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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.
| 1351. |
Find the quantum number `n` corresponding to the excited state of `He^(o+)` ion if on transition to the ground state that ion emits two photon in succession with wavelength `108.5` and `30.4 nm` |
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Answer» Gives `lambda_(2) = 30.4 xx 10^(-7) cm, lambda_(2) = 108.5 xx 10^(-7) cm` Let the excited state of `He^(Theta)` be `n_(2)` if comes from `n_(2)` to`n_(1)` and then `n_(1)` to `1` to emit two successive photon `(1)/(lambda_(2)) = R_(H)Z^(2)[(1)/(1^(2)) - (1)/(n_(1)^(2))]` `(1)/(30.4 xx 10^(-7)) = 109678 xx 4[(1)/(1^(2)) - (1)/(n_(1)^(2))]` `:. n_(1) = 2` `Now for lambda_(1) : n_(1) = 2 and n_(2) ,`? `(1)/(lambda_(1)) = R_(H)Z^(2)[(1)/(2^(2)) - (1)/(n_(2)^(2))]` `(1)/(108.5 xx 10^(-7)) = 109678 xx 4 [(1)/(2^(2)) - (1)/(n_(2)^(2))]` `:. n_(2) = 5` Thus excited state for lies is fifth orbit |
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| 1352. |
Assertion (A) : `Fe^(3+)` (g) ion is more stable than `Fe^(2+)(g)` ion. Reason (R) : `Fe^(3+)` ion has more number of unpaired electrons than `Fe^(2+)` ion.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 |
| Answer» Correct Answer - B | |
| 1353. |
Find the quantum number `n` corresponding to the excited state of `He^(Theta)` ion if on transition to the ground state that ion emits two photon in succession with wavelength `108.5 `and `30.4 nm` |
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Answer» `lambda_(1) = 30.4 xx 10^(-7) cm` `lambda_(2) = 108.5 xx 10^(-7) cm` Let the excited state of `He^(Theta)` be `n_(2)` it comes from `n_(2)` to `n_(1)` and then from `n_(1)` to `1` to emit two successive photons `(1)/(lambda_(2)) = R_(H)Z^(2)[(1)/((1)^(2)) - (1)/(n_(1))^(2)]` `:. n_(1) = 2` Now for `lambda_(2)n_(1) = 2,n_(2) = `? `(1)/(108.5 xx 10^(-7)) = 109678 xx (2)^(2)[(1)/((2)^(2))- (1)/((n_(2))^(2))]` `:. n_(2) = 3` Hence the excited state of `He^(Theta)` is fifth orbit |
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| 1354. |
The total allowed values of m for an given value of l are equal to ………….. |
| Answer» Correct Answer - `2l + 1` | |
| 1355. |
Match the List -I and List-II and select the correct set from the following sets given below: `{:((A)." The number of sub-energy levels in an energy level",(1)." "n^(2)),((B)." The number of orbitals in a sub-energy level",(2)." "3d),((C)." The number of orbitals in an energy level",(3)." "2l+1),((D)." n=3,l=2,m=0",(4)." "n):}`A. `{:((A),(B),"(C)","(D)"),("4","3","1","2"):}`B. `{:((A),(B),(C),(D)"),(3,1,2,4):}`C. `{:((A),(B),(C),(D)"),(1,2,3,4):}`D. `{:((A),(B),(C),(D)"),(3,4,1,2):}` |
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Answer» Correct Answer - A Number of orbitals in a shell `=n^(2)` Number of subshell in a shell=n Number of orbitals in a subshell=(2l+l) n=3,l=2,m=0 represents 3d |
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| 1356. |
Which of the following has maximum number of unpaired electrons?A. `Zn`B. `Fe^(2+)`C. `Ni^(3+)`D. `Cu^(+)` |
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Answer» Correct Answer - B `Fe^(2+) rarr 3d^(5) 4s^(0)` |
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| 1357. |
The work function of a metal is `4.2 eV` If radiation of `2000 Å`fall on the metal then the kinetic energy of the fastest photoelectrn isA. `1.6 xx 10^(-19) J`B. `16 xx 10^(10) J`C. `3.2 xx 10^(-19) J`D. `6.4 xx 10^(-10) J` |
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Answer» Correct Answer - C KE = ("Energy of radiation= Work function") `= (h xx (c )/(lambda) - 4.2)` `= ((6.6 xx 10^(-34) J s xx 3 xx 10^(8) M)/(2000 xx 10^(-10)m)) - (4.2 xx 1.602 xx 10^(-19) J` `= (9.9 xx 10^(-19) J) - (6.7 xx 10^(-19)J = 3.2 xx 10^(-19) J` |
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| 1358. |
Among the following transition in hydrogen and hydrogen-like spectrum, which one emits light of lngest wavelength ?A. `n = 2` "to" `n = 1` "for" `H`B. `n = 4` "to"` n = 3` "for"` Li^(2+)`C. `n = 4` "to" `n = 3` "for" `He^(o+)`D. `n = 5` "to" `n = 2` "for" `H` |
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Answer» Correct Answer - C Longest `lambda` or shortest E a. `(1)/(lambda) = R ((1)/(1) - (1)/(4)) = (3R)/(4) = 0.75 R` b.`(1)/(lambda) = R xx Z^(2)((1)/(3^(2)) - (1)/(4^(2))) = R xx 9 xx (7)/(114) = 0.4375 R` c.`(1)/(lambda) = R xx 2^(2)((1)/(3^(2)) - (1)/(4^(2))) = R xx 4 xx (7)/(114) = 0.194R` `(1)/(lambda) = R ((1)/(2^(2)) - (1)/(5^(2))) = R xx (21)/(100) = 0.21 R` Lowest value of `1//lambda` highest value of `lambda` is in c |
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| 1359. |
What is the maximum number of electron in an atom that can have the quantum numbers `n = 4, m_(l) =+1`?A. 4B. 15C. 3D. 6 |
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Answer» Correct Answer - D n=4, l=1,`m_(2)=0` `l=1,m_(e)=-1,0,+1` `l=2,m_(2)=-2,-1,0,+1,+2` `l=3,m_(e)=-3,-2,-1,0,+1,+2,+3` There are three orbitals having `m_(e)=+1`, thus maximum number of electrons in them will be 6. |
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| 1360. |
A hydrogen like species (atomic number Z) is present in a higher excited state of quantum number n. This excited atom can make a transitionn to the first excited state by successive emission of two photons of energies 10.20 eV and 17.0 eV respectively. Altetnatively, the atom from the same excited state can make a transition to the second excited state by successive of two photons of energy 4.25 eV and 5.95 eVv respectively. Determine the value of Z. |
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Answer» Correct Answer - 3 `Delta E = 13.6 xx z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))] rArr :. Z = 3` |
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| 1361. |
What is the shape `2s` orbital .Give two9 point of difference between `1s` and `2s` orbital |
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Answer» Correct Answer - A Spherical shape Difference between `1s` and `2s` a. `2s` orbital is bigger in size b. `2s` orbital has energy than `1s`1 orbital |
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| 1362. |
A photon of frequency n causes photoelectric emmission from a surfece with thereshold .The de Broglie wavelength `lambda ` of the photoelectrn emitted is given asA. `Delat n = (h)/(2m lambda)`B. `Delat n = (h)/(lambda)`C. `[(1)/(v_(0)) - (1)/(v)] = (mc^(2))/(h)`D. `lambda = sqrt((h)/(2m Delta n))` |
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Answer» Correct Answer - D `E_(1)= IE + KE` or `E_(1) = `Threshold E (or) work function `+ KE` `(hv = hv_(0) + (1)/(2) m u^(2))` or `hn = hn_(0) + (1)/(2) m u^(2)` `(1)/(2) m u^(2) = h(n - n_(0)) = h Delta n`…….(i) `(lambda = (h)/(mu) ,:. U = (h)/(m lambda))` Subsitute the value n in equation (i) `(1)/(2) m(h^(2))/(m^(2)lambda^(2)) = h DRelta n` `(h)/(2lambda^(2)m) = Delta n ` `:. lambda = sqrt((h)/(2m Delta n))` |
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| 1363. |
`Na^(Theta)` and Ne are ……. To each other |
| Answer» Correct Answer - Isoelectronic | |
| 1364. |
An isotone of `_(32)Ge^(36)` isi. `_(32)Ge^(77)` ii. `_(33)As^(77)` iii. `_(34)Se^(77)` iv. `_(34)Se^(78)`A. Only (i) and (ii)B. Only i(i) and (iii)C. Only (ii) and (iv)D. (ii),(iii) and (iv) |
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Answer» Correct Answer - C Isotones have same number of neutrons `_(32)Ge^(36) n = 76 - 32 = 44` i.`_(32)Ge^(77) n = 77 - 32 = 45` ii.`_(31)As^(77) n = 77 - 33 = 44` iii.`_(34)Se^(36) n = 77 - 34 = 43` iv.`_(34)Se^(36) n = 78 - 34 = 44` So anser is c |
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| 1365. |
How many electron in an atom may have the following quantum number ? A `n = 4, m_(s) = -(1)/(2)` b `n = 3,l = 0` |
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Answer» Correct Answer - A a. For `n = 4,,32(2 xx 4^(2))` electron are possible out of which `16` can have `m_(s) = - 1//2` b. `n = 3, l = 0 ` corresponding to `3 n` subshell. It can accommodate only `2` electron |
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| 1366. |
`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)` is not the electron configuration ofA. `Mn^(3+)`B. `Fe^(3+)`C. `Cr^(+)`D. `Co^(4+)` |
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Answer» Correct Answer - A `Mn^(3+)` has 23 electrons |
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| 1367. |
The number of d-electrons in `Fe^(2+)`(Z=26) is not equal to the number of electrons in which one of the following ?A. s-electrons in Mg (Z=12)B. p-electrons in Cl (Z=17)C. d-electrons in Fe (Z=26)D. p-electrons in Ne (Z=10) |
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Answer» Correct Answer - B (b) Electronic configuration of `Fe^(2+)` is [Ar] `3d^(6) 4s^(0)`. `:.` Number of electrons = 6 Mg - `1s^(2) 2s^(2)2p^(6)3s^(2)` (6s electrons) It matches with the 6d electrons of `Fe^(2+)` `Cl - 1s^(2) 2s^(2)2p^(6)3s^(2)3p^(5)` (11 p electrons) It does not match with the 6d electrons of `Fe^(2+)` Fe - [Ar] `3d^(6) 4s^(2)` (6d electrons) It does not match with the 6d electrons of `Fe^(2+)` Ne - `1s^(2) 2s^(2)2p^(6)` (6 p electrons) It matches with the 6d electrons of `Fe^(2+)` . Hence, Cl had 11p electrons which does not matches in number with 6d electrons of `Fe^(2+)`. |
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| 1368. |
Which quantum number is not related with Schrodinger equation :-A. PrincipalB. AzimuthalC. MagneticD. Spin |
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Answer» Correct Answer - D `hv=hv_(0)+(1)/(2)mv^(2)` `(hc)/(lambda)=(hc)/(lambda_(0))+(1)/(2) mv^(2)` `K.E. = hc ((lambda_(0)-lambda)/(lambdalambda_(0)))` `((h^(2))/(2mlambda_(e)^(2)))=hc((lambda_(0)-lambda)/(lambdalambda_(0)))" "(thereforelambda=(h)/(sqrt(2mK.E)))` `lambda_(e)^(2)=(lambdalambda_(0)h)/([lambda_(0)-lambda]2mc)` `lambda_(e)=[(hlambdalambda_(0))/(2mc[lambda_(0)-lambda])]^((1)/(2))` |
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| 1369. |
Energy density in the region between `1s` and `2s` orbital is ………… |
| Answer» Correct Answer - Zero | |
| 1370. |
What is the highest frequency of a photon that can be emitted from a hydrogen atom ? What is the wavelength of this this photon ? |
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Answer» Highest frequency photon is emitted an electron comes from infinity to the first energy level `E =- (13.6Z^(2))/(I^(2)) = -13.6 eV` or `13.6 xx 1.6 xx 10^(-19) J = E = hv` `:. V = (E )/(h) = (2.176 xx 10^(-18)J )/(6.626 xx 10^(-34) J s) = 0.328 xx 10^(13) Hz` `v = (c )/( lambda) :. lambda = (c )/(v)` `lambda = (3 xx 10^(8))/(0.328 xx 10^(16)) = 9.148 xx 10^(-8) m` |
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| 1371. |
The speed of a photon is one hundredth of the speed light in vacuum. What is the de Broglie wavalengths. Assume that one mole of protons has a mass equal to one gram. `h = 6.626 xx 10^-27 erg sec`.A. `3.31 xx 10^-3 Å`B. `1.33 xx 10^-3 Å`C. `3.13 xx 10^-2 Å`D. `1.31 xx 10^-2 Å` |
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Answer» Correct Answer - B (b) `m = (1)/(6.023 xx 10^23) g` `lamda = (h)/(mv)` =`(6.626 xx 10^-27)/(1 xx 3 xx 10^8 cm sec^-1) xx 6.023 xx 10^23` =`1.33 xx 10^-11 cm`. |
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| 1372. |
How many electron in a given atom can have the following quantum numbers- (a)`n=4.l=1` `n=2,l=1,m=-1`,`s=+^((1)//(2))` (c)`n=3 n=4,l=2,m=0` |
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Answer» (a) l=1 refers to p-subshell which has three orbits`(p_(x),p_(y) and p_(z))` each having two electrons.Therfore , total number of electrons are 6. (b) `l=1` refers top p -subhell, `m=-1` refers to `p_(x) or p_(y)` orbital where as, ,`s=+^((1)//(2))` indicate foe only 1 electron. (c) Number of electrons for any energy level is given by `2n^(2)i.e 2xx3^(2)` =18 electron. (d) `l=2` means d- subshell and m = 0 refer to `dz^(2)` orbital ` therefore` Number of electrons are 2. |
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| 1373. |
Which one is the correct outer configuration of chromium.A. B. C. D. |
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Answer» Correct Answer - C ( c) `Cr_24 = (Ar) 3 d^5 4 s^1` electronic configuration because half-filled orbital are more stable than other orbitals. |
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| 1374. |
Which of the following has maximum number of unpaired electron (atomic number of `Fe 26`)A. `Fe`B. `Fe(II)`C. `Fe(III)`D. `Fe(IV)` |
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Answer» Correct Answer - C Schrodinger equation gives only n, I and m quantum number, spin quantum number is not related to schrodinger equation. |
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| 1375. |
If uncertainty in position and momentum are equal then uncertainty in velocity is.A. `sqrt((h)/(pi))`B. `sqrt((h)/(2 pi))`C. `(1)/(2m) sqrt((h)/(pi))`D. None of these |
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Answer» Correct Answer - C ( c) `Delta x Delta p ge (h)/(4 pi)` `Delta p ge sqrt((h)/(4 pi)) ` when `Delta x = Delta p` `m Delta v ge sqrt((h)/(4 pi)) , Delta v^3 (1)/( 2m) sqrt((h)/(pi))`. |
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| 1376. |
Which of the following statements is correct ?A. An orbital describes the path of an electron in an atomB. An orbital is a region where the electron is not locatedC. An orbital is a function which gives the probabilities of finding the electron in a given regionD. All the above |
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Answer» Correct Answer - C `psi^(2)` represent probability of finding `e^(-)` |
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| 1377. |
If the mass of the electron is reduced to half the Rudberg constantA. Remains unchangedB. Becomes halfC. Becomed doubleD. Becomes one fourth |
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Answer» Correct Answer - B `R = (2pi^(2)mZ^(2)e^(4))/(Ch^(3)), E prop m` |
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| 1378. |
Which one is the correct outer configuration of chromium.A. B. C. D. |
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Answer» Correct Answer - C ( c) `Cr_24 = (Ar) 3 d^5 4 s^1` electronic configuration because half-filled orbital are more stable than other orbitals. |
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| 1379. |
Which of the following scientists demonstrated the wave nature of electron ?A. DavissonB. HeisenbergC. de- BroglieD. Schrodinger |
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Answer» Correct Answer - C Electron behaves as a material as well as wave, this was proposed by de-Broglie `lambda = h/(mv)` |
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| 1380. |
The electron of hydrogen atom is excited to certain level. When the electron returns to the first Bohr orbit, the wavelength of line emitted if the energy difference is 11.0 eV would beA. `11.25xx10^(-7)m`B. `2.2xx10^(-6)m`C. `9.1176xx10^(-8)m`D. `1.22xx10^(-7)m` |
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Answer» Correct Answer - A `E_(2)-E_(1)=DeltaE=(hc)/lambda` `lambda=(hc)/(DeltaE)=(6.62xx10^(-34)xx3xx10^(8))/(11.0xx1.6xx10^(-19))` `lambda =11.25xx10^(-7)`m |
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| 1381. |
The number of `d` electrons in `Fe^(2+)` (atomic number of `Fe = 26`) is not equal to that of the.A. p-electrons in Ne (At. No. =10)B. s-electrons in Mg (At. No. =12)C. d-electrons in FeD. p-electrons in `Cl^-` (At. No. Of Cl =17) |
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Answer» Correct Answer - D `._26Fe=[Ar]3d^6, 4s^2 rArr ` No. of d electron =6 `Fe^(2+)=[Ar]3d^6 , 4s^0 rArr` No. of d electron =6 `._10Ne=1s^2 2s^2 2p^6 rArr ` No. of p electron =6 `._12Mg=1s^2 2s^2 2p^6 3s^2 rArr` No. of s electron = 6 `Cl^(-)=1s^2 2s^2 2p^6 3s^2 3p^6 rArr` No. of p electron =12 Hence the ans. is `Cl^-` |
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| 1382. |
The number of `d` electrons in `Fe^(2+)` (atomic number of `Fe = 26`) is not equal to that of the.A. d-electrons in Fe(Z = 26)B. p-electrons in Ne(Z = 10)C. s-electrons in Mg(Z = 12)D. p-electrons in Cl (Z = 17) |
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Answer» Correct Answer - D (d) Number of d-electrons is `Fe^(2+) = 6` Number of p-electrons in `Cl = 11`. |
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| 1383. |
The number of `d` electrons in `Fe^(2+)` (atomic number of `Fe = 26`) is not equal to that of the.A. p-electrons in Ne (At. No. = 10)B. s-electrons in Mg(At. No = 12)C. d-electric in FeD. p-electron in `Cl^(-)` (At. No. of Cl =17). |
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Answer» Correct Answer - D (d) `._26 Fe = [AR]3d^6, 4s^2` `Fe^(2+) = [Ar] 3d^6, 4s ^(0)d^(6)` Number of d-electrons `= 6` `._17 Cl = [Ne]3s^2, 3 p^5` `Cl^- = [Ne]3s^2 , 3p^6` Number of p-electrons `= 6`. |
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| 1384. |
Which of the following statement is correct ?A. An orbital containing an electron having quantum number `n = 2,l= 0,s = + 1//2` is sphericalB. All photon have the same energyC. The frequancy of X-rays is less than that of radiowavesD. As intensity of light increases the frequancy increases |
| Answer» `l= 0` or `s` orbital the shape of s spherical | |
| 1385. |
The electron density between 1s and 2s isA. HighB. LowC. ZeroD. Abnormal |
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Answer» Correct Answer - C radial node is present |
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| 1386. |
Calculate the energy of an electron in second orbit of an excited hydrogen atom the energy of the electron in the first Bohr orbit the energy of the electron in the Bohr orbit is `-2.18 xx 10^(-11) erg` |
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Answer» We know that `E_(n) = (E_(1))/(n^(2))= (2.18 xx 10^(-11))/((2)^(2))` ` = - 0.545 xx 10^(-11) erg` |
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| 1387. |
Calculate the iongest wavelength transition in the paseches series of `He^(o+)`. |
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Answer» `bar v = R_(H) xx Z6*(2) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` For He `Z = 2` For paschen series , `n_(1) = 3` For the longest wavelength, `n_(2) = 4` `(1)/(lambda) = 109678 xx (2)^(2) xx [(1)/(3^(2)) - (1)/(4^(2))]` `= 10978 xx 4 xx [(1)/(9) - (1)/(16)]` `= 109678 xx 4 xx (7)/(144)` `or lambda = 4589 Å` |
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| 1388. |
The orbital with `n=3` and `l=2` isA. 5dB. 3dC. 4dD. 5s |
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Answer» Correct Answer - B 3d orbital `[(l=0,1,2,3),(=s,p,d,f)]` |
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| 1389. |
(i) What is the energy and wavelenght of phpotons of frequncey 3.4MHz? (ii) Also calculate the energy per mole of photons of the same wavelength. |
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Answer» `v=3.4MHz=3.4xx10^(6)HZ=3.4xx10^(6)s^(-1).` `lambda=(c)/v=(3xx10^(8)ms^(-1))/(3.4xx10^(6)s^(-1))`=88.2m Energy, `E=hv=6.626xx10^(-34)xx3.4xx10^(6)J` =`2.253xx10^(-27)J` Enegry per mole of photon = `6.02xx10^(23)xx2.53xx10^(-27)J mol^(-1)` `=1.356xx10^(-3) J mol^(-1)` |
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| 1390. |
Which one of the following statements is most appropriate ?A. Electron spins around its own axis onlyB. Electron moves around the nucleus in spherical orbitsC. Electron moves around the nucleus in elliptical orbitsD. Electrons moves around the necleus in spherical or elliptical orbits spins around its own axis |
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Answer» Correct Answer - D Electron moves around the nucleus in spherical or elliptical orbits and spins around its own axis |
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| 1391. |
The energy equivalent of 2.0 mg mass defect isA. `1.8 xx 10^4` ergB. `9xx 10^(-19)` ergC. `1.5 xx 10^20` ergD. `1.8 xx 10^18` erg |
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Answer» Correct Answer - D `E=mc^2` `c=3xx10^10` cm/s, m=2mg = `2xx10^(-3) g` `E=2xx10^(-3) (3xx10^10)^2` `=2xx10^(-3) xx 9xx 10^20` `E=1.8xx10^18` erg |
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| 1392. |
Which of the following statement is//are not correct ?A. The shape of an atomic orbit depends on the azimothal quantum numberB. The orienation of an atomic orbit depends an the magnitic quantum numberC. The energy of an electron in an atomic orbit of a multi electron atom depends as on the principal quantum numberD. The number of depends atomic orbital of one type depends on the values of principal azimothad and magnetic number |
| Answer» In a multi electron and `H^(o+)` has electron, the panli exclassion principal does not apply to them.However `H^(o+)` has two electron hence this principal applies on it | |
| 1393. |
Suggest the angular and spherical nodes in the followingA. `3p`B. `3d`C. `2s`D. `3s` |
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Answer» Angular nodes `= 1` (azimathal quantum number) Spherical node `= n - l- 1` a.`1.1 `b. `2.0` c.`0.1` d. `0.2` |
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| 1394. |
The set of quantum numbers, n = 3, l = 2, `m_(l) = 0`A. Describes and electron in a 2s orbitalB. Is not allowedC. Describes an electron in a 3p orbitalD. Describes one of the five orbitals of a similar type |
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Answer» Correct Answer - D represent `3dz^(2)` orbital. |
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| 1395. |
Which of the following statement(s) is/are correct about angular nodesA. They are independent from the radial wave functionB. They are directional in natureC. The number of angular nodes of orbital is equal to azimuthal quantum numberD. All are correct |
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Answer» Correct Answer - D All are correct |
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| 1396. |
The spectrum of `He` is expected to be similar to.A. `Li^(+)`B. `He`C. `H`D. `Na` |
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Answer» Correct Answer - C `Li^(+2) & He^(+)` both have same no. of electron so spectrum pattern will be similar. |
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| 1397. |
The spectrum of `He` is expected to be similar to that ofA. HB. `Li^+`C. NaD. `He^+` |
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Answer» Correct Answer - B Both He and `Li^+` contain 2 electrons each . |
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| 1398. |
Values of magnetic orbital quantum for an electron of M-shell can be:A. 0,1,2B. `-2,-1,0,+1,+2`C. `0,1,2,3`D. `-1,0,+1` |
| Answer» Correct Answer - B | |
| 1399. |
The maximum number of electrons that can be taken by a subshell with `l=3` isA. 8B. 14C. 10D. 12 |
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Answer» Correct Answer - B When `l=3 m=-3 -2 -1 0,1,1,2,3` Each m can have electrons `=2xx7=14` |
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| 1400. |
The spectrum of `He` is expected to be similar to that ofA. hydrogenB. `Li^(+)`C. NaD. He |
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Answer» Correct Answer - A Both H and `He^(+)` have one electron |
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