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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.
| 2301. |
In Bohr series of lines of hydrogen spectrum, third line from the red end corresponds to which one of the following inner orbit jumps of electron for Bohr orbit in atom in hydrogen :A. ` 4rarr 1`B. `2rarr 5`C. `3rarr2`D. `5rarr2` |
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Answer» Correct Answer - D The emission is in visible (VIBGYOR) region . The Balmer seies i.e., either ( c) or (d) . In ( c) only one emiwsion . The thrd line will be in the jupmp from ` 5 ` to ` 2`. |
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| 2302. |
In Bohr series of lines of hydrogen spectrum, third line from the red end corresponds to which one of the following inner orbit jumps of electron for Bohr orbit in atom in hydrogen :A. `4 rarr 1`B. `2 rarr 5`C. `3 rarr 2`D. `5 rarr 2` |
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Answer» Correct Answer - D (d) The emission is in visible `(VIBGYOR)` region. Thus Balmer series i.e., either `( c)` or `(d)`. In ( c) only one emission. The third line will be in the junp from `5` to `2`. |
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| 2303. |
In Bohr series of lines of hydrogen spectrum, third line from the red end corresponds to which one of the following inner orbit jumps of electron for Bohr orbit in atom in hydrogen :A. `4to1`B. `2to5`C. `3to2`D. `5to2` |
| Answer» Correct Answer - D | |
| 2304. |
In Bohr series of lines of hydrogen spectrum, third line from the red end corresponds to which one of the following inner orbit jumps of electron for Bohr orbit in atom in hydrogen :A. `5 rarr 2`B. `4 rarr 1`C. `2 rarr 5`D. `3 rarr 2` |
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Answer» Correct Answer - A Balmer series - visible region for `3^(rd)` line `n_(1) = 2, n_(2) = 5` |
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| 2305. |
What are the values of the orbital angular momentum of an electron in the orbitals `1s,3s,3d` and `2p`:-(a). `0,0sqrt(6h),sqrt(2h)`(b). `1,1sqrt(4h),sqrt(2h)`(c). `0,1sqrt(6h),sqrt(3h)`(d). `0,0sqrt(20h),sqrt(6)`A. `0,0,sqrt6ħ,sqrt(2)ħ`B. `1,1,sqrt(6)ħ,sqrt(2)ħ`C. `0,1,sqrt(6)ħ,sqrt(3)ħ`D. `0,0,sqrt(20)ħ,sqrt(6)ħ` |
| Answer» Correct Answer - 1 | |
| 2306. |
Which of the following electronic configurations represents the atom of aluminium ?A. 2,8,2B. 2,8,4C. 2,8,3D. 2,8,7 |
| Answer» Correct Answer - B::D | |
| 2307. |
`1.8 g` hydrogen atoms are excited by a radiation. The study of species indicates that `27%` of the atom are in third energy level and `15 %` of atom in second energy level and the rest in ground state. If IP of H is `13.6 eV`, calculate a. Number of atoms present in first and third energy levels b. Total energy involved when all the atoms return to the ground state. |
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Answer» Correct Answer - `292.68xx10^(21)` atom.`162.60xx10^(21)` atoms,`832.50kJ` `1.8` mole = `(1.8Na)` atoms `27%` =IIIrd energy level `=1.8xxNaxx0.27` `15%` =IInd energy level `=1.8xxNaxx0.15` `DeltaE = overset(DeltaE_(1))(3rarr1)+overset(DeltaE_(2))(2rarr1)=1.8xxN_(A)xx0.27xxIE[(1)/(9)-(1)/(1)]+1.8xxN_(A)xx0.15xxIE[(1)/(4)-(1)/(1)]=292.68xx10^(21)`atom |
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| 2308. |
In an atom, an electron is moving with a speed of `600m//s` with an accuracy of `0.005%` certainity with which the positive of the electron can be located is `[h = 6.6 xx 10^(-34)Js,m = 9.1 xx 10^(-31)kg]`A. `1.52 xx 10^(-4)m`B. `5.1 xx 10^(-3)m`C. `1.92 xx 10^(-3)m`D. `3.84 xx 10^(-3)m` |
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Answer» Correct Answer - C `Delta x.m. Delta x ge (h)/(4p), Delta x ge (h)/(4pi.m.Deltav)` `Deltax ge = (6.6256 xx 10^(-34))/(4 xx 3.14 xx 9.1 xx 10^(-31) xx (600 xx 0.005)/(100))` |
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| 2309. |
The valency of the element having atomic number 9 isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A `X(9)to 1s^2 2s^2 2p^5` Thus it accept 1 electron to make noble gas configuration. Therefore its valency is one (l) |
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| 2310. |
What is the valency of the element with atomic number 36 and mass number 84 ?A. 2B. 4C. 0D. 3 |
| Answer» Correct Answer - C | |
| 2311. |
Three laser guns labelled as I,II&III have power 2 watt, 3watt and 5 watt (not necessarily in same order) are used to produce photocurrent from nmental plate as show in diagram. Number of photons emitted by laser guns are `4xx10^(18),5xx10^(18)` and `9xx10^(18)` per sec. metal plante have thershold energy `4.5` `xx10^(-19) J`. Neither the power nor the number of photon emitted b a particlar laser gun is known and it is know that call all capable emits a photonelectron. Minimum photocurrent which must be passed thought the circuit.A. 2.88B. 1.44C. 2.08D. 0.64 |
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Answer» Correct Answer - D For minnimum photocurrenth the combination is follows is as follows (a) `2 wattrArr 5xx10^(18)` photons/sec. ltbgt (b) `3 wattrArr 9xx10^(18)` photons/sec. (c) `5 watt rArr 4xx10^(18)` photons /sec. Combination (a) &(b) does not given any photoelectron only combination (c) will produce [photoelectron . `therefore` photocurrent =`4xx10^(18)xx1.6xx10^(-19)` Amp. ` =0.64` Amp. |
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| 2312. |
Three laser guns labelled as I,II&III have power 2 watt, 3watt and 5 watt (not necessarily in same order) are used to produce photocurrent from nmental plate as show in diagram. Number of photons emitted by laser guns are `4xx10^(18),5xx10^(18)` and `9xx10^(18)` per sec. metal plante have thershold energy `4.5` `xx10^(-19)` J. Neither the power nor the number of photon emitted b a particlar laser gun is known and it is know that call all capable emits a photonelectron. A. `(5)/(4)`B. `(9)/(5)`C. `(9)/(4)`D. `(9)/(2)` |
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Answer» Correct Answer - D For maxiumum photocurrent the combination is as follows (a) 2 watt `rArr 4xx10^(18)` photons/sec. ltbgt (b) 3 watt `rArr 5xx10^(18)` photons/sec. (c) 5 watt `rArr 9xx10^(18)` photons /sec. photoelectrons will be emitted from comcination (a),(b)and(c) all `therefore` Photocurrent =`(4+5+9)xx10^(18)xx1.6xx10^(-19)` Amp =2.88 ` therefore` Ratio`=(92.88)/(0.64)=4.5(18)/(4)rArr(9)/(2) ` |
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| 2313. |
Nucleons are:A. Protons and neutronsB. neutrons and electronsC. protons and electronsD. proton, neutrons and electrons. |
| Answer» Correct Answer - A | |
| 2314. |
Ultraviolet light of `6.2 eV` falls on an aluminium surface (work function `= 4.2 eV`). The kinetic energy (in joule) of the fastest electron emitted is approximately.A. `3 xx 10^-21`B. `3 xx 10^-19`C. `3 xx 10^-17`D. `3 xx 10^-15` |
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Answer» Correct Answer - B (b) `hv = hv_0 + KE` `KE = hv - hv_0 hv = 6.2 eV` =`6.2 - 4.2 hv_0 = 4.2 e V` =`2.00 xx 1.6 xx 10^-19` `KE = 3.2 xx 10^-19 J`. |
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| 2315. |
The number of spherical nodes in `3p` orbital areA. oneB. threeC. twoD. None of these |
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Answer» Correct Answer - A (a) The number of spherical nodes in any orbital` (=n-l-1)` For 3p-orbital, n=3 and l=1 `:.` Number of spherical nodes `= n-l-1` `3-1-1=3-2=1` node |
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| 2316. |
Calculate the energy of hydrogen atom in the ground state given that the after Bohr orbit og hydrogen is `5xx10^(-11)m` and electronic charge is `1.6xx10^(-19) C`. |
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Answer» Correct Answer - `-2.304xx10^(-18) J` |
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| 2317. |
Find the atomic structure of `._(12)^(24)Mg`. What is its valency? |
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Answer» Correct Answer - No. of protons or electrons `=12` No. fo neutraons `=12; 1s_(1)1s_(2)2s_(1)2s_(2)2p_(1)2p_(2)….2p_(6)3s_(1)3s_(2);` valency `=2` |
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| 2318. |
If uncertainty in possition of electron is zero, then the uncertainty in its momentum would be ………. |
| Answer» Correct Answer - Infinity | |
| 2319. |
Atomic mass of an element is not necessarily a whole number becauseA. it contians electrons, protons and neutronsB. it contains isotopesC. it contains allotropesD. all of the above |
| Answer» Correct Answer - B | |
| 2320. |
In an atom an electrron is momving with a speed fo ` 600 m//s` with an accuracy fo ` 0.005%` . Cenrtatinty with which the position fo the elelcron can be locatied si : ` ( h = 6.6 xx 10^(-34) kg m^2 s^(-1)` , mass of electron ` (e_m) = 9. 1 xx10^(-31) kg)`.A. ` 1. 52 xx 10^(-4) m`B. ` 5. 10 xx 10^(-3) m`C. ` 2.5 nm`D. ` 14. 0 nm` |
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Answer» Correct Answer - C ` :. U = (600 xx 0.005)/( 100) = 0. 30 m sec^(-1)` ` Now Delta x . M. Delta u = h/( 4pi ) ` ` Delta x = ( 6.6xx 10^(-34))/( 4 xx 3.14 xx 9. xx 10^(34) xx 0. 30) = 1 . 92 xx 10^(-3) m`. |
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| 2321. |
In an atom , an electron is moving with a speed of 600 m/s with an accuracy of `0.05%` . The certainty with which the position of the electron can be located is (h = `6.6 xx 10^(-34) kgm^(2)s^(-1)` , mass of electron , `e_(m) = 9.1 xx 10^(-31)` kg ):A. `5.10 xx 10^(-3)`mB. `1.75 xx 10^(-3)` mC. `3.83 xx 10^(-3)` mD. `1.52 xx 10^(-4)`m |
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Answer» Correct Answer - B |
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| 2322. |
A 3p-orbital has :A. Two non spherical nodesB. Two spherical nodesC. One spherical and one non spherical nodesD. One spherical and two non spherical nodes |
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Answer» Correct Answer - C Total nodes =n-l=2 No. of radial nodes =n-l=1 For 3p, n=3, l=1 `rArr` radial nodes =3-1-1 =1 Angular nodes =l=1 |
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| 2323. |
An atom has a mass of `0.02kg` and uncertainty in its velocity is `9.218xx10^(-0)m//s` then uncertainly in position is `(h=6.626xx10^(-34)Js)`A. `2.86xx10^(-28)m`B. `2.86xx10^(-32)m`C. `1.5xx10^(27)m`D. `3.9xx10^(10)m` |
| Answer» Correct Answer - A | |
| 2324. |
Given: The mass of electron is `9.11 × 10^(–31)`Kg Planck constant is `6.626 ×10^(–34)`Js, the uncertainty involved in the measurement of velocity within a distance of 0.1Å is:-A. `5.79xx10^8 "ms"^(-1)`B. `5.79xx10^5 "ms"^(-1)`C. `5.79xx10^6 "ms"^(-1)`D. `5.79xx10^7 "ms"^(-1)` |
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Answer» Correct Answer - C `Deltax.mDetlav=h/(4pi)` `0.1xx10^(-10) xx 9.11xx10^(-31)xx Deltav=(6.626xx10^(-34))/(4xx3.14)` `Deltav=(6.626xx10^(-34)) /(0.1xx10^(-10)xx 9.11xx10^(-31)xx4xx3.14)` `5.79xx10^6` m |
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| 2325. |
Calculate the speed of the electron in the first Bohr orbit given that `h=6.6xx10^(-34) Js, m=9.11xx10^(-31) kg` and `e=1.603xx10^(-19) C`. |
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Answer» Correct Answer - `2.2xx10^(6) m s^(-1)` |
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| 2326. |
The orbital diagram in which the Aufbau principle is violated isA. B. C. D. |
| Answer» Correct Answer - B | |
| 2327. |
Atomic mass of an element is not neccessurity a whole number becauseA. It contains electrons ,photons and neutronsB. It excists in allotropic formsC. It containts isotopesD. Atom are no longer indivisible |
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Answer» Correct Answer - C Atomic number ` = ((% "Abundance" ) xx ("isotope") + % "of" ("isotope")_(2))/(100)` |
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| 2328. |
Uncertainty in position of a 0.25 kg particle is `10^(-5)` , uncertainty of velocity is `(h=6.6xx10^(-34) Js)`A. `1.2xx10^(34)` m/secB. `2.1xx10^(-29)` m/secC. `1.6xx10^(-20)` m/secD. `1.7xx10^(-9)` m/sec |
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Answer» Correct Answer - B According to `Deltax xx mxx Deltav =h/(4pi) , Deltav=h/(Deltax xx m xx 4pi)` `=(6.6xx10^(-34))/(10^(-5)xx0.25xx3.14xx4)=2.1xx10^(-29)` m/s |
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| 2329. |
In an atom, an electron is moving with a speed of ` 600 m//s` with an accuracy of ` 0.005%` . Certainty with which the position of the electron can be localized is : ` ( h = 6.6 xx 10^(-34) kg m^2 s^(-1)` , mass of electron ` (e_m) = 9. 1 xx10^(-31) kg)`.A. `1.52xx10^(-4) ` mB. `5.10xx10^(-3)` mC. `1.92xx10^(-3)`mD. `3.84 xx 10^(-3)` m |
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Answer» Correct Answer - C `Deltax xx Deltap =h / (4pi)` `Deltax xx [mDeltav] =h/(4pi) , Deltav=(600xx0.005)/100 =0.03` So, `Deltax[9.1xx10^(-31)xx0.03] =(6.6xx10^(-34))/(4xx3.14)` `Deltax=(6.6xx10^(-34))/(4xx3.14xx9.1xx0.03xx10^(-31))=1.92xx10^(-3)` m |
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| 2330. |
The orbital angular momentum of an electron of an electron in `2s` orbitals isA. 4B. 1C. 0D. `(h)/(2pi)` |
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Answer» Correct Answer - C The orbital angular momentum of an electri=on is calculated as `sqrt(l(l + 1)) (h)/(2pi)` `sqrt(0( 0 + 1)) (h)/(2pi) = 0 ` |
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| 2331. |
Which of the following sets of quantum numbers represents the highest energy of an atom?A. `n=3,l=0,m=0,s=+1//2`B. `n=3,l=1,m=1,s=+1//2`C. `n=3,l=2,m=1,s=+1//2`D. `n=4,l=0,m=0,s=+1//2` |
| Answer» Correct Answer - C | |
| 2332. |
Which of the following sets of quantum numbers is correct for an electron in 3d-orbital?A. `n=3,l=2,m=-3,s=+(1)/(2)`B. `n=3,l=3,m+3,s=-(1)/(2)`C. `n=3,l=2,m=-2,s+(1)/(2)`D. `n=3,l=2,m=-3,s=-(1)/(2)` |
| Answer» Correct Answer - C | |
| 2333. |
The orbital angular momentum of an electron in `2s`-orbital isA. `+(1)/(2).(h)/(2pi)`B. zeroC. `(h)/(2pi)`D. `sqrt(2).(h)/(2pi)` |
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Answer» Correct Answer - B |
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| 2334. |
The orbital angular momentum of an electron in `2s`-orbital isA. `+1/2. h/(2pi)`B. ZeroC. `h/(2pi)`D. `sqrt2. h/(2pi)` |
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Answer» Correct Answer - B The expression of orbital angular momentum is `L=sqrt(l(1+1)(h//2pi))` For 2s orbital , l=0 , Hence , l=0 |
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| 2335. |
The orbital angular momentum of an electron in `2s` orbital is ……… |
| Answer» Correct Answer - Zero | |
| 2336. |
Which set of quantum number is not consistent with the quatum mechanical theory.A. `n = 2, l = 1, m = 1, s = 1//2`B. `n = 4, l = 3, m = 2, s = - 1//2`C. `n = 3, l = 2, m = 3, s = + 1//2`D. `n = 4, l = 3, m = 3, s = + 1//2` |
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Answer» Correct Answer - C ( c) There is a well-defined co-relation between `n, 1 m` and `s` quantum numbers in an atom. If `n = n` then `1 = (n - 1), m` will be `+- 1` and `s` will always be `+- 1//2`. For `n = 3` there cannot be `m = 3`. Highest value `m` will be `2`. |
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| 2337. |
Which sets of quantum number are consitent with the theory ?A. `n = 2,l = 1,m = 0,s = -1//2`B. `n = 4,l = 3,m = -2,s = -1//2`C. `n = 3,l = 2,m = -3,s = +1//2`D. `n = 4,l = 3,m = -3,s = +1//2` |
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Answer» Correct Answer - A::B::C In `d,n = l i.e. , 3 but l = n` |
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| 2338. |
The energy of an electron in the Bohr orbit for hydrogen is `-13.6 eV` .Which of the following is a possible excited state for electron in Bohr orbit of hydrogen atom ?A. `-3.4 eV`B. `-6.8 eV`C. `-1.7 eV`D. `13.6 eV` |
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Answer» Correct Answer - A `E_(n)` "for" `H_(2)^(o+) = (E_(1) for H xx Z^(2))/(n^(2))` `E_(1) = - 13.6 eV` The possible excited state value are `E_(2) = (-13.6)/(4) = - 3.4 eV` `E_(3) = (-13.6)/(9) = - 1.5 eV` `E_(4) = (-13.6)/(16) = - 0.85 eV` So, the value is only `-3.4 eV` |
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| 2339. |
The orbital angular momentum of an electron in`2s`orbital isA. `+ (1)/(2)(h)/(2pi)`B. ZeroC. `(h)/(2pi)`D. `sqrt(2)(h)/(2pi)` |
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Answer» Correct Answer - B The orbital angular momentum is `(h)/(2pi) sqrt(l(l + 1))` The orbital angular momentum for an electron in s orbital `(l = 0) is 0` |
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| 2340. |
The first use of quantum theory to explain the structure of atom was made byA. HeisenbergB. BohrC. PlankD. Einstein |
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Answer» Correct Answer - B The first use of quantum theory to explain the structure of atom was made by Bohr |
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| 2341. |
The enrgy of an electron in the first Borh orbit of H atom is `-13.6 eV` The possible energy values (s)of the excited state (s) for electron in bohr orbitsw of hydrogen is (are)A. `-3.4 eV`B. `-4.2 eV`C. `-6.8 eV`D. `+ 6.8 eV` |
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Answer» Correct Answer - A The energy of an electron in Bohr orbit of hydrogen atom is given by the expeeression `E_(n) = - (Constant)/(n^(2))` When n takes only value for the first Bohr orbit `n = 1`and it is gigev that `E_(1) = -13.6 eV` Hence `E_(n) = (13.6 eV)/(n^(2))` Of the given value of energy only `-13.6 eV` can be obtained by substating `n = 2` in the above expression |
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| 2342. |
The ionisation enrgy of H-atom is ` 13. 6 eV`. What will be ionisation energy ofg ` He^(2+)`. Ions ? |
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Answer» We know, ` E_1 " for " He^+ = E_1 " or " H xx z^2` ` = 13 . 6 xx 4 = 54 .4 eV` ` E_1 " for " Li^(2+) = E_1 "or " H xx Z^2` ` = 13 .6 xx 9= 122 . 4 eV`. |
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| 2343. |
The bromine atom posseses `3s` electrons It contain six electron in `2p` orbits six electron in `3p` orbit and five electron `4p` orbits .Which of these electrons experience the lower effective nuclear charge ? |
| Answer» `4p` electron having the value of was farthest from the nucleus and hence experience the lower effective nucleus charge for orbit of the same type of sub -shelll, the greater the value of n the smaller the effective nuclear charge | |
| 2344. |
In an atom, two electrons move around nucleus in circular orbits of radii ( R) and ( 4R) . The ratio of the time taken by them to complete one revolution is :A. `1:4`B. `4:1`C. `1:8`D. `8:1` |
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Answer» Correct Answer - C |
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| 2345. |
Order of no. of revolutions/sec `gamma_(1),gamma_(2),gamma_(3)` and `gamma_(4)` for I, II, III and IV orbits is:A. `gamma_(1) gt gamma_(2) gt gamma_(3) gt gamma_(4)`B. `gamma_(4) gt gamma_(3) gt gamma_(2) gt gamma_(1)`C. `gamma_(1) gt gamma_(2) gt gamma_(4) gt gamma_(3)`D. `gamma_(2) gt gamma_(3) gt gamma_(4) gt gamma_(1)` |
| Answer» Correct Answer - A | |
| 2346. |
The electrons identifted by quantum nymbers ( n) and (l) : can be placed in order fo increasing enrgy as :A. ` (4) lt (2) lt (3) lt (1)`B. `(2) gt (40) gt (1) gt(3)`C. `(1) gt (3) lt (2) lt(4)`D. `(3) lt (4) lt (2) lt (1)` |
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Answer» Correct Answer - A (1) `n=4, l=1 rArr 4p` (2) ` n=4, l=0 rArr 2s` (3) ` n=3 , l=2 rArr 3d` (4) ` n=3 , l=1 rArr 3p`. Increasing order of energy is ` 3 p lt 4 s lt 3d lt 4 p` ` (4) lt (2) lt (3) lt (1)` . |
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| 2347. |
The energy of hydrogen atom is its ground state is `-13.6 eV` The enrgy of the level corresponding to the quantum number `n = 5` isA. `-0.54 eV`B. `-0.50 eV`C. `-0.85 eV`D. `-2.72 eV` |
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Answer» Correct Answer - A `E_(n) = (E for H xx Z^(2))/(n^(2)) = (-13.6 xx 1^(2))/(25) = - 0.54 eV` |
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| 2348. |
`IE_1` for (H) and De are 13.6 eV 1 and 24.6 eV respectively . Thus enrgy liberated during the formation fo He by ` He^(2+) + 2e rarr H3,` is:A. ` 54.4 eV`B. ` 49 . 2 eV`C. ` 0. 27 , 4 eV`D. ` 13. 6 lambda` |
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Answer» Correct Answer - D `Hrarr H^+ + e , 13 . 6 eV` ` Herarr He ^+ + e, 24. 6 eV` ` IE_2" of "He = IE _1" of "HE^+` ` +E_(1H) xx Z^2 = 13 . 6 xx 4 = 54 . 4 eV` ` He^+ rarr He^(2+) , 54. 4 eV` ` :. Herarr He^(2+) + 2e, 24 . 6 + 54.4 = 79. 0 eV`. |
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| 2349. |
how many orbits, orbitals and electrons are there in an atom having atomic mass 24 and atomic number 12? |
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Answer» Atomic number=No. of protons=No. of electrons=12 Electrons configuration=2,8,2 No. of orbits=(K,L and M) No. of orbitals on which electrons are present `=(one 1s+ one 2s+three 2p+one 3s). |
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| 2350. |
Differentiate between the terms orbits and orbitals |
| Answer» Correct Answer - A | |