=r//4xx9" "r_(3)=2.25R` |
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| 306. |
Statement : For ` n=3,l` may be ` 0,1` and ` 2,` and (m) may be ` 0 , 0, +- 1,` and ` 0, +- 1` and `+- 2`. Explanation : For each value of (n) there are ` 0` to ( n-1) possible values of `l ,` and for each value of `l` there are (0) to ` +- 1` values of (m).A. S is correct but E is wrongB. S is wrong but E is correct.C. Both S and E are correct and E is correct explanation of SD. Both S and E are correct but E is not correct explanation of S |
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Answer» Correct Answer - C Explanation is correct reason for statement . |
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| 307. |
Write the electronic configuration and the atomic number of the atom which becomes stable by gaining 3 electrons in fifth shell. |
| Answer» Hence, the electronic configuration of the atom which becomes stable after gaining 3 electrons in the fifth shell is `2,8,18,18,5` and the atomic numebr is 51. | |
| 308. |
Statement : The tendency of a atom to reach a stable electronic arrangement may be satisfied by the transfer of electrons form one atom to another. Explanation : Loss and gain of electron constitute reduction and oxidation respectively .A. S is correct but E is wrongB. S is wrong but E is correct.C. Both S and E are correct and E is correct explanation of SD. Both S and E are correct but E is not correct explanation of S |
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Answer» Correct Answer - D It is a fact . |
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| 309. |
`[Cr(H_2 O)_6]Cl_3` (at no. of Cr = 24) has a magnetic moment of `3.83 B.M`. The correct distribution of `3d` electrons the chromium of the complex.A. ` 3d_(xy)^1, 3d_(yz)^1, 3_(xz)^1`B. ` 3d_(xy)^1, 3d_(yz)^1, 3_(z^2)^1`C. `(3d^1`._(x^(2) - y^2)), 3d^1 ._(z^2),3d^1._(xz)`D. `3d^1_(xy),(3d^1 _(x^2 - y^2)), 3d^1_(yz)` |
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Answer» Correct Answer - A (a) `Cu^(3+) = 1 s^2 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^3`. The `3 d_(xy)^1 3 d_(xz)^1 3 d_(yz)^1` has lower energy. |
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| 310. |
The orientation of an atomic orbital is governed by :A. azimuthal quantum numberB. spin quantum numberC. Magnetic quantum numberD. principal quantum number |
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Answer» Correct Answer - C ( c) The orientation of an atomic orbital is governed by magnetic quantum number. |
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| 311. |
` psi_(310)` has :A. `1` aredial node and `1` angualr nodeB. `1` aredial node and `1` angualr nodeC. `1` aredial node and `1` angualr nodeD. `1` aredial node and `1` angualr node |
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Answer» Correct Answer - A `psi_(310)` is 3p-prbital , Radial bnode ` =n -l -1` , angular node `=l`. |
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| 312. |
A particle of chagre equal to that of an electron and mass 208 times the mass of the electron moves in a circular orbit around nucleus of chagre `+3e`. Assuming that the Bohar of the atom is applicable to this system ,(a) dervie am expression for the radius of the radius of the nth bohar orbit ,(b) find the value of n for which the radius wavelenght of the raditatio emitted when the revolving particle jumps from the thrid orbit to the first. |
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Answer» Correct Answer - `r_(n)=(n^(2)h^(2))/(4Kpi^(2)xx3e^(2)xx208m_(e))n=25;55.2` pm Centrifgal force =Electrostatic force (a) `(1)/(4lambdaepsilon_(0)).(3e.e)/(r^(2))=(208me.(V^(2))/(r))" ".......(i)` `mvr=(nh)/(2pi)" "....(2)` `r=(208)/(3e^(2)).(4piepsilon_(0))(v^(2)r^(2))` `=(n^(2)h^(2))/(k.4pi^(2).3e^(2).208me)` (b) `r=r_(0).n^(2)=6.9xx106n^(2)=0.53xx10^(-10)m` `n^(2)~~625` `n=25` (c) `DeltaE=(1)/(2)m(V_(3)^(2)-V_(1)^(2))=(hc)/(lambda)` putting the value of `V_(3)` and `V_(1)` from eq.(1) and (2) `lambda=55.2xx10^(-12)m` |
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| 313. |
Two ` 1. 0 g ` carbon disks ` 1.00 cm` apart have opposite charges of equal magnitude such that there is a ` 1.00 xx 10^(-5) N` force between then . Calculate the ration of excess electron between them . Calculate the ration charged disk. |
| Answer» ` 4.2 xx 10^(-14) ,` | |
| 314. |
A hydrogen atom and a `Li^(2+)` ion ae both in the second excited state . If `l_H` and ` l_(li)` are their respective angular momentum in an orbit and `E_H` and ` E_(Li)` are their respective enrgies, then :A. `l_H gtl_(Li)` and `E_H gtE_(Li`B. `l_H gtl_(Li)` and `E_H gtE_(Li`C. `l_H gtl_(Li)` and `E_H gtE_(Li`D. `l_H gtl_(Li)` and `E_H gtE_(Li)` |
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Answer» Correct Answer - B `E_(2Li^(2+)) = E_(2H) xx Z^2` `:. E_(2Li^(2+)) =E_(2h) xx Z^2` `:. E_(2Li^(2+)) gt E_(2H^+)` `= sqrt (l (l + 1)) h/(2pi)` (`:. L= 0`for both ) `:. l_(H^+)= L_(Li^(2+))`. |
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| 315. |
The angular momentum of an electron in a particular orbit of H atom is `5.27 3 10^(-34) kg m^2//s` . Calculate the radius of the particular orbit of lithium. |
| Answer» When an electron in an atom gains energy, it gets | |
| 316. |
calculate the angular momentum of the electron in third orbit of hydrogen atom,if the angular momentum in the second orbit of hydrogen atom is L.A. `L`B. `3L`C. `3/2L`D. `2/3L` |
| Answer» Correct Answer - C | |
| 317. |
In which of the following transition, the wavelength will be minimum `:`A. `n=6 ` to `n=4`B. `n=4` to `n=2`C. `n=3` to `n=1`D. `n=2` to `n=2` |
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Answer» Correct Answer - 3 `lambda=(hc)/(E)" "rArr" "lambdaprop(1)/("Energy gap")` |
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| 318. |
What is the orbit angular momentum of a D ELECTRON ? |
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Answer» Orbit angular momentum `= sqrt(1(1 + 1))(h)/(2pi)` For d electron , `1= 2` Orbit angular momentum `= sqrt(2(2+1))(h)/(2pi) = sqrt(6)(h)/(2pi)` |
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| 319. |
Calculate the threshold frequency of metal if the binding energy is `180.69 kJ mol^(-1)` of electron. |
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Answer» Correct Answer - `4.5xx10^(14)s^(-1)` `B.E.=180.69kJ//"mole"impliesw=hv_(0)` `(180.69)/(96.368)eV//"atom"=hv_(0)` `(180.69)/(96.368)xx1.6xx10^(-19)=6.6xx10^(-34)xx v_(0)` `v_(0) = 6.626xx10^(-34)` |
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| 320. |
If the average life time of an excited state of H atom is of order `10^(-8)` sec, estimate how many orbits an `e^(-)` makes when it is in the state `n=2` and before it suffers a transition to `n=1` state. |
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Answer» Correct Answer - `8xx10^(6)` No.of revolutions `=("Total time")/("Time taken in one revolution ")` `10^(-8)/=((2pir)/V)=(10^(-8)xxV_(0)Z^(2))/(2pir_(0)n^(3))=(10^(-8)xx2.18xx10^(6)xx1^(3))/(2pixx.529xx10^(-10)xx2^(3))=8xx10^(6)` |
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| 321. |
What is the value of the spin quantum number of the last electron e^(9) configuration ? |
| Answer» The value of the spin quantum number (s) can be `+1//2` or `- 1//2` becqause an electron can rotate on its axis oither in clockwise or in unit-clockwise direction .But one quantum number depicts one electron and thus its velue will be `-1//2 ` for `d^(9)` configuration | |
| 322. |
The binding energy of electron in a metal is `193 kJ mol^(-1)` .Find the threshold frequancy of the metal |
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Answer» Correct Answer - A::C::D E(binding energy per atom) `= (193 xx 10^(3) J mol ^(-1))/(6.023 xx 10^(23) mol^(-1))` `= 3.20 xx 10^(-19) J" atom"^(-1)` Threshold frequency of the metal `v = (E)/(h) = (3.20 xx 10^(-19) J )/(6.626 xx 10^(-34) J s)- 4.83 xx 10^(14) s^(-1)` |
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| 323. |
What should be the value of the apin quantum number of the electron in d ? Configuration ? |
| Answer» The value of spin quantum num,ber (s) can be `+1//2` or `- 1//2` becqause in electron rotate on its axis either in clockwise or uin unit-clockwise direction .But one quantum number depicts one electron and this its velue will be `-1//2 for d^(9)` configuration | |
| 324. |
What is the difference is the angular momentum associated with the electron in two successive whits of a hydrogen atom ? |
| Answer» The angular momentum of an electron in an orbit is an integral of `h//2pi` .Therefore the difference In the angular momentum associated with the electronm in two sucessive orbit of hydrogen atom is `h//2pi` | |
| 325. |
The circumference of the second orbit of an atom or ion having single electron is 4 nm the de Broglie wavelength of electron (in nm) revolving in this orbit si : |
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Answer» Correct Answer - 2 |
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| 326. |
Calculate the momentum of a particle which has a de Broglie wavelength of `10 nm` |
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Answer» Correct Answer - A::B `1 J = 1 kg m^(2) s^(2) = 10^(7)erg` `1 eV = 1.60 xx 10^(-19) J` `p = (h)/(lambda) = 6.62 xx 10^(-34)( js)10 xx 10^(-9)(m)` `= 6.626 xx 10^(-26) J s m^(-1)` `= 6.626 xx 10^(-26) kg m s^(-1)` |
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| 327. |
Hydrogen when subjected to photon disocation, yieds one normal atom and atom possessing `1.97 eV` more energy than normal atom .The bond dissociation energy of hydrogen molecule into normal atom is `103 `kcal `mol^(-1)` . Campate the wavelength of effective photon for photo dissociation of hydrogen molecule in the given case |
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Answer» `H_(2) rarr H + H^(o+)` where H is atom and `H^(o+)` is excited H atom so the energy required to dissociate `H_(2)` in this marmer will be greater then the bond energy of `H_(2)` + extra energy of `E_(("absorbed"))` = dissociate energy of `H_(2)` + extra exrtra energy of excuted atom Energy required to dissociation in normal mather `= 103 xx 10^(3) cal` `= (103 xx 10^(3) xx (4.18))/(6 xx 10^(23))` `= 7.716 xx 10^(-19) J atom ^(-1)` The extra energy possessed by the excited atomk in `1.97 eV= 1.97 xx 1.6 xx 10^(-19) + 3.152 xx 10^(-19)` `E_(("abesorbed")) = 7.17 xx 10^(-19) + 3.152 xx 10^(-19) = 1.080668 xx 10^(-19) J` Now calculate the wavelength of photon corresponding this energy `lambda= (hc)/(E ) = (6.63 xx 10^(-34) xx 3xx 1100^(8))/(1.0868 xx 10^(-18))` `= 1.830 xx 10^(-7)m = 1830 Å` |
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| 328. |
Find the wavelength of light emitted when an electron drops from `n=3` level to ground state in `He^(+)` ion Given Rydberg constant `=10^(7)m^(-1)` |
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Answer» `bar(upsilon)=(1)/(lamda)=Z^(2)R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]implies bar(upsilon)=(2)^(2).(10^(7)).[(1)/(1)-(1)/(9)]implies bar(upsilon)=4xx10^(7)[(8)/(9)]` `implies lambda = (9)/(32)xx10^(-7)m=0.2813xx10^(-7)m=281.3Å` |
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| 329. |
A photon of energy `12 eV` can break three molecules of `A_2` into atoms which has bond dissociation energy of `4 eV//"molecule"`. Total energy is conserved and interaction is always one to one between photon and molecule.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 - D (d) Since interaction between a photon and a molecule is always one to one, so a photon of energy `12 eV` can break only one molecule of `A_2` into atoms and remaining `8 eV` energy becomes kinetic energy of atoms. |
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| 330. |
A photon of energy `12 eV` can break three molecules of `A_2` into atoms which has bond dissociation energy of `4 eV//"molecule"`. Total energy is conserved and interaction is always one to one between photon and molecule. |
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Answer» Correct Answer - d |
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| 331. |
Photon having wavelength 310 nm is used to break the bond of `A_(2)` molecule having bond energy 288 kJ `mol^(-1)` then `%` of energy of photon converted to the KE is : [hc = `12400 eV"Å"` , 1 e V = 96 kJ/mol]A. 25B. 50C. 75D. 80 |
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Answer» Correct Answer - A |
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| 332. |
Minimum accelerating potential (volts) needed to accelerate an electron to produce a yellow line (`lambda= 310` nm) in the spectrum , in an electron tube containing some vapours is :A. 1.2B. 4C. `6.4 xx 10^(-19)`D. `2 xx 10^(-15)` |
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Answer» Correct Answer - B |
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| 333. |
Suppose a particle has four quantum numbers such that the permited values are those as given below: n,1,2,3... `l: (n-1), (n-3),(n-5)`...but no negative number `J:(//+(1)/(2))` or `(//-(1)/(2))` if the latter is not negative. m : J in integral steps to +J Thus, how many particles could be fitted into that `n = 2` shell? |
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Answer» Correct Answer - 6 `l = 1, j = (3)/(2),(1)/(2), m =- (3)/(2), -(1)/(2), +(1)/(2),+(3)/(2), -(1)/(2), +(1)/(2)` |
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| 334. |
Give the notation for the sab-shell denoted by the following quantum number: |
| Answer» a.`5d` b. `6d` c. `4s` d. `5g` | |
| 335. |
If the principal quantum has a value of `3` what are the permited values of the quantum number l ? |
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Answer» Correct Answer - A `1` will have value `0,1` and `2` |
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| 336. |
If the quantum number has a value of `2` what are the permited values of the quantum number m ? |
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Answer» Correct Answer - A If `l = 2` then permited values of m are `-2 , -1,0,+1,+2` |
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| 337. |
Give the set of quantum number that describe an electron in a `3p` orbital. |
| Answer» In a `3p` orbital , the principal quantum number (n) is `3` , the azimuthal quantum number (l) is `1` , the magnetic momentum are `m_(1) = + 1,0, - 1`, the quantium number `(m_(1)) is + 1//2 or 1//2` | |
| 338. |
In two element `._(Z1)A^(M1)` and `._(Z2)B^(M2)` `M_(1)"ne" M_(2)` and `Z_(1)"ne"Z_(2)` but `M_(1)-Z_(1) =M_(2)-Z_(2)`. These elements areA. IsotonicB. `IsoharicC. IsotopleD. Isoprotonic |
| Answer» `M_(1) =` atomic weight `Z_(1) = `Atomic number in isohares `M_(1) = M_(2)` and in isotopes `Z_(1) = Z_(2)` IN isonce (isoncatromic element) `M_(1) -Z_(1) = M_(2) - Z_(2)` | |
| 339. |
How many sub-shell are there in N shell ? How many orbitals are there in d sub-shell ? |
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Answer» For N shell principle quantum number is ` n = 4` ,Number of sub-shell in an energy level`n`, :. Number of sub-shells in N shell `= 4` ,For d subshell, `1 = 2 ` Number of orbitals in a sub-shell` = 2l+ 1` gt Number of orbitals in a sub-shell` = 2xx 2+ 1= 5` |
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| 340. |
The number of orbitals in 2p sub-shell isA. 6B. 2C. 3D. 4 |
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Answer» Correct Answer - C It has 3 orbitals `p_x,p_y,p_z` |
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| 341. |
The number of orbitals in d sub-shell isA. 1B. 3C. 5D. 7 |
| Answer» Correct Answer - C | |
| 342. |
Number of orbitals in h sub-shell isA. 11B. 15C. 17D. 19 |
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Answer» Correct Answer - A `{:(s,p,d,f,g,h),(l=0,1,2,3,4,5):}` Number of orbitals = 5 x 2 +1 =11 |
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| 343. |
How many orbitals are possible in a. 4th energy level b. `5f` sub-shell |
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Answer» Correct Answer - A::B a.`16` b. `7` |
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| 344. |
(a) An electron is in 5f-orbital. What possible values of quantum numbers n,lm and s can it have? (b). What designation is given to an orbital having (i) n=2, l=1 and (ii). N=3,l=0? |
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Answer» (a). For an electron in 5f-orbital, quantum numbers are: n=5, l=3, m=-3,-2,-1,0,+1,+2,+3 and s=either`+(1)/(2)` or `-(1)/(2)` (b) (i) 2p, (ii) 3s. |
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| 345. |
An isotone of `._(32)^(76)Ge` is-A. `._(32)^(77)Ge`B. `._(33)^(77)As`C. `._(34)^(77)Se`D. `._(34)^(78)Se` |
| Answer» Correct Answer - B::D | |
| 346. |
In a single isolated atom an electron make transition from 5th excited state to 2nd state then maximum number of different types of photons observed is |
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Answer» Correct Answer - 3 In a single isolated atom, no of spectral lines formed `=n_(2)-n_(1)` |
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| 347. |
In a sample of three H-atoms , all in the 5th excited state, electrons make a transition to 1st excited state. The maximum number of different spectral lines observed will be : |
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Answer» Correct Answer - 7 |
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| 348. |
The number of lines of Balmer series of H-atom that belong to visible region. |
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Answer» Correct Answer - 4 |
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| 349. |
The maximum kinetic energy of photoelectrons ejected from a metal, when it is irradiated with radiation of frequency `2xx10^(14)s^(-1)` is `6.63xx10^(-20)`J. the threshold frequency of the metal is:A. `2xx10^14 s^(-1)`B. `3xx10^14 s^(-1)`C. `2xx10^(-14) s^(-1)`D. `1xx10^(-14) s^(-1)` |
| Answer» Correct Answer - D | |
| 350. |
For a given value of quantum number l , the number of allowed values of m is given byA. l+2B. 2l+2C. 2l+1D. l+1 |
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Answer» Correct Answer - C m=-1 to +1 including zero |
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