InterviewSolution
Saved Bookmarks
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.
| 1201. |
If W is atomic weight and N is the atomic number of an element, thenA. Number of `e^(-1)` =W-NB. Number of `._0n^1` =W-NC. number of `._1H^1`=W-ND. Number of `._0n^1`=N |
|
Answer» Correct Answer - B No. of neutrons =mass number - no. of protons . =W-N |
|
| 1202. |
Six protons are found in the nucleus ofA. BoronB. LithiumC. CarbonD. Helium |
|
Answer» Correct Answer - C Atomic no. of C =6 so the number of protons in the nucleus =6 |
|
| 1203. |
Which of the following is/are correct energy order for H-atom?A. `1s tl 2s lt 2p lt 3s lt 3p`B. `1s lt 2s = 2p lt 3s = 3p`C. `1s lt 2p lt 3d lt 4s`D. `1s tl 2s lt 4s lt 3d` |
|
Answer» Correct Answer - B::C In H-atom, all are degenerate orbitals in a given main shell |
|
| 1204. |
Atomic number of an element represents:A. Number of neutrons in the nucleusB. Number of protons in the neucleusC. Atomic weight of elementD. Valency of element |
|
Answer» Correct Answer - B Atomic number is defined as the number of protons in the nucleus |
|
| 1205. |
The spectrum with all wavelengths may beA. Absorption spectrumB. Emission spectrumC. Continuous spectrumD. Discontinuous spectrum |
|
Answer» Correct Answer - C Confirceoces spectrum |
|
| 1206. |
Cyclotron is not capable of acceleratingA. NeutronsB. ProtonsC. DeutronsD. `alpha`-particles |
|
Answer» Correct Answer - A Cyclotron is used for accelerating charged paticles and not neutral particles |
|
| 1207. |
An atom has 26 electrons and its atomic weight is 56. The number of neutrons in the nucleus of the atom will beA. 26B. 30C. 36D. 56 |
|
Answer» Correct Answer - B `._26X^56` A =P+N =Z+N =E+N , For neutral atom N=A-E=56-26=30 |
|
| 1208. |
STATEMENT-1: The kinetic energy of photo-electrons increases with increase in frequency of incident light were `vgtv_(o).` STATEMENT-2: Whenever intenksity of light is increased the number of photo-electron ejected always increases.A. If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
|
Answer» Correct Answer - B `hv = hv_(0) +KE` |
|
| 1209. |
In which one of the following pairs of experimental observations and phenomemnon does the experimental observation correctly account for phenomenon does the experimental observation correctly account for phenomenonA. `{:("Experimental observation","Phenomenon"),("X-ray spectra","Charge on the nucleus"):}`B. `{:("Experimental observation","Phenomenon"),(alpha"-particle scattering","Quantized electron orbit"):}`C. `{:("Experimental observation","Phenomenon"),("Emission spectra","The quantization of energy "):}`D. `{:("Experimental observation","Phenomenon"),("The photoelectric effect","The nuclear atom"):}` |
|
Answer» Correct Answer - C Emission spectra of different `lambda` accounts for quantisation of energy |
|
| 1210. |
p-orbitals fo an atom in presence fo magnetic field are :A. Three fold degeneratB. Tow fold degenrateC. Non-degeneratD. None fo the above |
|
Answer» Correct Answer - C In avbsence fo magentic field p-prbitals have same enrgy level and are diegenrate orbitals . |
|
| 1211. |
A completely filled d-orbital `(d^10)` isA. Spherically symmetricalB. has octahedral symmetryC. Has tetrahedral symmetryD. Depends on the atom |
| Answer» Correct Answer - B | |
| 1212. |
Write the electronic configuration whose atomic number are `6,16,26,36,56`. |
|
Answer» `1s^(2)2s^(2)2p^(2)` `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(4)` `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(2) 3d^(6)` `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6) 4s^(2)3d^(10)4p^(6)` `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)4p^(6)5s^(2)4d^(10)5p^(6)6s^(2)` |
|
| 1213. |
Which of the following is isoelectronic ?A. `CO_2, NO_2`B. `NO_2^-, CO_2`C. `CN^-, CO`D. `SO_2, CO_2` |
|
Answer» Correct Answer - C ( c) Both `CN^-` and `CO` gave `14` electrons. |
|
| 1214. |
p-orbitals of an atom in presence of magnetic field are :A. Two fold degenerateB. Non degenerateC. Three fold degenerateD. none of these |
| Answer» Correct Answer - C | |
| 1215. |
Maximum number of electrons in a shell with principal quantum number n is given byA. nB. 2nC. `n^2`D. `2n^2` |
| Answer» Correct Answer - D | |
| 1216. |
Satement-1: Spin quantum number can have two values `+1/2 and -1/2.` Statement-2: + ve and -ve signs signify the positive and negative wave functions. |
|
Answer» Correct Answer - c |
|
| 1217. |
In hydrogen atom an whit has a diemeter of about `16.92Å` .What in the maximum number of electron that can be accommodated ?A. 8B. 32C. 50D. 72 |
|
Answer» Correct Answer - B |
|
| 1218. |
The maximum number of electron that can be accommodated in `4^(th)` shell is.A. `8`B. `16`C. `32`D. `50` |
|
Answer» `[C]" "n^(th)` shell principal quantum number `{:(,K,,L,,M,,N,),(,n=1,,2,,3,,4,):}` Possible value of `l` for `n=4` `l=0,1.2,3` no. of sub shells =4 [number of subshell in a shell = n] `{:(l=0,,,-,,,s),(l=1,,,-,,,p),(l=2,,,-,,,d),(l=3,,,-,,,f):}` number of orbitals in a subshell `-2l+1` `{:(s-1 " orbital",,,d-5 " orbital"),(p-3 " orbital",,,f-7 " orbital"):}` Total number of orbital `=n^(2)=4^(2)=16` in a shell Total number of electron in each orbital `=2` implies Total electron accomodated in N shell `=16xx2=32` |
|
| 1219. |
The number of electrons that can be accommodated in `dz^2` orbital isA. 10B. 1C. 4D. 2 |
| Answer» Correct Answer - D | |
| 1220. |
The energy corresponding to one of the lines in the paschen series of H-atom is `18.16xx10^(-20)`J. Find the quantum numbers for the transition which produce this line. |
|
Answer» Correct Answer - 6 `DeltaE=2.18xx10^(-18)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` `18.16xx10^(-20)=2.18xx10^(-18)[(1)/(9)-(1)/(n^(2))]` On solving, n=6. |
|
| 1221. |
What is the ratio of the Rydberg constant for helium to Hydrogen atom?A. `1//2`B. `1//4`C. `1//8`D. `1//16` |
|
Answer» Correct Answer - C `R=(-2pi^(2)mZ^(2)e^(4))/(ch^(3))` `(R_(He))/(R_(H))=(2xx2^2)/(1xx1^(2))=8` `therefore(R_(H))/(R_(He))=(1)/(8)`. |
|
| 1222. |
Rydberg constant isA. Same for all elementsB. Different for different elementsC. A universal constantD. Is different for lighter elements but same for heavier elements |
|
Answer» Correct Answer - B `R alpha Z^(2)` |
|
| 1223. |
The value of the total energy of an electron in the hydrogen atom is given byA. `mv^(2)`B. `1//2mv^(2)`C. `-e^(2)//2r`D. `-mv//r^(2)` |
|
Answer» Correct Answer - C Energy of electron `(-ze^(2))/(2r)` |
|
| 1224. |
Assertion (A) : Ionisation potential of Be (atomic number `4` ) is than B (atomic number `5)` Reason (R ) : The first electron released fromm Be is of p orbitals but that from B is of a orbitals.A. If both (A) and (R ) correct and (R ) is the correct explanation for (A)B. If both (A) and (R ) correct and (R ) is the correct explanation for (A)C. If (A) is correct but (R ) is incorrectD. If both (A) and (R ) are incorrect |
| Answer» Ionisation potential of Be is more than B fuirst elkectron released from Be is s orbital, white B is of p orbital | |
| 1225. |
Which of the following statement are correct ?A. The electronic configuration of `Cr` is `[Ar]3d^(5), 4s^(1)` (atomic number of Cs = 24)B. The magnitic quantum number may have a negative valueC. In silver atom `23` electron have spin of one type and `24` of the opposite type .(Atomic number of `Ag = 47)`D. The oxidation state of nitrogen in `NH_(3)` is `-3` |
| Answer» Correct Answer - A::B::C | |
| 1226. |
Assertion(A): Wavelength of limiting line of lyman series is less less than wavelength of limiting line of Balmer series. Reason(R): Rydberg constant value is same for all elementsA. 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 - C H-spectrum can be both emission and absorption |
|
| 1227. |
Assertion (A) : Limiting line is the balmer series ghas a wavelength of `364.4 nm` Reason (R ) : Limiting line is obtained for a jump electyron from `n = prop`A. If both (A) and (R ) correct and (R ) is the correct explanation for (A)B. If both (A) and (R ) correct and (R ) is the correct explanation for (A)C. If (A) is correct but (R ) is incorrectD. If (A) is incorrect but (R ) is correct |
|
Answer» Correct Answer - A Limiting line corresponding to transition from `n = prop `to lower energy level |
|
| 1228. |
A dispositive ion has `12` protons .What is the number of electrons in the intrapositive ion ? |
|
Answer» A dipositive ion `X^(2+)` has `12` protons and the number of electron is `10` in tetrapositive ion `X^(4+)` , the number of electrons is `8` |
|
| 1229. |
The limiting line Balmer series will have a frequency ofA. `32.29 xx 10^(15) s^(-1)`B. `3.65 xx 10^(15) s^(-1)`C. `-8.22 xx 10^(15) s^(-1)`D. `8.22 xx 10^(15) s^(-1)` |
|
Answer» Correct Answer - C The limiting line of Balmer series refers to the transition from `= omega 2nd` orbit `v= 3.29 xx 10^(13)((1)/(n_(1)^(2)))- ((1)/(n_(2)^(2)))s^(-1) (n_(1) = 2, n = prop)` `= 3.29 xx 10^(15) xx (1)/(4) = 8.22 xx 10^(16) s^(-1)` |
|
| 1230. |
The values of `n_(1)` and `n_(2)` in the pfund spectral series of hydrogen atom are…….. And …….. Respectively. |
| Answer» Correct Answer - `5,(6,7,…….)` | |
| 1231. |
Write different isotopes of oxygen , carbon and chlorine. |
|
Answer» The isotopes of oxygen are: `._(8)O^16,._(8)O^17,._(8)O^18` The isotopes of chlorine are: `._(17)Cl^35,._(17)cl^37` The isotopes of carbon are: `._(6)C^12,._(6)C^13,._(6)C^14` |
|
| 1232. |
The transition of electron in if atom that will emit maximum energy isA. `n_(3) rarr n_(2)`B. `n_(4) rarr n_(3)`C. `n_(2) rarr n_(4)`D. `n_(6) rarr n_(5)` |
|
Answer» Correct Answer - A `n_(1) rarr n_(2)` |
|
| 1233. |
An atom of element is represented as `._(Z)C^A`. After the emission of a `beta`-particle , another element Y is formed. Represent Y with atomic number and mass number. |
| Answer» `._(z)X^A to ._(z+1)Y^(A)+beta` particle. | |
| 1234. |
Calculate the specific charges (e/m) of the following particles and then arrange the particles in the asceding order of their specific charges. (a) Electron (b) Proton (c) `alpha`-particle. |
|
Answer» Specific charge of an electron `=(1.6xx10^-19)/(9.1xx10^-31) " coulomb//kg"` `=0.176xx10^12=17.6xx10^10 " coulomb /kg"` Specific charge of a proton `=(1.6xx10^(-19))/(1.67xx10^(-27))"coulomb /kg" = 0.96xx10^8 "coulomb/kg"` Specific charge of `alpha`-particle `=(2xx1.6xx10^(-19))/(2xx10^(-27)(1.67+1.72))=(1-6xx10^8)/(3.39)=0.472xx10^8 "coulomb/kg"` Hence, ascending order of specific charges of electron, proton and `alpha`-particles is `alpha-"particle" lt "proton" lt "electron"`. |
|
| 1235. |
When there are two el,ectron is the same orbitals , they have ……………spins |
| Answer» Correct Answer - Antiparallel | |
| 1236. |
If there were three possible values `(-1//2,0,+1//2)` for the spin magnetic quantum number , ` m_s` how many elements would there be in the ` 4 th` period of periodic table in ` 4_s`, 3p, 4d` respectively ? |
|
Answer» ` n= 4, l = 0, m_l =0, m_s =- 1/2 , 0, + 1/2` for each ` m_l i.e, 3` in `4s` ` l = 1, m_l = -1, 0, +1 i.e, 9` in `4 p` ` l= 2, m _l = +- 2 , +- 1, 0 i.e., 15` in `3 d` Thus ` 4 s^3 , 4 p^9` and ` 3d^(15) ` i.e., in all ` 27` elemnts would have been in ` 4th` period . |
|
| 1237. |
Assertion: Atomic orbital in an atom is designated by n,l,`m_1` and `m_s` Reason: These are helpful in designating electron present in an orbital.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 assertion is false but reason is true |
|
Answer» Correct Answer - D Assertion is false but reason is true. Atomic orbital is designated by n,l and `m_1` while state of an electron in an atom in specified by the four quantum numbrs n,l,`m_1`, and `m_s`. |
|
| 1238. |
Which of the following should be the posible sub-shell for `n +1 = 7`?A. `7s 6p 5d 4f`B. `4f 5p 6s 4d `C. `7s 6p 5d 6d `D. `4s 5d 6p 7s` |
|
Answer» `n + 1 = 7` 7 + 0 = 7s ,` 6 + 1 = 6p,` `5 + 2 = 5d ,` `4 + 3 = 4f` |
|
| 1239. |
Minimum de-Broglie wavelength is associated withA. ElectronB. protonC. `CO_2` moleculeD. `SO_2` molecule |
|
Answer» Correct Answer - D `lambda=h/"mv"`.For same velocity `lambda prop 1/m` `SO_2` molecule has least wavelength because their molecular mass is high. |
|
| 1240. |
How long would it take a radio wave of frequency `6xx10^(3) sec^(-1)` to travel from mars to the earth, a distance of `8xx10^(7)km` ? |
|
Answer» Correct Answer - `2.66xx10^(2)sec` Distance to be travelled from mars to earth `=8xx10^(7-)km` `because` Velocity `= 3xx10^(8) m//sec` `therefore` Time `=D//V=(8xx10^(10))/(3xx10^(8))=2.66xx10^(2)`sec. |
|
| 1241. |
The minimum energy required to eject an electron from an atom is calledA. Kinetic energyB. Electrical energyC. Chemical energyD. Work function |
|
Answer» Correct Answer - D Minimum energy is work function |
|
| 1242. |
The energy of an electron in the first Bohr orbit of H atom is `-13.6 eV` The potential energy value (s) of excited state(s) for the electron in the Bohr orbit of hydrogen is(are)A. `-3.4eV`B. `4.2eV`C. `-6.8 eV`D. `+6.8eV` |
|
Answer» Correct Answer - A `E_(n) =- (13.6)/(n^(2))ev//"atom"` |
|
| 1243. |
Which of the following statements is wrong? The probability of finding the electron in `p_(x)` orbital isA. maximum on two opposite sides of the nucleus along x-axisB. zero at the nucleusC. same on all the sides around the nucleusD. zero on the z-axis |
|
Answer» Correct Answer - A::B::D `P_(x)-` orbital has dumb bell shape and s situated along x-axis |
|
| 1244. |
If ionising energy of H atom is 13.6 eV, then the second ionising energy of He should beA. 27.2 eVB. 40.8 eVC. 54.4 eVD. 108.8 eV |
|
Answer» Correct Answer - C `E_H=(-13.6)/n^2 , (E_1)_H=-13.6 eV` `(E_1)He^(+)=-13.6Z^2/n^2=-13.6xx2^2/1=-54.4` eV The second I.E. =`E_oo-(E_1)_(He^+)` `=0-(-54.4) eV=54.4 eV` |
|
| 1245. |
In photoelectric effect, the number of photoelectrons emitted is proportional to :A. Intensity of incident beamB. Frequency of incident beamC. Wavelength of incident beamD. All |
|
Answer» Correct Answer - A each photon can emit one electron only |
|
| 1246. |
The colour of sky is due toA. Absorption of light by atmospheric gasesB. Transmission of lightC. Wavelength of scattered lightD. All of these |
|
Answer» Correct Answer - C Scattering of light `prop 1/lambda^4` (Rayleigh scattering law). The colour of the sky is blue due to least scattering of blue light, whose wavelength is maximum in the visible range of light |
|
| 1247. |
In photoelectric effect, the kinetic energy of photoelectrons increases linearly with theA. Wavelength of incident lightB. Frequency of incident lightC. Velocity of incident lightD. Atomic mass of an element |
|
Answer» Correct Answer - B The kinetic energy of photoelectrons increases linearly with the frequency of the incident light used |
|
| 1248. |
the wavelength range of the visible spectrum extends from violet ( 400 nm) to red ( 750 nm). Express these wavelengths in frequencies (Hz) . (1nm =` 10 ^(-9)m`) |
|
Answer» Frequency of violet light `v = (c)/(lambda) = (3.00 xx 10^(8)ms^(-1))/(400 xx 10^(-9)m) = 7.50 xx 10^(14)Hz` frequenct of red light `v = (c)/(lambda) = (3.00 xx 10^(8)ms^(-1))/(750 xx 10^(-9)m) = 4.00 xx 10^(14)Hz` The range of visible spectrum is from `4.0 xx 10^(14)` to `7.5 xx 10^(14)Hz` in terms of frequency |
|
| 1249. |
When there are two electron is the same orbitals , they have ……………spins. |
|
Answer» Correct Answer - opposite |
|
| 1250. |
Photoelectric effect shows :A. Particle - like behaviour of lightB. Wave-like behaviour fo lightC. Both wave -like and particle -like behaviour of lightD. Neither wave-like nor particle -like behaviour of light |
|
Answer» Correct Answer - A It is an experimental eveience for particle nature of elctron . |
|