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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.
| 1451. |
According to Aufbau principle , the `19^(th)` electron in an atom goes into the :-A. 4s - orbitalB. 3d - orbitalC. 4p - orbitalD. 3p - orbital |
| Answer» Correct Answer - 1 | |
| 1452. |
An electron is in one of this electrons |
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Answer» Correct Answer - A `n = 3, l = 2, m_(1) = + 2, + 1 0, -1 , -2 ` |
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| 1453. |
`psi^(2)` means.A. radial probability densityB. probability densityC. always positive valueD. 2 & 3 |
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Answer» Correct Answer - D de Brogle concept is applicable to very small particles like photon, electron |
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| 1454. |
The kinetic energy of thephotoelectrons does not depends uponA. Intensity of incident radiationB. Frequency of incident radiationC. Wavelengthof incident radiationD. Wave number of incident radiation |
| Answer» Correct Answer - A | |
| 1455. |
The angular momentum of an electron in 2p-orbital is :A. ` h/(2pi)`B. ` h/(2pi)`C. ` h/(2pi)`D. None of these |
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Answer» Correct Answer - B Angural momentum in an orbital `= h/( 2pi) sqrt (l(l+1)) (l=1)`. |
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| 1456. |
According to aufbau principle the electron has a tendency to occupy that subshell which has.....energy.A. LowestB. HighestC. No energyD. Both 1 and 2 |
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Answer» Correct Answer - A Aufbau rule |
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| 1457. |
What transition in the hydrogen spetrum would have the same wavelength as the balmer transition `n = 4` to `He^(o+)` spectrum ? |
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Answer» `bar v = (1)/(lambda) = ((1)/(2^(2)) - (1)/(4^(2))) RZ^(2) = (4)/(4) R` In H spectrom for the same `bar v ` or `lambda` as `Z = 1, n = 1` |
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| 1458. |
The havest subatomic particle isA. NeutronB. PositronC. ElectronD. Proton |
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Answer» Correct Answer - A Neutron |
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| 1459. |
The electronic configuration of the pllotoelectrons does not depends uponA. 32B. 42C. 30D. 34 |
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Answer» Correct Answer - C Welectronic configuration of `M^(2+) = 2 + 8 + 14 = 24 e^(bar)` Since two `e ^(bar)`has been lost in forming `M^(2+)` ion from metal . So total number of proton should be `= 24 + 2 = 2b` Hence number of neutrons `= 56 - 26 = 30` |
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| 1460. |
If the largest value of`m_(1) `for an electron is `+ 3` in what type of subshell the electron may be present ? |
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Answer» Correct Answer - A The largest value of `m_(1)` can be `+3` when `l = 3,l = 3` corresponding to f subshell .Therefore , the electron must be present in f subshell |
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| 1461. |
Which of the following variant of hydrogen spectrum results from nuclear spin interaction with that of electron?A. Fine spectrumB. Stark effectC. Zemann effectD. Hyperfine spectrum |
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Answer» Correct Answer - D `lambda = (h)/(sqrt(2mq.v)), (lambda_(alpha))/(lambda_(p)) = sqrt((m_(p)(1).V)/(4m_(p).(2).x)), x = (v)/(8)` |
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| 1462. |
What transition of `Li^(+2)` spectrum will have same wavelength as that of second line of Balmer series in `He^(+)` spectrum ? |
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Answer» `(1)/(lambda_(He^(+))) = (1)/(lambda_(Li^(+)))` `RZ^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))] = RZ^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` `2^(2) [(1)/(2^(2))-(1)/(4^(2))] = 3^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` `4[(1)/(4)-(1)/(16)] = 9 [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` `(4)/(9) xx (3)/(16) = [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` `:. (1)/(n_(1)^(2)) -(1)/(n_(2)^(2)) = (1)/(12) n_(1) = 3, n_(2) = 6` `(3 rarr 6)` will have same wavelength as that of second line of Balmer series in `He^(+)` spectrum. |
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| 1463. |
The hydrogen -like species `Li^(2+)` is in a spherically sysmmetric state `S_(1)` with one node ,Upon ansorbing light , the ion undergoes transition to a state `S_(2)` The state `s_(2)` has one radial node and its energy is equal is to the ground state energy of the hydrogen atom Energy of the state `S_(1)` in units of the hydrogen atom ground state enegy isA. 0.75B. 1.5C. 2.25D. 4.5 |
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Answer» Correct Answer - C `E_(n) = (-13.6)/(n^(2))Z^(2)` For ground state of H atom `E_(H) = (-13.6)/(1^(2)) xx 1^(2)` For `2s` of `Li^(2+)` `E_(Li^(2+)) = (-13.6)/(2^(2)) xx 3^(2)` `E_(Li^(2+))/(E_(H)) = ((-13.6)/(4) xx 9)/(-13.6) = 2.29` or `E_(Li^(2+)) = 2.25E_(H)` |
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| 1464. |
The hydrogen -like species `Li^(2+)` is in a spherically sysmmetric state `S_(1)` with one node ,Upon ansorbing light , the ion undergoes transition to a state `S_(2)` The state `s_(2)` has one radial node and its energy is equal is to the ground state energy of the hydrogen atom The orbital momentum number of teh state `S_(2)` is |
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Answer» Correct Answer - B `S_(2)` has one radial node `E_(S_2) = E_(H) (given)` `or (-13.6)/(n^(2)) xx 2^(2) = (-13.6)/(1^(2)) xx 1^(2)` `or n = 3` Radial nodes ` = n - l- 1 = 1` `or 3 - 1 - l- 1= 1` `or l = 1` This means `S_(2)` is `3p` The orbital angular momentum quantum number means azimuthal quantum numner which is `1` in this case |
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| 1465. |
The ratio of the enrgy of the electron in groun state of hydrogen to the elecron in first excited state of `Be^(2+)` is :A. ` 1:4`B. `1:8`C. ` 1:16`D. ` 16 :1` |
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Answer» Correct Answer - A `E_(2_((Be^(3+))) = E_(2(H)) xx Z^2` , Also ` E_(2 (H)) = E_(1(H0))/2^2` ` ,. E_(2(Be^(3+0)) = (E_(1(H)))/2^2 xx 4^2 = 4 xx E_(1(H))`. |
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| 1466. |
The number of d-electrons in ` Fe^(2+) (Z= 26 ) ` is not equal to that of :A. p-electrons in `Ne (Z=10)`B. p-electrons in `Me (Z=12)`C. p-electrons in `Fe (Z=26)`D. p-electrons in `Cl (Z=17)` |
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Answer» Correct Answer - D `Fe^(2+)` has six d-electrons : Cl has `12` p-electrons . |
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| 1467. |
What are the possible values of `l` for an electron with `n=3` ? |
| Answer» Correct Answer - `l` can be `0,1` or `2` | |
| 1468. |
The line spectrum of two elements is not identical becauseA. They do not have same number of nuctronsB. They have dissimilar mass numberC. They have different energy level schemesD. They have different numebr of valence electron |
| Answer» Correct Answer - C | |
| 1469. |
Line spectrum of `Li^(+2)` and `He^(+)` are identical due toA. Both are cationsB. Both have same no of protonsC. Isoelectronic species produce identical spectrumD. none |
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Answer» Correct Answer - C H atom, `He^(+),Li^(2+),Be^(3+)` have 1 electron only |
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| 1470. |
Which is the correct order of probability of being found close to the nucleus is.A. `s gt p gt d gt f`B. `f gt d gt p gt s`C. `p gt d gt f gt s`D. `d gt f gt p gt s` |
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Answer» Correct Answer - A (a) The order of penetration power of `s, p, d` and `f` electrons are in order `s gt p gt d gt f`. |
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| 1471. |
What are the possible values of `m_(l)` for an electron with `l=2` ? |
| Answer» Correct Answer - The value of `m_(l)` can very from `-l` to `+l`; in this case `m_(l)` can have values `-2,-1,0,+1` and `+2` | |
| 1472. |
Explain , giving reason , which of the following sets of quantum number are not possible `{:(a,n= 0 ,l = 0, m_(1) = 0, m_(s) = +1//2),(b,n= 1 ,l = 0, m_(1) = 0, m_(s) = -1//2),(c,n= 1 ,l = 0, m_(1) = 0, m_(s) = +1//2),(d,n= 2 ,l = 1, m_(1) = 0, m_(s) = -1//2),(e,n= 3 ,l = 3, m_(1) = -3, m_(s) = +1//2),(f,n= 3 ,l = 1, m_(1) = 0, m_(s) = +1//2):}` |
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Answer» Correct Answer - A a. Not possible because n correct be zero c.Not possible because the value of l in this case is equal to n, which is not allowed e.Not possible because the value of l in this case is equal to n, which is not allowed |
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| 1473. |
The line spectrum of two elements is not identical becauseA. The elements do not have the same number of neutronsB. They have different mass numbersC. Their outermost electrons are at different energy levelsD. All of the above |
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Answer» Correct Answer - C Energy difference between orbits is different. |
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| 1474. |
For principle quantum number `n = 4` the total number of orbitals having `l = 3`.A. 3B. 7C. 5D. 9 |
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Answer» Correct Answer - B (b) `n = 4 rarr 1 s^2 2 s^2 2 p^6, 3 s^2, 3p^6, 3 d^10, 4 s^2 ,4 p^6, 4 d^10, 4 f^14` So `f = (n - 1) = 4 - 1 = 3` which is `h` orbit contains `7` orbital. |
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| 1475. |
`._(6)^(14)C ` and `._(8)^(16) O` are ……….. |
| Answer» Correct Answer - Oisotones | |
| 1476. |
For each value of l the possible value of m_(1) are……… |
| Answer» Correct Answer - `(2l + 1)` | |
| 1477. |
When an electron makes a transition from `(n + 1)` state to n state the frequency of emitted radiation is related to n according to `(n gtgt 1)`A. `v prop n^(-3)`B. `v prop n^(2)`C. `v prop n^(3)`D. `v prop n^(2//3)` |
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Answer» Correct Answer - A |
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| 1478. |
When an electron makes a transition from `(n+1)` state of nth state, the frequency of emitted radistions is related to n according to `(n gt gt 1)`:A. `v = (2CZ^(2)R_(H))/(n^(3))`B. `v =(CZ^(2)R_(H))/(n^(4))`C. `v =(CZ^(2)R_(H))/(n^(2))`D. `v = (2CZ^(2)R_(H))/(n^(2))` |
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Answer» Correct Answer - A `v = Cbar(v) = CR_(H)z^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` |
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| 1479. |
If an electron is revolving is revolving around the hydrogen nucleus at a distance of 0.1 nm,what should be its speed?A. `2.188times10^(6)m//s`B. `1.094times10^(6)m//s`C. `4.376times10^(6)m//s`D. `1.59times10^(6)m//s` |
| Answer» Correct Answer - D | |
| 1480. |
When an electron makes a transition from `(n + 1)` state to n state the frequency of emitted radiation is related to n according to `(n gtgt 1)`A. `v prop n^(-3)`B. `v prop n^(2)`C. ` v prop n^(3)`D. `v prop n^((2)/(3))` |
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Answer» Correct Answer - A `v = (c )/(lambda) = RcZ^(2) ((1)/(n^(2)) - (1)/((n + 1)^(2))) = RcZ^(2) ((1 + 2n)/(n^(2)(n +1)^(2)))` `RcZ^(2)((2n))/(n^(2)) prop n^(-3)` |
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| 1481. |
For principle quantum number `n = 4`, the total number of orbitals having `l =3` isA. 3B. 7C. 5D. 9 |
| Answer» Correct Answer - 2 | |
| 1482. |
The frequency of light emitted for the transition `n = 4` to `n =2` of `He^+` is equal to the transition in `H` atom corresponding to which of the following ?A. `n = 3 "to" n = 1`B. `n = 2 "to" n = 1`C. `n = 3 "to" n = 2`D. `n = 4 "to" n = 3` |
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Answer» Correct Answer - B (b) `Delta E = hu = (2 pi^2 mZ^2 e^4 k^2)/(h^2) [(1)/(n_1^2)-(1)/(n_2^2)]` If electron falls from `n_2 - "level to" n_1 - "level"` `:.` In `He^+, "for" n_2 = 4 "to" n_1 = 2 transition` `v_(He^+) = "constant" (4) [(1)/(2^2) -(1)/(4^2)][because Z_(He^2) = 2]` =`"constant" xx 4 [(3)/(16)] = (3)/(4) xx "constant"` `v(H) = "constant" (1) 2 [(1)/(n_1^2) - (1)/(n_2^2)]` =`"constant" xx [(1)/(n_1^2) - (1)/(n_2^2)]`. (a) For `n_2 = 3` and `n_1 = 1` `v(H) = "constant" [(1)/(1) -(1)/(9)]` =`(8)/(9) xx "constant" ne (3)/(4) xx "constant"` (b) For `n_2 = 2` and `n_1 = 1` `v(H) = "constant" xx [(1)/(1) -(1)/(4)]` =`(3)/(4) xx "constant" = v(He^+)` ( c) For `n=3` and `n_1 = 2` `v(H) = "constant" xx [(1)/(2^2) -(1)/(3^2)]` =`(5)/(36) xx "constant" ne v(He^+)` (d) For `n_2 = 4` and `n_1 = 3` `v(H) = "constant" xx [(1)/(3^2) - (1)/(4^2)]` =`(7)/(144) xx "constant" ne v(He^+)`. |
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| 1483. |
In which transition minimum energy is emitted :-A. `ooto1`B. `2to1`C. `3to2`D. `nto(n-1)[nge4]` |
| Answer» Correct Answer - 4 | |
| 1484. |
In each of the following atoms or ions, electronic transition form `n=4 to n=1`take place. Frequency of the radiation emitted out will be minimum forA. hydrogen atomB. deuterium atomC. `He^(+)ion`D. `Li^(2+)ion` |
| Answer» Correct Answer - A | |
| 1485. |
In which transition minimum energy is emitted :-A. `oorarr1`B. `2rarr1`C. `3rarr2`D. `nrarr(n-1)(nge4)` |
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Answer» Correct Answer - D `n_(2)=4,n_(1)=3` , `n_(2)=5 " "n_(1)=4` , `n_(2)=6 " "n_(1)=5` , `n rarr (n-1)(n ge 4)` |
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| 1486. |
A hydrogen like species with atomic number Z is present in a higher excited state (n) . This electron can make transition to the first excited level by successively emitting two photons of energy 2.64 eV and 48.36 eV . This electron can also make transition to third excited state by emitting three photons of energy 2.64 eV , 2. 66 eV and 4.9 eV . Identify the hydrogen like species involved .A. `He^(+)`B. `Li^(2+)`C. `Be^(+3)`D. `B^(+4)` |
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Answer» Correct Answer - C |
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| 1487. |
`E_(n) =` total energy , `l_(n)` angular momentum `K_(n) = K.E., V_(n) = P.E.,` `T_(n) =` time period , `r_(n)=` radius of nth orbit |
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Answer» Correct Answer - (a-s);(b-s);(c-p);(d-q,r) |
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| 1488. |
Angular wave function of `P_(x)` orbital is : (Given : 0 is angle from z-axis)A. `((3)/(4pi))^(1//2) "sin 0 sin" phi`B. `((3)/(4pi))^(1//2) "sin 0 cos" phi`C. `((3)/(4pi))^(1//2) "cos0"`D. `((15)/(4pi))^(1//2) "sin 0 cos0" cos phi` |
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Answer» Correct Answer - B |
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| 1489. |
The ratio of `(E_(2) - E_(1))` to `(E_(4) - E_(3))` for `He^(+)` ion is approximately equal to (where `E_(n)` is the energy of nth orbit ):A. 10B. 15C. 17D. 12 |
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Answer» Correct Answer - B |
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| 1490. |
The ratio of energy of the electron in group state of the hydrogen to electron in first excited state of `He^+` is.A. `1 : 4`B. `1 : 1`C. `1 : 8`D. `1 : 16` |
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Answer» Correct Answer - B (b) `E_n prop (-Z^2)/(n^2)` `z_1 = 1` `n_1 = 1` `z_2 = 1` `n_2 = 2` `(E_1(H))/(E_2(He^(+))) = (1)/(1) = 1 : 1`. |
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| 1491. |
If first ionisation energy of hydrogen be `E`, then the ionisation energy of `He^+` would be :A. EB. 2EC. 0.5ED. 4E |
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Answer» Correct Answer - D `I_(2)(He^(2+))=Z^(2)I_(1)(H)` `=2^(2)xxE=4E` |
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| 1492. |
If `E_(1) , E_(2) "and" E_(3)` represent respectively the kinetic energies of an electron , an alpha particle and a proton each having same de Broglie wavelength then :A. `E_(1) gt E_(3) gt E_(2)`B. `E_(2) gt E_(3) gt E_(1)`C. `E_(1) gt E_(2) gt E_(3)`D. `E_(1) = E_(2) = E_(3)` |
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Answer» Correct Answer - A |
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| 1493. |
If `E_(1) , E_(2) "and" E_(3)` represent respectively the kinetic energies of an electron , an alpha particle and a proton each having same de Broglie wavelength then :A. `E_(e) = E_(alpha) = E_(p)`B. `E_(e) gt E_(alpha) gt E_(p)`C. `E_(alpha) lt E_(P) lt E_(e)`D. `E_(e) = E_(P) lt E_(n)` |
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Answer» Correct Answer - C `lambda = (h)/(sqrt(2mkE))` |
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| 1494. |
In Hydrogen atom electron is present in the N shell. If it loses energy, a spectral line many be observed in the regionA. Infra-redB. VisibleC. Ultra-violetD. All the above |
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Answer» Correct Answer - D Can produce lines in lyman, balmer, pasctem series |
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| 1495. |
The number of spectral lines that are possible when electrons in 7th shell in different hydrogen atoms return to the 2nd shell is: |
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Answer» `sum (n_(2)-n_(1)) = sum 7 -2 sum 5` `= 5 +4 +3 +2 +1 = 15` |
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| 1496. |
The number of spectral lines that are possible when electrons in 7th shell in different hydrogen atoms return to the 2nd shell is:A. 12B. 15C. 14D. 10 |
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Answer» Correct Answer - B Number of spectral lines `=((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2)` `=((7-2)(7-2+1))/(2)=15` |
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| 1497. |
For a d electron the orbital angular momentum isA. ` sqrt 6 h`B. ` sqrt 2 h`C. `h`D. ` 2 h` |
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Answer» Correct Answer - A Angular momentum in 3d-orbital ltbRgt ` = h/( 2 pi) sqrt ( l(1l + 1)) = h/( 2 pi) sqrt ( 2( 2 + 1))` ` = sqrt(6)h/( 2 pi) = sqrt(6 h) ( h = (h)/( 2pi0))` . |
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| 1498. |
The ionisation enthalpy of hydrogen atom is `1.312xx10^(6)J*mol^(-1)`. The energy required to excited the electron in the atom from n=1 to n=2 is-A. `8.51 xx 10^5 J mol^-1`B. `6.56 xx 10^5 J mol^-1`C. `7.56 xx 10^5 J mol^-1`D. `9.84 xx 10^5 J mol^-1` |
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Answer» Correct Answer - D (d) `Delta E = E_2 - E_1 = - (E_1)/(2^2) + (E_1)/(l^2)` =`-(1.312 xx 10^6)/(2^2) + 1.312 xx 10^6` =`9.84 xx 10^5 J mol^-1`. |
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| 1499. |
de-Broglie equation describes the relationship of wavelength associated with the motion of an electron and itsA. MassB. EnergyC. MomentumD. Charge |
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Answer» Correct Answer - C According to de-Broglie equation `lambda=h/"mv" "or" h/p "or" h/"mc"` |
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| 1500. |
Calculate the longest wavelength that can an electron from the first bohr given `E_(1) = 13.6 eV` |
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Answer» The photon capable of removing an electron from the first bohr orbit posses energy `= 13.6eV` `= 13.6 xx 1.602 xx 10^(-19) J` `= 21.787 xx 10^(-19) J` `," E = (hc)/(lambda)` `21.787 xx 10^(-19) = (6.625 xx 10^(-34) xx 3.0 xx 10^(8))/(lambda)` `:. lambda = 912.24 xx 10^(-10)m = 912.24 Å` Energy of photon `E = Nhv = (Nhv)/(lambda)` :. Number of photon electrons emitted per minut `(N) = (E lambda)/(hc)` `= (6000 J xx 300 xx 10^(-9)m)/(6.626 xx 10^(-34) J s xx 3 xx 10^(8) ms^(-1))` `= 9.05 xx 10^(22)`photons |
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